I have an array of three element like [31,23,12] and I want to find the second largest element and its related position without rearranging the array.
Example :
array = [21,23,34]
Second_largest = 23;
Position is = 1;
Make a clone of your original array using .slice(0) like :
var temp_arr = arr.slice(0);
Then sor it so you get the second largest value at the index temp_arr.length - 2 of your array :
temp_arr.sort()[temp_arr.length - 2]
Now you could use indexOf() function to get the index of this value retrieved like :
arr.indexOf(second_largest_value);
var arr = [23, 21, 34, 34];
var temp_arr = [...new Set(arr)].slice(0); //clone array
var second_largest_value = temp_arr.sort()[temp_arr.length - 2];
var index_of_largest_value = arr.indexOf(second_largest_value);
console.log(second_largest_value);
console.log(index_of_largest_value);
Using ES6 Set and Array.from
const secondLargest = (arr) => Array.from([...new Set(arr)]).sort((a,b) => b-a)[1]
Above function removes duplicate elements using Set and returns the second largest element from the sorted array.
I tried to make the answer as simple as possible here, you can it super simple
function getSecondLargest(nums) {
var flarge = 0;
var slarge = 0;
for (var i = 0; i < nums.length; i++) {
if (flarge < nums[i]) {
slarge = flarge;
flarge = nums[i];
} else if (nums[i] > slarge) {
slarge = nums[i]
}
}
return slarge;
}
Its fully logical ,there is no array sort or reverse here, you can also use this when values are duplicate in aray.
function getSecondLargest(nums) {
nums.sort(function(x,y){
return y-x;
});
for(var j=1; j < nums.length; j++)
{
if(nums[j-1] !== nums[j])
{
return nums[j];
}
}
}
getSecondLargest([1,2,3,4,5,5]);
OUTPUT: 4
This method will also take care of the multiple occurrence of a number in the array. Here, we are first sorting the array and then ignoring the same number and returning our answer.
You could create a copy of the original array using spread and sort() it. From you'd just get the second to last number from the array and use indexOf to reveal it's index.
const array = [21,23,34];
const arrayCopy = [...array];
const secondLargestNum = arrayCopy.sort()[arrayCopy.length - 2]
console.log(array.indexOf(secondLargestNum));
Alternatively you can use concat to copy the array if compatibility is an issue:
var array = [21, 23, 34];
var arrayCopy = [].concat(array);
var secondLargestNum = arrayCopy.sort()[arrayCopy.length - 2]
console.log(array.indexOf(secondLargestNum));
This way is the most verbose, but also the most algorithmically efficient. It only requires 1 pass through the original array, does not require copying the array, nor sorting. It is also ES5 compliant, since you were asking about supportability.
var array = [21,23,34];
var res = array.reduce(function (results, curr, index) {
if (index === 0) {
results.largest = curr;
results.secondLargest = curr;
results.indexOfSecondLargest = 0;
results.indexOfLargest = 0;
}
else if (curr > results.secondLargest && curr <= results.largest) {
results.secondLargest = curr;
results.indexOfSecondLargest = index;
}
else if (curr > results.largest) {
results.secondLargest = results.largest;
results.largest = curr;
results.indexOfSecondLargest = results.indexOfLargest;
results.indexOfLargest = index;
}
return results;
}, {largest: -Infinity, secondLargest: -Infinity, indexOfLargest: -1, indexOfSecondLargest: -1});
console.log("Second Largest: ", res.secondLargest);
console.log("Index of Second Largest: ", res.indexOfSecondLargest);
I recently came across this problem, but wasn't allowed to use looping. I managed to get it working using recursion and since no one else suggested that possibility, I decided to post it here. :-)
let input = [29, 75, 12, 89, 103, 65, 100, 78, 115, 102, 55, 214]
const secondLargest = (arr, first = -Infinity, second = -Infinity, firstPos = -1, secondPos = -1, idx = 0) => {
arr = first === -Infinity ? [...arr] : arr;
const el = arr.shift();
if (!el) return { second, secondPos }
if (el > first) {
second = first;
secondPos = firstPos;
first = el;
firstPos = idx;
} if (el < first && el > second) {
second = el;
secondPos = idx;
}
return secondLargest(arr, first, second, firstPos, secondPos, ++idx);
}
console.log(secondLargest(input));
// {
// second: 115,
// secondPos: 8
// }
Hope this helps someone in my shoes some day.
Simple recursive function to find the n-largest number without permutating any array:
EDIT: Also works in case of multiple equal large numbers.
let array = [11,23,34];
let secondlargest = Max(array, 2);
let index = array.indexOf(secondlargest);
console.log("Number:", secondlargest ,"at position", index);
function Max(arr, nth = 1, max = Infinity) {
let large = -Infinity;
for(e of arr) {
if(e > large && e < max ) {
large = e;
} else if (max == large) {
nth++;
}
}
if(nth==0) return max;
return Max(arr, nth-1, large);
}
Just to get 2nd largest number-
arr = [21,23,34];
secondLargest = arr.slice(0).sort(function(a,b){return b-a})[1];
To get 2nd largest number with index in traditional manner-
arr = [20,120,111,215,54,78];
max = -Infinity;
max2 = -Infinity;
indexMax = -Infinity;
index2 = -Infinity;
for(let i=0; i<arr.length; i++) {
if(max < arr[i]) {
index2 = indexMax;
indexMax = i;
max2 = max;
max = arr[i];
} else if(max2 < arr[i]) {
index2 = i;
max2 = arr[i];
}
}
console.log(`index: ${index2} and max2: ${max2}`);
I have tried to solve without using the inbuilt function.
var arr = [1,2, -3, 15, 77, 12, 55];
var highest = 0, secondHighest = 0;
// OR var highest = arr[0], secondHighest = arr[0];
for(var i=0; i<arr.length; i++){
if(arr[i] > highest){
secondHighest = highest;
highest = arr[i];
}
if(arr[i] < highest && arr[i] > secondHighest){
secondHighest = arr[i];
}
}
console.log('>> highest number : ',highest); // 77
console.log('>> secondHighest number : ',secondHighest); // 55
var arr = [21,23,34];
var output = getSecondLargest(arr);
document.getElementById("output").innerHTML = output;
function getSecondLargest(nums) {
if (nums.length == 0){
return undefined;
}
nums.sort((a,b) => b-a);
var newArr = [...new Set(nums)];
return newArr[1];
}
<p id="output"></p>
function getSecondLargest(nums) {
const sortedArray = new Set(nums.sort((a, b) => b - a)).values();
sortedArray.next();
return sortedArray.next().value;
}
console.log(getSecondLargest([1, 2, 4, 4, 3]));
//Suggest making unique array before checking largest value in the array
function getSecondLargest(arr) {
let uniqueChars = [...new Set(arr)];
let val=Math.max(...uniqueChars);
let arr1 = arr.filter(function(item) {
return item !== val;
})
let num=Math.max(...arr1);
return num;
}
function main() {
const n = +(readLine());
const nums = readLine().split(' ').map(Number);
console.log(getSecondLargest(nums));
}
Here the code will give the second largest number and the index of it
const a = [1, 2, 3, 4, 6, 7, 7, 8, 15]
a.sort((a,b)=>a-b) //sorted small to large
const max = Math.max(...a)
const index = a.indexOf(max)
const s = {secondLargest:a[index-1],index:index-1}
console.log(s)
var elements = [21,23,34]
var largest = -Infinity
// Find largest
for (var i=0; i < elements.length; i++) {
if (elements[i] > largest) largest = elements[i]
}
var second_largest = -Infinity
var second_largest_position = -1
// Find second largest
for (var i=0; i < elements.length; i++) {
if (elements[i] > second_largest && elements[i] < largest) {
second_largest = elements[i]
second_largest_position = i
}
}
console.log(second_largest, second_largest_position)
function getSecondLargest(nums) {
let arr = nums.slice();//create a copy of the input array
let max = Math.max(...arr);//find the maximum element
let occ = 0;
for(var i = 0 ; i < arr.length ; i++)
{
if(arr[i] == max)
{
occ = occ +1;//count the occurrences of maximum element
}
}
let sortedArr =arr.sort(function(x, y) { return x > y; } );//sort the array
for(var i = 1 ; i <= occ ; i++){
sortedArr.pop()//remove the maximum elements from the sorted array
}
return Math.max(...sortedArr);//now find the largest to get the second largest
}
I write the most simple function with O(n) complexity using two variables max and secondMax with simple swapping logic.
function getSecondLargest(nums) {
let max = 0, secondMax = 0;
nums.forEach((num) => {
if (num > max) {
secondMax = max;
max = num;
} else if (num != max && num > secondMax) secondMax = num;
});
return secondMax;
}
here you can also deal with if the second largest or largest number is repeated
var nums =[2,3,6,6,5];
function getSecondLargest(nums) {
let secondlargets;
nums.sort(function(a, b){return a - b});
// all elements are in the accesindg order
// [1,2,3,5,6,6]
var highest;
// that is the last sorted element
highest = nums[nums.length-1];
nums.pop();
// through above statment we are removing the highest element
for(let i =0;i<nums.length-1;i++){
if(nums[nums.length-1]==highest){
/* here we remove gives this as conditon because might be the hiesht
had more indecis as we have in this question index(5) &index(6)
so remove the element till all positon have elemnt excepts the highest */
nums.pop()
}
else{
return nums[nums.length-1]
/* our array is already sorted and after removing thew highest element */
}
}
}
Please find a simple solution, without using inbuild functions:
Time complexity is O(n)
function secondLargest(arr) {
let prev = [0]
let i =1;
let largest =0;
while(i<arr.length){
let current = arr[i];
if(current > largest ) {
largest = current;
prev = arr[i-1];
} else if (current > prev && current < largest) {
prev = current
}
i++;
}
return prev;
}
let arr = [1,2,3,41,61,10,3,5,23];
console.log(secondLargest(arr));
Here is a simple solution using .sort() and new Set()
const array = [21, 23, 34, 34];
function getSecondLargest(arr){
return list = [...new Set(arr)].sort((a, b) => b - a)[1]
};
getSecondLargest(array);
In this case, if there are repeated numbers then they will be removed, and then will sort the array and find the second-largest number.
let arr=[12,13,42,34,34,21,42,39]
let uniqueArray=[...new Set(arr)]
let sortedArray=uniqueArray.sort((a,b)=>b-a)
console.log(sortedArray[1])
/* we can solve it with recursion*/
let count = 0; /* when find max then count ++ */
findSecondMax = (arr)=> {
let max = 0; /* when recursive function call again max will reinitialize and we get latest max */
arr.map((d,i)=>{
if(d > max) {
max = d;
}
if(i == arr.length -1) count++;
})
/* when count == 1 then we got out max so remove max from array and call recursively again with rest array so now count will give 2 here we go with 2nd max from array */
return count == 1 ? findSecondMax(arr.slice(0,arr.indexOf(max)).concat(arr.slice(arr.indexOf(max)+1))) : max;
}
console.log(findSecondMax([1,5,2,3]))
function getSecondLargest(nums) {
// Complete the function
var a = nums.sort();
var max = Math.max(...nums);
var rev = a.reverse();
for(var i = 0; i < nums.length; i++) {
if (rev[i] < max) {
return rev[i];
}
}
}
var nums = [2,3,6,6,5];
console.log( getSecondLargest(nums) );
Find Second Highest Number (Array contains duplicate values)
const getSecondHighestNumber = (numbersArry) => {
let maxNumber = Math.max( ...numbersArry);
let arrFiltered = numbersArry.filter(val => val != maxNumber);
return arrFiltered.length ? Math.max(...arrFiltered) : -1;
}
let items = ["6","2","4","5","5","5"];
const secondHighestVal = getSecondHighestNumber(items);
console.log(secondHighestVal); // 5
const arrData = [21, 23, 34];
const minArrValue = Math.min(...arrData);
const maxArrValue = Math.max(...arrData);
let targetHighValue = minArrValue;
let targetLowValue = maxArrValue;
for (i = 0; i < arrData.length; i++) {
if (arrData[i] < maxArrValue && arrData[i] > targetHighValue) {
targetHighValue = arrData[i];
}
if (arrData[i] > minArrValue && arrData[i] < targetLowValue) {
targetLowValue = arrData[i];
}
}
console.log('Array: [' + arrData + ']');
console.log('Low Value: ' + minArrValue);
console.log('2nd Lowest Value: ' + targetLowValue);
console.log('2nd Highest Value: ' + targetHighValue);
console.log('High Value: ' + maxArrValue);
Notice that if the max number appears multiple times in your array (like [6, 3,5,6,3,2,6]), you won't get the right output. So here is my solution:
function getSecondLargest(nums) {
// Complete the function
const sortedNumbers = nums.sort((a, b) => b - a);
const max = sortedNumbers[0];
const secondMax = sortedNumbers.find(number => number < max);
return secondMax;
}
const findSecondlargestNumber = (data) => {
let largest = null;
let secondlargest = null;
data.forEach(num => {
if(!largest) {
largest = num;
}
else if(num > largest) {
secondlargest = largest;
largest = num;
}
else if((!secondlargest && num !== largest) || (secondlargest)) {
secondlargest = num;
}
})
return secondlargest;
}
console.log(findSecondlargestNumber([11, 11, 3, 5, 6,2, 7]))
Here's the simplest way to get the second largest number and it's respective position from an array without rearranging the array or without using sorting method.
function findSecondLargest(arr) {
const largest = Math.max.apply(null, arr);
arr[arr.indexOf(largest)] = -Infinity;
const secondLargest = Math.max.apply(null, arr);
const position = arr.indexOf(secondLargest);
return { secondLargest, position };
}
console.log(findSecondLargest([3, 5, 7, 9, 11, 13])); //{ secondLargest: 11, position: 4 }
-Infinity is smaller than any negative finite number.
function getSecondLargest(nums) {
const len = nums.length;
const sort_arr = nums.sort();
var mynum = nums[len-1];
for(let i=len; i>0; i--){
if(mynum>nums[i-1]){
return nums[i-1];
}
}
}
Related
I have an array like this
[-2,4,5,6,7,8,10,11,15,16,17,18,21]
Is anyone know, how to make the output from that array become integer like this
-2,4-8,10-11,15-18,21
The output will take the consecutive number become one number
This things is new for me, any help would be appreciated, thanks
Below I created function for replacing a sequence in an array with a string containing its range. There are three functions.
getConsectiveCount will take array and index as arguments and will get the count of consecutive numbers after that.
replaceFirstConsective will take array and will replace only first sequence in the array.
replaceAllConsectives will replace all the sequences in an array.
const arr = [-2,4,5,6,7,8,10,11,15,16,17,18,21];
const getConsectiveCount = (arr, index) => {
let count = 0;
for(let i = index; i < arr.length; i++){
if(arr[i + 1] === arr[index] + (i - index) + 1){
count++;
}
}
return count;
}
console.log(getConsectiveCount(arr, 1));
const replaceFirstConsective = (arr) => {
for(let i = 0; i < arr.length; i++){
let count = getConsectiveCount(arr,i);
if(count){
return [...arr.slice(0, i), `${arr[i]}-${arr[i + count]}`, ...arr.slice(i + count + 1)]
}
}
return arr;
}
const replaceAllConsectives = (arr) => {
for(let i = 0; i < arr.length;i++){
arr = replaceFirstConsective(arr)
}
return arr;
}
console.log(JSON.stringify(replaceAllConsectives(arr)))
const inp = [-2,4,5,6,7,8,10,11,15,16,17,18,21];
let res = [];
for(let i=0;i<inp.length;i++){
let b = inp[i];
let j = i+1;
while(j<inp.length){
if(b+1 == inp[j]){
b = inp[j++];
continue;
}
break;
}
if(i == j-1){
res.push(inp[i]);
}
else{
res.push(inp[i]+"-"+inp[j-1]);
i=j-1;
}
}
console.log(res);
Check this if it helps.
I have done that :
const arr1 = [-2,4,5,6,7,8,10,11,15,16,17,18,21]
const arr2 = arr1.reduce((a,c,i,{[i+1]:nxt})=>
{
if (!a.s1) a.s1 = c.toString(10)
if ( (c+1) !== nxt )
{
a.s1 += a.s2 ? `_${a.s2}` : ''
a.r.push(a.s1)
a.s1 = a.s2 = ''
}
else a.s2 = nxt.toString(10)
return (nxt===undefined) ? a.r : a
},{r:[],s1:'',s2:''})
console.log(JSON.stringify( arr2 ))
.as-console-wrapper { max-height: 100% !important; top: 0; }
It's an interesting problem. The way I solved it was to iterate over the array and then find the index of the last consecutive number from the current one. We can either write the single number into the result array or write the range string in there and continue from the next number after the range.
function lastConsecutive(arr, start)
{
let ind = start;
while(ind < arr.length && (arr[ind] + 1) == arr[ind + 1])
{
ind++;
}
return ind;
}
function consecCollapse(nums)
{
let i = 0;
const result = [];
while (i < nums.length)
{
let n = lastConsecutive(nums, i);
result.push((n == i) ? nums[n]+"" : nums[i]+"-"+nums[n]);
i = n + 1;
}
return result;
}
console.log(consecCollapse([-2,4,5,6,7,8,10,11,15,16,17,18,21]));
const arr = [-2,4,5,6,7,8,10,11,15,16,17,18,21];
const _newArray = [];
let start = arr[0];
let end = start;
for(let i=1; i<=arr.length; i++) {
let elem = arr[i];
if (elem === end+1) {
end = elem; // update the end value (range)
}else {
if (end !== start) {
_newArray.push(`${start}-${end}`);
} else {
_newArray.push(start);
}
start = elem;
end = start;
}
}
console.log(_newArray.join(','))
If a sorted array with unique numbers, this is how I find the range of numbers with dash:
function findRange(arr) {
const results = []
for (let i = 0; i < arr.length; i++) {
// only more than 2 consecutive numbers can be form a range
if (arr[i + 1] === arr[i] + 1 && arr[i + 2] === arr[i] + 2) {
// store the first number of a range
results.push(arr[i])
// loop until meet the next one is not consecutive
while (arr[i] + 1 === arr[i + 1]) {
i++
}
// store the last number of a range with '-' in between
results[results.length - 1] = results[results.length - 1] + '-' + arr[i]
} else {
// if only 2 consecutive number or not consecutive at all
results.push(arr[i])
}
}
return results
}
console.log(findRange([1, 2, 3, 4, 6, 7, 8, 9]))
console.log(findRange([1, 2, 4, 6, 7, 8, 9]))
console.log(findRange([-2,4,5,6,7,8,10,11,15,16,17,18,21]))
I encountered this question:
Find longest decrease sequence in an array
So basically, if [3,1,4,3,2,1,2,3] is given, it will return [4,3,2,1] as the longest sequence within it. I was able to solve this question with O(n^2) time complexity but I feel like I can do O(n).
I know I am probably making a giant logic mistake but this is how I approach with 2 pointers method.
function findDecreaseSubArray(arr) {
let subList = [arr[0]];
// let maxLength= 0;
let left = 0;
for (let right = 1; right < arr.length; right++) {
// maxLength = (maxLength, windowEnd - windowStart + 1);
if (arr[left] > arr[right]) {
subList.push(arr[right]);
}else{
subList.shift();
left = right-1;
}
}
}
What I am trying to accomplish is moving left and right pointers if left element is larger than right, and if so, push both elements into sublist array.
My brain starts giving 404 error after this, so there are 2 subarrays that are decreasing, one of them is [3,1] and the other one is [4,3,2,1].
How could I track one subarray is larger than the other subarray? Or there is a better way to approach this solution?
Any hints, hands, code snippet would highly appreciated. (And sorry about my code snippet, I know it is shitty but I just wanted to display some of my thoughts on code)
You just have to iterate over the array once, but keep track of the start and length of the longest sequence as you progress through it:
var arr = [3,1,4,3,2,1,2,3];
function findDecreaseSubArray(arr) {
let startIndex = 0;
let length = 1;
let longestSequence = {
startIndex: 0,
length: 1
}
arr.forEach((element, index, arr) => {
if (index === 0) return;
if (element < arr[index -1]) {
length += 1;
} else {
length = 1;
startIndex = index;
}
if (length > longestSequence.length) {
longestSequence.length = length;
longestSequence.startIndex = startIndex;
}
})
return longestSequence;
}
console.log(findDecreaseSubArray(arr));
This approach supports VLAZ's comment and uses only the indices of the array.
function longestDecreasing(array) {
var longest,
start = 0,
i;
for (i = 0; i < array.length; i++) {
if (!i || array[i] < array[i - 1]) continue;
if (!longest || longest[1] - longest[0] < i - start) longest = [start, i];
start = i;
}
return array.slice(...longest && longest[1] - longest[0] > i - start
? longest
: [start, i]);
}
console.log(longestDecreasing([3, 1, 4, 3, 2, 1, 2, 3]))
It would probably be easier to iterate normally, from the first index to the end, while pushing sequential sequences to an array, which gets reset when a number greater than the last is found. When resetting, assign the resulting sequence array to a more permanent variable if it's longer than the array in that permanent variable:
const findDecreaseSubArray = (arr) => {
let longestSoFar = [];
let currentSequence = [];
const reset = (newItem) => {
if (currentSequence.length > longestSoFar.length) {
longestSoFar = currentSequence;
}
currentSequence = [newItem];
};
for (const item of arr) {
if (currentSequence.length && item > currentSequence[currentSequence.length - 1]) {
reset(item);
} else {
currentSequence.push(item);
}
}
reset();
return longestSoFar;
};
console.log(findDecreaseSubArray([3,1,4,3,2,1,2,3]));
Here is my code.
function getLongestDecreaseSequence(arr) {
if (
Object.prototype.toString.call(arr) !== "[object Array]" ||
arr.length == 0
) {
return arr;
}
let longestArr = [];
let tmpArr = [],
lastTmpIndex = -1;
arr.forEach((value, i) => {
if (arr[lastTmpIndex] < value) {
tmpArr = [];
}
// no matter what, I will put it in tmpArray
lastTmpIndex = i;
tmpArr.push(value);
if (longestArr.length < tmpArr.length) {
longestArr = tmpArr;
}
});
return longestArr;
}
console.log(getLongestDecreaseSequence([3, 1, 4, 3, 2, 1, 2, 3]));
This looks like a reducing job.
var a = [3,1,4,3,2,1,2,3],
r = a.reduce( function(r,e){
var l = r[r.length-1];
return l[l.length-1] > e ? ( l.push(e)
, r
)
: ( r.push([e])
, r
);
}
, [[]]
)
.reduce((p,c) => p.length > c.length ? p : c);
console.log(r);
Given an array of integers, find the pair of adjacent elements that has the largest product and return that product.
and here is my code
function adjacentElementsProduct(inputArray) {
var arr = inputArray;
var x=0;
var y=0;
var p=0;
for(var i=0;i<arr.length;i++){
x=arr[i];
y=arr[i+1];
if(x*y>p){
p=x*y;
};
};
return p;
};
the problem is all the tests works fine but except the array with the negative product as it shown in the attached photo
can anyone help .. and thanks in advance
You could start with a really large negative value, instead of zero.
var p = -Infinity;
You are initializing the variable p to zero. That means any multiplication values smaller than that are not accepted. Rather set it to the smallest possible integer value:
var p = Number.MIN_SAFE_INTEGER;
function adjacentElementsProduct(inputArray) {
var arr = inputArray;
var x = 0;
var y = 0;
var p = Number.MIN_SAFE_INTEGER;
for (var i = 0; i < arr.length; i++) {
x = arr[i];
y = arr[i + 1];
if (x * y > p) {
p = x * y;
};
};
return p;
};
console.log(adjacentElementsProduct([-23, 4, -3, 8, -12]));
This is quite simple actually
function adjacentElementsProduct(inputArray) {
let max = -Infinity;
for (let i = 1; i < inputArray.length; i++) {
max = Math.max(inputArray[i] * inputArray[i - 1], max);
}
return max;
}
This is quite simple actually
const solution = (inputArray) => Math.max(...inputArray.slice(0, -1).map((n, index) => n * inputArray[index + 1]))
console.log(solution([3, 6, -2, -5, 7, 3]))
function solution(inputArray: number[]): number {
var max = -Infinity;
for(var i=0; i+1<inputArray.length; i++)
{
if(max<(inputArray[i]*inputArray[i+1])){
max=inputArray[i]*inputArray[i+1];
}
}
return max;
}
console.log(solution([2,3,6]))
I had the same problem at first, defining the first max as 0. Then i came up with this:
function solution(inputArray) {
let products = inputArray.map(function(x, index){
return inputArray[index+1] != undefined? x *inputArray[index+1] : -Infinity;
})
return Math.max(...products);
}
Problem:
Given an array of integers, find the pair of adjacent elements that has the largest product and return that product. #javascript #arraymethods
function solution(inputArray) {
let productsArr = []; // to hold the products of adjacent elements
let n = 0;
for (let i = 0; i < inputArray.length; i++) {
if (i < inputArray.length - 1)
{
productsArr[n] = inputArray[i] * inputArray[i + 1];
n++;
}
}
return productsArr.reduce((aggr, val) => Math.max(aggr, val)); // to find out the biggest product
}
Here's a very simple implementation without using any additional variables (actually less), and no special values. Just simple logic.
function adjacentElementsProduct(inputArray) {
var c =inputArray[0]*inputArray[1];
var p = c;
for(var i=1;i<inputArray.length;i++){
console.log(c);
var c=inputArray[i]*inputArray[i+1];
if(c > p){
p=c;
};
};
return p;
};
console.log("minimum product = " + adjacentElementsProduct([-23,4,-3,8,-12]));
What I did was, initialize a variable c (current product) with the product of first two elements of the array. And then I declared the variable p and initialize it to c. This way, all other products are compared to this product. Rest is simple.
Hope it helps. :)
you can try to initialize a integer as negative infinity value -math.inf and then use the python ternary operator var=true if condition else false to find the maximum value
code in python
def adjacentarray(a):
maximum=-math.inf
for i,in range(0,len(a)-1):
maximum=a[i]*a[i+1] if a[i]*a[i+1]>maximum else maximum
return maximum
code in javascript
function adjacentElementsProduct(a) {
var maximum=-Infinity;
for (var i=0;i<a.length-1;i++){
maximum= a[i]*a[i+1]>maximum?a[i]*a[i+1]:maximum;
}
return maximum;
}
function solution(inputArray) {
let first, second, sum = []
inputArray.map((arr,index)=>{
first = arr;
second = inputArray[index+1]
if(second == undefined){
return second
}
return sum.push(first * second)
})
let last = sum.sort().reduce((pre,next)=> {
return pre > next ? pre : next
})
return last;
}
//Kotlin
fun solution(inputArray: MutableList<Int>): Int {
var result: Int = Int.MIN_VALUE
for (i in 0..inputArray.size - 2) {
if (inputArray[i] * inputArray[i + 1] > result)
result = inputArray[i] * inputArray[i + 1]
}
return result
}
import 'dart:math';
int solution(List<int> inputArray) {
//assumption for highest number
int highestNumber = inputArray[0] * inputArray[1] ;
//we'll go through the array to campare the highestNumber
//with next index
for(var i = 1 ; i < inputArray.length ; i++){
highestNumber = max(highestNumber, inputArray[i] * inputArray[i - 1]);
}
return highestNumber;
}
In Javascript, you could use the reduce method from an array to avoid iterating in a for loop, just like this.
function solution(inputArray) {
let maxProd = []
inputArray.reduce((accumulator, currentValue) => {
maxProd.push(accumulator*currentValue)
return currentValue
},
);
return Math.max(...maxProd)
}
Once you have in the maxProd array the products, you use the spread operator to get the numbers and using Math.max() you get the largest
python solution
You can make a loop from 1 to end of your list and do the following arithmetic operations
def solution(inputArray):
list1 =[]
for i in range(1,len(inputArray)):
list1.append(inputArray[i]*inputArray[i-1])
return max(list1)
Here is a solution in PHP that is quite simple.
function solution($inputArray) {
$largest = null;
$pos = null;
for($i = 0; $i < count($inputArray) -1; $i++){
$pos = ($inputArray[$i] * $inputArray[$i+1]);
if($largest < $pos){
$largest = $pos;
}
}
return $largest ?? 0;
}
You can try to create a new array of length (arr.length-1) inside the function and append the products of adjacent numbers to this new array. Then find the largest number in the array and return it. This will solve the problem with negative product.
function adjacentElementsProduct(inputArray) {
var arr = inputArray;
var prodArr[];
var p;
for (var i = 0; i < arr.length-1; i++) {
prodArr[i] = arr[i]*arr[i+1];
};
for (j=prodArr.length; j--){
if (prodArr[j] > p) {
p = prodArr[j];
};
return p;
};
console.log(adjacentElementsProduct([-23, 4, -3, 8, -12]));
The var p which saves the max product should be initialized as small as possible instead of a 0. So that when the product is negative, it will still meet the if condition and save the value.
Here is a C# solution:
static void Main(string[] args)
{
int[] arr = { 1, -4, 3, -6, -7, 0 };
Console.WriteLine(FindMaxProduct(arr));
Console.ReadKey();
}
static int FindMaxProduct(int[] arr) {
int currentProduct = 0;
int maxProduct = int.MinValue;
int a=0, b = 0;
for (int i = 0, j = i + 1; i < arr.Length - 1 && j < arr.Length; i++, j++)
{
currentProduct = arr[i] * arr[j];
if (currentProduct>maxProduct) {
a = arr[i];
b = arr[j];
maxProduct = currentProduct;
}
}
Console.WriteLine("The max product is {0}, the two nums are {1} and {2}.",maxProduct,a,b);
return maxProduct;
}
function solution(inputArray) {
let f, s, arr = []
for(let i=0; i<inputArray.length; i++){
f = inputArray[i]
s = inputArray[i+1]
arr.push(f*s)
}
let max = arr.sort((a, b) => b - a)
return max[0]
}
console.log(solution([3, 6, -2, -5, 7, 3]))
This should help, wrote it in python. Concept: Pass an empty list, for every consecutive product keep storing it in the list. Then just return the max value.
def consecutive_product_max(a):
lst2 = []
for i in range(0, len(a)-1):
x = a[i] * a[i+1]
lst2.append(x)
return max(lst2)
var store = ['1','2','2','3','4'];
I want to find out that 2 appear the most in the array. How do I go about doing that?
I would do something like:
var store = ['1','2','2','3','4'];
var frequency = {}; // array of frequency.
var max = 0; // holds the max frequency.
var result; // holds the max frequency element.
for(var v in store) {
frequency[store[v]]=(frequency[store[v]] || 0)+1; // increment frequency.
if(frequency[store[v]] > max) { // is this frequency > max so far ?
max = frequency[store[v]]; // update max.
result = store[v]; // update result.
}
}
Solution with emphasis to Array.prototype.forEach and the problem of getting more than one key if the max count is shared among more items.
Edit: Proposal with one loop, only.
var store = ['1', '2', '2', '3', '4', '5', '5'],
distribution = {},
max = 0,
result = [];
store.forEach(function (a) {
distribution[a] = (distribution[a] || 0) + 1;
if (distribution[a] > max) {
max = distribution[a];
result = [a];
return;
}
if (distribution[a] === max) {
result.push(a);
}
});
console.log('max: ' + max);
console.log('key/s with max count: ' + JSON.stringify(result));
console.log(distribution);
arr.sort();
var max=0,result,freq = 0;
for(var i=0; i < arr.length; i++){
if(arr[i]===arr[i+1]){
freq++;
}
else {
freq=0;
}
if(freq>max){
result = arr[i];
max = freq;
}
}
return result;
Make a histogram, find the key for the maximum number in the histogram.
var hist = [];
for (var i = 0; i < store.length; i++) {
var n = store[i];
if (hist[n] === undefined) hist[n] = 0;
else hist[n]++;
}
var best_count = hist[store[0]];
var best = store[0];
for (var i = 0; i < store.length; i++) {
if (hist[store[i]] > best_count) {
best_count = hist[store[i]];
best = store[i];
}
}
alert(best + ' occurs the most at ' + best_count + ' occurrences');
This assumes either there are no ties, or you don't care which is selected.
Another ES6 option. Works with strings or numbers.
function mode(arr) {
const store = {}
arr.forEach((num) => store[num] ? store[num] += 1 : store[num] = 1)
return Object.keys(store).sort((a, b) => store[b] - store[a])[0]
}
If the array is sorted this should work:
function popular(array) {
if (array.length == 0) return [null, 0];
var n = max = 1, maxNum = array[0], pv, cv;
for(var i = 0; i < array.length; i++, pv = array[i-1], cv = array[i]) {
if (pv == cv) {
if (++n >= max) {
max = n; maxNum = cv;
}
} else n = 1;
}
return [maxNum, max];
};
popular([1,2,2,3,4,9,9,9,9,1,1])
[9, 4]
popular([1,2,2,3,4,9,9,9,9,1,1,10,10,10,10,10])
[10, 5]
This version will quit looking when the count exceeds the number of items not yet counted.
It works without sorting the array.
Array.prototype.most= function(){
var L= this.length, freq= [], unique= [],
tem, max= 1, index, count;
while(L>= max){
tem= this[--L];
if(unique.indexOf(tem)== -1){
unique.push(tem);
index= -1, count= 0;
while((index= this.indexOf(tem, index+1))!= -1){
++count;
}
if(count> max){
freq= [tem];
max= count;
}
else if(count== max) freq.push(tem);
}
}
return [freq, max];
}
//test
var A= ["apples","oranges","oranges","oranges","bananas",
"bananas","oranges","bananas"];
alert(A.most()) // [oranges,4]
A.push("bananas");
alert(A.most()) // [bananas,oranges,4]
I solved it this way for finding the most common integer
function mostCommon(arr) {
// finds the first most common integer, doesn't account for 2 equally common integers (a tie)
freq = [];
// set all frequency counts to 0
for(i = 0; i < arr[arr.length-1]; i++) {
freq[i] = 0;
}
// use index in freq to represent the number, and the value at the index represent the frequency count
for(i = 0; i < arr.length; i++) {
freq[arr[i]]++;
}
// find biggest number's index, that's the most frequent integer
mostCommon = freq[0];
for(i = 0; i < freq.length; i++) {
if(freq[i] > mostCommon) {
mostCommon = i;
}
}
return mostCommon;
}
This is my solution.
var max_frequent_elements = function(arr){
var a = [], b = [], prev;
arr.sort();
for ( var i = 0; i < arr.length; i++ ) {
if ( arr[i] !== prev ) {
a.push(arr[i]);
b.push(1);
} else {
b[b.length-1]++;
}
prev = arr[i];
}
var max = b[0]
for(var p=1;p<b.length;p++){
if(b[p]>max)max=b[p]
}
var indices = []
for(var q=0;q<a.length;q++){
if(b[q]==max){indices.push(a[q])}
}
return indices;
};
All the solutions above are iterative.
Here's a ES6 functional mutation-less version:
Array.prototype.mostRepresented = function() {
const indexedElements = this.reduce((result, element) => {
return result.map(el => {
return {
value: el.value,
count: el.count + (el.value === element ? 1 : 0),
};
}).concat(result.some(el => el.value === element) ? [] : {value: element, count: 1});
}, []);
return (indexedElements.slice(1).reduce(
(result, indexedElement) => (indexedElement.count > result.count ? indexedElement : result),
indexedElements[0]) || {}).value;
};
It could be optimized in specific situations where performance is the bottleneck, but it has a great advantage of working with any kind of array elements.
The last line could be replaced with:
return (indexedElements.maxBy(el => el.count) || {}).value;
With:
Array.prototype.maxBy = function(fn) {
return this.slice(1).reduce((result, element) => (fn(element) > fn(result) ? element : result), this[0]);
};
for clarity
If the array contains strings try this solution
function GetMaxFrequency (array) {
var store = array;
var frequency = []; // array of frequency.
var result; // holds the max frequency element.
for(var v in store) {
var target = store[v];
var numOccurences = $.grep(store, function (elem) {
return elem === target;
}).length;
frequency.push(numOccurences);
}
maxValue = Math.max.apply(this, frequency);
result = store[$.inArray(maxValue,frequency)];
return result;
}
var store = ['ff','cc','cc','ff','ff','ff','ff','ff','ff','yahya','yahya','cc','yahya'];
alert(GetMaxFrequency(store));
A fairly short solution.
function mostCommon(list) {
var keyCounts = {};
var topCount = 0;
var topKey = {};
list.forEach(function(item, val) {
keyCounts[item] = keyCounts[item] + 1 || 1;
if (keyCounts[item] > topCount) {
topKey = item;
topCount = keyCounts[item];
}
});
return topKey;
}
document.write(mostCommon(['AA', 'AA', 'AB', 'AC']))
This solution returns an array of the most appearing numbers in an array, in case multiple numbers appear at the "max" times.
function mode(numbers) {
var counterObj = {};
var max = 0;
var result = [];
for(let num in numbers) {
counterObj[numbers[num]] = (counterObj[numbers[num]] || 0) + 1;
if(counterObj[numbers[num]] >= max) {
max = counterObj[numbers[num]];
}
}
for (let num in counterObj) {
if(counterObj[num] == max) {
result.push(parseInt(num));
}
}
return result;
}
var store = ['1','2','2','3','4'];
I want to find out that 2 appear the most in the array. How do I go about doing that?
I would do something like:
var store = ['1','2','2','3','4'];
var frequency = {}; // array of frequency.
var max = 0; // holds the max frequency.
var result; // holds the max frequency element.
for(var v in store) {
frequency[store[v]]=(frequency[store[v]] || 0)+1; // increment frequency.
if(frequency[store[v]] > max) { // is this frequency > max so far ?
max = frequency[store[v]]; // update max.
result = store[v]; // update result.
}
}
Solution with emphasis to Array.prototype.forEach and the problem of getting more than one key if the max count is shared among more items.
Edit: Proposal with one loop, only.
var store = ['1', '2', '2', '3', '4', '5', '5'],
distribution = {},
max = 0,
result = [];
store.forEach(function (a) {
distribution[a] = (distribution[a] || 0) + 1;
if (distribution[a] > max) {
max = distribution[a];
result = [a];
return;
}
if (distribution[a] === max) {
result.push(a);
}
});
console.log('max: ' + max);
console.log('key/s with max count: ' + JSON.stringify(result));
console.log(distribution);
arr.sort();
var max=0,result,freq = 0;
for(var i=0; i < arr.length; i++){
if(arr[i]===arr[i+1]){
freq++;
}
else {
freq=0;
}
if(freq>max){
result = arr[i];
max = freq;
}
}
return result;
Make a histogram, find the key for the maximum number in the histogram.
var hist = [];
for (var i = 0; i < store.length; i++) {
var n = store[i];
if (hist[n] === undefined) hist[n] = 0;
else hist[n]++;
}
var best_count = hist[store[0]];
var best = store[0];
for (var i = 0; i < store.length; i++) {
if (hist[store[i]] > best_count) {
best_count = hist[store[i]];
best = store[i];
}
}
alert(best + ' occurs the most at ' + best_count + ' occurrences');
This assumes either there are no ties, or you don't care which is selected.
Another ES6 option. Works with strings or numbers.
function mode(arr) {
const store = {}
arr.forEach((num) => store[num] ? store[num] += 1 : store[num] = 1)
return Object.keys(store).sort((a, b) => store[b] - store[a])[0]
}
If the array is sorted this should work:
function popular(array) {
if (array.length == 0) return [null, 0];
var n = max = 1, maxNum = array[0], pv, cv;
for(var i = 0; i < array.length; i++, pv = array[i-1], cv = array[i]) {
if (pv == cv) {
if (++n >= max) {
max = n; maxNum = cv;
}
} else n = 1;
}
return [maxNum, max];
};
popular([1,2,2,3,4,9,9,9,9,1,1])
[9, 4]
popular([1,2,2,3,4,9,9,9,9,1,1,10,10,10,10,10])
[10, 5]
This version will quit looking when the count exceeds the number of items not yet counted.
It works without sorting the array.
Array.prototype.most= function(){
var L= this.length, freq= [], unique= [],
tem, max= 1, index, count;
while(L>= max){
tem= this[--L];
if(unique.indexOf(tem)== -1){
unique.push(tem);
index= -1, count= 0;
while((index= this.indexOf(tem, index+1))!= -1){
++count;
}
if(count> max){
freq= [tem];
max= count;
}
else if(count== max) freq.push(tem);
}
}
return [freq, max];
}
//test
var A= ["apples","oranges","oranges","oranges","bananas",
"bananas","oranges","bananas"];
alert(A.most()) // [oranges,4]
A.push("bananas");
alert(A.most()) // [bananas,oranges,4]
I solved it this way for finding the most common integer
function mostCommon(arr) {
// finds the first most common integer, doesn't account for 2 equally common integers (a tie)
freq = [];
// set all frequency counts to 0
for(i = 0; i < arr[arr.length-1]; i++) {
freq[i] = 0;
}
// use index in freq to represent the number, and the value at the index represent the frequency count
for(i = 0; i < arr.length; i++) {
freq[arr[i]]++;
}
// find biggest number's index, that's the most frequent integer
mostCommon = freq[0];
for(i = 0; i < freq.length; i++) {
if(freq[i] > mostCommon) {
mostCommon = i;
}
}
return mostCommon;
}
This is my solution.
var max_frequent_elements = function(arr){
var a = [], b = [], prev;
arr.sort();
for ( var i = 0; i < arr.length; i++ ) {
if ( arr[i] !== prev ) {
a.push(arr[i]);
b.push(1);
} else {
b[b.length-1]++;
}
prev = arr[i];
}
var max = b[0]
for(var p=1;p<b.length;p++){
if(b[p]>max)max=b[p]
}
var indices = []
for(var q=0;q<a.length;q++){
if(b[q]==max){indices.push(a[q])}
}
return indices;
};
All the solutions above are iterative.
Here's a ES6 functional mutation-less version:
Array.prototype.mostRepresented = function() {
const indexedElements = this.reduce((result, element) => {
return result.map(el => {
return {
value: el.value,
count: el.count + (el.value === element ? 1 : 0),
};
}).concat(result.some(el => el.value === element) ? [] : {value: element, count: 1});
}, []);
return (indexedElements.slice(1).reduce(
(result, indexedElement) => (indexedElement.count > result.count ? indexedElement : result),
indexedElements[0]) || {}).value;
};
It could be optimized in specific situations where performance is the bottleneck, but it has a great advantage of working with any kind of array elements.
The last line could be replaced with:
return (indexedElements.maxBy(el => el.count) || {}).value;
With:
Array.prototype.maxBy = function(fn) {
return this.slice(1).reduce((result, element) => (fn(element) > fn(result) ? element : result), this[0]);
};
for clarity
If the array contains strings try this solution
function GetMaxFrequency (array) {
var store = array;
var frequency = []; // array of frequency.
var result; // holds the max frequency element.
for(var v in store) {
var target = store[v];
var numOccurences = $.grep(store, function (elem) {
return elem === target;
}).length;
frequency.push(numOccurences);
}
maxValue = Math.max.apply(this, frequency);
result = store[$.inArray(maxValue,frequency)];
return result;
}
var store = ['ff','cc','cc','ff','ff','ff','ff','ff','ff','yahya','yahya','cc','yahya'];
alert(GetMaxFrequency(store));
A fairly short solution.
function mostCommon(list) {
var keyCounts = {};
var topCount = 0;
var topKey = {};
list.forEach(function(item, val) {
keyCounts[item] = keyCounts[item] + 1 || 1;
if (keyCounts[item] > topCount) {
topKey = item;
topCount = keyCounts[item];
}
});
return topKey;
}
document.write(mostCommon(['AA', 'AA', 'AB', 'AC']))
This solution returns an array of the most appearing numbers in an array, in case multiple numbers appear at the "max" times.
function mode(numbers) {
var counterObj = {};
var max = 0;
var result = [];
for(let num in numbers) {
counterObj[numbers[num]] = (counterObj[numbers[num]] || 0) + 1;
if(counterObj[numbers[num]] >= max) {
max = counterObj[numbers[num]];
}
}
for (let num in counterObj) {
if(counterObj[num] == max) {
result.push(parseInt(num));
}
}
return result;
}