I am completing the hackerrank's 10 days of javascript challenge. The question:
write a function to take an array as an argument and then return the second largest element in the array.
I have written the code but my code is returning the largest element and not the second largest as asked.
function getSecondLargest(nums) {
// Complete the function
var largest=nums[0];
for(let i=1;i<nums.length;++i)
{
if(nums[i]>largest)
largest=nums[i];
}
var large=nums[0];
for(let j=1;j<nums.length;++j)
{
if(large<nums[j]&&large<largest)
large=nums[j];
}
return large;
}
When input array nums={2,3,6,6,5} the result is coming 6 while expected output is 5. Please help and point out the errors in the function code below.
should not initialize large with first value var large=nums[0]; because it may appear the biggest value and won't work
should use nums[j]<largest instead of large<largest as mentioned above
I think don't need second loop as all checks can be done in first loop, and you can assign prev largest to large whenever you change it:
function getSecondLargest(nums) {
var largest = nums[0];
var large;
for (let i = 1; i < nums.length; ++i) {
if (nums[i] > largest) {
large = largest;
largest = nums[i];
} else if (nums[i] > large || typeof large === 'undefined') {
large = nums[i]
}
}
return large;
}
console.log(getSecondLargest([5,1-2,3]))
console.log(getSecondLargest([-5,1,-2,3]))
GET SECOND LARGEST
first, I create new array with unique values.
let arr = [...new Set(nums)];
second, sort value using built-in function .sort().
note : by default .sort() always sorts asciibetically, but for some testcase, it doesn't work. So, I put (a, b) => { return a - b } to make sure it will work properly.
arr = arr.sort((a, b) => { return a -b });
third, get the value from arr
let result = arr[arr.length - 2] || arr[0];
finally, return the result
return result
function getSecondLargest(nums) {
let arr = [...new Set(nums)];
//Javascript's array member method .sort( always sorts asciibetically.
arr = arr.sort((a, b) => { return a - b });
let result = arr[arr.length - 2] || arr[0];
return result
}
Just one minor change:
Use nums[j]<largest instead of large<largest in the second for loop
function getSecondLargest(nums) {
// Complete the function
var largest=nums[0];
for(let i=1;i<nums.length;++i)
{
if(nums[i]>largest)
largest=nums[i];
}
var large;
//To ensure that the selected number is not the largest
for(let j=0;j<nums.length;++j)
{
if (nums[j] !== largest){
large = nums[j];
break;
}
}
for(let j=1;j<nums.length;++j)
{
if(large<nums[j]&&nums[j]!=largest)
large=nums[j];
else
console.log(large)
}
return large;
}
var secondLargest = getSecondLargest([6,3,6,6,5]);
console.log("Second largest number", secondLargest);
If you want to avoid using library functions like #ifaruki suggests, this line
if(large<nums[j]&&large<largest)
should read
if (large<nums[j] && nums[j] < largest)
Sorting and picking the second or second-to-last value fails when there are duplicates of the highest value in the input array.
Another easiest logic is to remove duplicates from the array and sort.
let givenArray = [2, 3, 6, 6, 5];
let uniqueArray = [...new Set(givenArray)];
console.log("The second largets element is", uniqueArray.sort()[uniqueArray.length - 2]);
I know you had your question answered, just thought I would provide my solution for any future users looking into this.
You can use reduce to go through the array while remembering the two largest numbers so far.
You just make a simple reduction function:
function twoMax(two_max, candidate)
{
if (candidate > two_max[0]) return [candidate,two_max[0]];
else if (candidate > two_max[1]) return [two_max[0],candidate];
else return two_max;
}
And then you use it for example like this:
let my_array = [0,1,5,7,0,8,12];
let two_largest = my_array.reduce(twoMax,[-Infinity,-Infinity]);
let second_largest = two_largest[1];
This solution doesn't require sorting and goes through the array only once.
If you want to avoid using **sort method. I think here's the easiest logic to do that, which will also work in arrays where there's duplicates of largest integer exists.
function getSecondLargest(arr) {
const largest = Math.max.apply(null, arr);
for (let i = 0; i < arr.length; i++) {
if (largest === arr[i]) {
arr[i] = -Infinity;
}
}
return Math.max.apply(null, arr);
}
console.log(getSecondLargest([5, 7, 11, 11, 11])); //7
Related
I'm working with arrays and I have made the code so that it prompts the user to enter an array of their choice of size. It lets them input each number one by one. I would like so that once this is done the program will then find and print the second largest number in the array but I'm not sure how to go about it.
I would highly like to avoid using infinity or splices just wanted to do something that a newbie like me can understand.
CODE:
<html>
<head>
<script type="text/javascript">
//<!--
function show_prompt() {
var n = document.frm.text1.value;
if (n == "") {
alert("Please enter the array size");
} else {
var a = new Array();
var temp;
for (i = 0; i < n; i++) {
var num = prompt("Please enter a number", " ");
document.frm.arrayTxt.value = document.frm.arrayTxt.value + num + "\n";
a[i] = parseInt(num);
}
document.frm.text1.value = "";
document.frm.arrayTxt.value = "";
}
}
//-->
</script>
</head>
<body>
<form name="frm">
<center>
<hr color="red"> Enter the array size : <input type="text" name="text1"><br><br>
<input type="button" onclick="show_prompt()" value="Submit"> <br><br>
<textarea name="arrayTxt" rows="10" cols="4"></textarea><br>
</center>
</form>
</body>
</html>
You can use a simple for loop, storing the largest and second largest elements, in O(n) time. There is no need to sort the array, which is very inefficient for larger arrays.
let largest = Number.MIN_SAFE_INTEGER, largest2 = largest;
for(const num of a){
if(num > largest){
largest2 = largest;
largest = num;
} else if(num > largest2){
largest2 = num;
}
}
console.log(largest2);
If you need the strictly second largest element, you can remove the duplicates from the array using Set.
a = [...new Set(a)];//add this before the loop
The easiest option is probably to simply sort the array and select the second element. JavaScript Arrays nowadays actually support sorting an array by passing a comparison function.
[2, 5, 3, 1, 4].sort((a, b) => b - a); // [5, 4, 3, 2, 1]
Simply speaking, if the function returns a value < 0, a is inserted before b, if it returns a value > 0 it a is insterted behind b. If it returns 0 nothing is changed. b - a means "if a > b, move it to the front", resulting in an array sorted highest to lowest. Annoyingly, if you don't pass in a comparison function, the elements are converted to strings and then sorted low to high ([1, 5, 10, 20] would be sorted [1, 10, 20, 5]).
To select the second hightest element you would simply select the second index:
a.sort((a, b) => b - a)[1];
Keep in mind, that after calling sort(), the original array will be changed as well.
A different approach would be more hands on, similar to finding the max value.
let max = 0, secondMax = 0;
for (let i = 0; i < a.length; i++) {
if (a[i] >= max) {
secondMax = max;
max = a[i];
} else if (a[i] > secondMax) {
secondMax = a[i];
}
}
This can be slightly more efficient when dealing with large arrays. But since you fill your array with user input, I would choose the first version. It is much shorter and more readable.
NOTE:
The so called "arrow notation" (a, b) => b - a is simply a shorthand for function (a, b) { return b - a; }. There is more to it, namely arrow functions inheriting this from the creation context, but this has no impact in this specific example.
You could just sort the array and return the second number. This should work fine:
var a = [5,3,1,6,8,3,9];
var sorted = a.sort((a, b) => b - a);
alert(a[1])
You could try this:
var numArray = [21, 81, 3847, 218];
numArray.sort((a, b) => b - a); // [3847, 218, 81, 21]
numArray = [...new Set(numArray)]; // Removes duplicates
console.log(numArray[1]); // Returns 218
This basically sorts the array in descending order, and selects the second number via index.
This is mine:
var first = 0;
var second = 0;
array.forEach(number => {
if (number > first) {
second = first;
first = number;
}
else if (number > second) second = number;
});
I believe this is more efficient than a sort but I'm not sure. It might be a more readable solution but it's definitely more over-engineered.
Here's a simple way to do this without using short method. This one will also work if there's even duplicate numbers of largest integer.
function getSecondLargest(arr) {
const largest = Math.max.apply(null, arr);
for (let i = 0; i < arr.length; i++) {
if (largest === arr[i]) {
arr[i] = -Infinity;
}
}
return Math.max.apply(null, arr);
}
console.log(getSecondLargest([2, 3, 4, 4, 4])); //3
I have this code that pairs same elements in an array, with the expectation that the array will have an odd length and it should return the only element that couldn't get a pair. So I wrote the code just well, and it works fine for smaller arrays, but with very large big integers of over 1 billion, the time complexity became O(N**2) and then the need to refactor my code to get a much better performance for large arrays and large array elements. Here is my code below;
function solution(A) {
if(!Array.isArray(A)) return 0;
var temp = new Array(A.length);
var position = 0;
for(let i=0; i<A.length; i++){
if(temp.includes(A[i])){
position = temp.indexOf(A[i]);
index = A.indexOf(A[i]);
delete temp[position];
delete A[index];
delete A[i];
}else{
temp[i] = A[i];
}
}
for(let j=0; j<A.length; j++){
if(A[j] !== undefined) return A[j];
else continue;
}
}
To test it, source data can look like [2,3,6,7,3,5,5,6,2] and it will give an output of 7. But when the array is so large up to [1,2,....] with length n = n=999,999, or n = 5000,000,000, the time complexity increases exponentially.
You might use Object to store non-paired elements only.
Please note that you don't need to store all the array elements and their counts in the Object and then filter by count (like #StepUp does).
Everything's been done in a single loop.
The function returns Array of all non-paired elements:
const solution = A => Array.isArray(A) ?
Object.keys(
A.reduce((r, k) => {
r[k] = r[k] || 0;
if (++r[k] > 1) delete r[k];
return r;
}, {})
) : [];
console.log(solution([2, 3, 6, 7, 3, 5, 5, 6, 2]))
We can try to find odd occurrences for one iteration by using great features of object. Object is key - value pair. So access to object key is O(1). So when we meet the same element, then we just increment value:
const hashMap = arr.reduce((a, c)=> {
a[c] = a[c] || 0;
a[c] += 1;
return a;
},{})
const result = Object.keys(hashMap).filter(key => hashMap[key] === 1);
An example:
let arr = [2, 3, 6, 7, 3, 5, 5, 6, 2];
const hashMap = arr.reduce((a, c)=> {
a[c] = a[c] || 0;
a[c] += 1;
return a;
},{})
const result = Object.keys(hashMap).filter(key => hashMap[key] === 1);
console.log(result);
My two 100% JavaScript solutions with optimized time complexity. The first one is using Set:
function solution(A) {
const pairs = new Set();
for (const num of A) {
if (pairs.has(num)) {
pairs.delete(num);
} else {
pairs.add(num);
}
}
const [unpaired] = pairs;
return unpaired;
}
The second one is using bitwise XOR:
function solution(A) {
let unpaired;
for (const num of A) {
unpaired ^= num;
}
return unpaired;
}
I'm studing a solution of this lesson:
https://app.codility.com/programmers/lessons/4-counting_elements/perm_check/
I headed up of this solution made my a github user.
https://github.com/daraosn/codility/tree/master/02-CountingElements/02-PermCheck/javascript
I did understand everything of the code below:
function solution(A) {
var N = A.length;
var sum = (N * (N+1)) / 2;
var tap = [];
for (var i in A) {
sum-=A[i];
if(tap[A[i]]) {
return 0;
}
tap[A[i]] = true;
}
return +(sum==0);
}
with exception of these code lines below:
if(tap[A[i]]) {
return 0;
}
tap[A[i]] = true;
What is its purppose? I didn't understand.
I did a test deleting these code lines from the answer in the
codility interface and it returned 75% right instead of 100% when I had these lines
function solution(A) {
const set = new Set(A)
const max = Math.max(...A)
return set.size === max && set.size === A.length ? 1:0
}
That section checks to see if the number being iterated over has been found before, and per the instructions, duplicates are forbidden:
A permutation is a sequence containing each element from 1 to N once, and only once.
On every normal iteration, the current number being iterated over is assigned to a property of tap:
tap[A[i]] = true;
Then, on subsequent iterations, that test checks to see if the new number being iterated over has already been used:
if(tap[A[i]]) {
return 0;
}
This helps to invalidate inputs like [2, 2, 2], while permitting inputs like [1, 2, 3].
That said, there are two major red flags with this. First, for..in shouldn't be used to iterate over arrays. Instead:
for (const num of A) {
// use num
}
Also, sparse arrays are a very bad idea - it would make much more sense to use an object:
var tap = {};
or a Set:
var tap = new Set();
for (const num of A) {
sum -= num;
if (tap.has(num)) {
return 0;
}
tap.add(num);
}
return +(sum == 0);
Array solution is not so proper way such above explaining. But I will put the solution(O(n)) in case you want :)
const solution = A => ~~(A.sort((a,b) => a-b).every((a,i) => a === i+1));
I have a very large array of arrays (on the order of 960,799 entries or possibly much larger). I need to process it into a new array such that:
Each sub-array contains no duplicates.
The main array contains no duplicate sub-arrays.
The problem is that "duplicate sub-arrays" must include arrays with the same values in a different order. In other words, if I had these sub-arrays:
[[1,2,3], [1,2,3], [3,1,2]]
They would all be considered duplicates and only one would be kept (any of them, it doesn't matter; I've been just keeping the first one; it's also fine if the order of the selected sub-array doesn't actually match, i.e. if the order of elements in the sub-array changes during processing).
My attempted solution has been to map all the sub-arrays into strings based on de-duping the sub-array, sorting it, and joining it with a delimiter. Then I de-dupe that final array, then map them back to arrays with a split. It works, but the process is extremely slow. It takes over 30 seconds for a single pass, and since the array I end up processing can grow exponentially larger, this is not acceptable. I need a more efficient algorithm.
Here's the code I'm using now that's slow (ret is the input array):
const stringList = ret.map(list => {
return [...new Set(list)].sort().join('|');
});
const hashSet = new Set(stringList);
const output = [...hashSet].map(str => str.split('|'));
Can anyone help me get the same result more efficiently? Thanks.
EDIT
To elaborate, I'm getting these massive input arrays by calculating what is essentially the power set of some input of strings. This is the code; if it's possible to stop it from producing duplicate entries in the first place, that would work well, too, I think:
// Calculate the Cartesian product of set s
function cart(s) {
return s.reduce((acc, val) => {
return acc.map((x, i) => {
return val.map(y => {
return x.concat([y]);
});
}).flat();
}, [[]]);
}
// Use the Cartesian product to calculate the power set of set s
function pset(s) {
let ret = [];
for (let i = 0; i < s.length; ++i) {
const temp = [];
for (let j = 0; j <= i; ++j) {
temp.push([].concat(s));
}
ret = ret.concat(cart(temp));
}
return ret;
}
You could generate the power set without duplicates.
function pset(array) {
function iter(index, temp) {
if (index >= array.length) {
temp.length && result.push(temp);
return;
}
iter(index + 1, temp.concat(array[index]));
iter(index + 1, temp);
}
var result = [];
iter(0, []);
return result;
}
console.log(pset(['a', 'b', 'c']));
.as-console-wrapper { max-height: 100% !important; top: 0; }
Given that I'm not able to perform a benchmark with real data, I can't verify how much faster this approach is for your use case, but by using basic for loops and avoiding functional code as much as conveniently possible, I've come up with the following:
const ret = [[1, 2, 3], [1, 2, 3], [3, 1, 2], [1, 4, 5], [4, 1, 5]];
function ascending (a, b) {
// works for strings and numbers
return -(a < b) || +(a > b);
}
function ascending2d (a, b) {
const aLength = a.length;
const bLength = b.length;
const length = Math.min(aLength, bLength);
for (let i = 0; i < length; ++i) {
const difference = ascending(a[i], b[i]);
if (difference !== 0) return difference;
}
return aLength - bLength;
}
for (let i = 0; i < ret.length; ++i) {
ret[i].sort(ascending);
}
ret.sort(ascending2d);
const output = [ret[0]];
for (let i = 1; i < ret.length; ++i) {
const value = ret[i];
if (ascending2d(ret[i - 1], value) !== 0) output.push(value);
}
console.log(output);
Let me know if this is an improvement over your current approach. You can always improve performance further by profiling your code and looking for bottlenecks that can be re-written.
Performance Benchmark
I've published a benchmark using the test data in my example here, comparing your original solution, my solution, and Andrew's solution. I couldn't include Nina's for comparison because hers doesn't perform deduplication on ret, instead it modifies the generation of ret.
EDIT: Nevermind, my implementation had no benchmarks. It is slower. Due to the underlying implementation of JSON.parse, JSON.stringify, and the default algorithm for Array#sort.
Since you're looking for bleeding edge performance, it's hard to get an elegant solution. If you instantiate an object with Object.create(null) you minimize the overhead for O(1) insertion. It creates a POJO with no prototype. You also don't need to check in the for in loop for Object.hasOwnProperty, because there's no prototype to search.
const ret = [[], [1, 2, 3], [3, 1, 2], [1, 4, 5], [4, 1, 5]];
const hashMap = Object.create(null)
function createUniqArraysOfPrimitiveArrays(ret) {
for (let i = 0; i < ret.length; i++) {
const currEl = ret[i]
if (currEl.length === 0) {
hashMap['[]'] = null
} else if (currEl.length === 1) {
hashMap[`[${currEl[0]}]`] = null
} else {
hashMap[JSON.stringify(currEl.sort())] = null
}
}
const outputArray = []
for (const array in hashMap) {
outputArray.push(JSON.parse(array))
}
return outputArray
}
console.log(createUniqArraysOfPrimitiveArrays(ret))
I have this:
var arr = [0, 21, 22, 7];
What's the best way to return the index of the highest value into another variable?
This is probably the best way, since it’s reliable and works on old browsers:
function indexOfMax(arr) {
if (arr.length === 0) {
return -1;
}
var max = arr[0];
var maxIndex = 0;
for (var i = 1; i < arr.length; i++) {
if (arr[i] > max) {
maxIndex = i;
max = arr[i];
}
}
return maxIndex;
}
There’s also this one-liner:
let i = arr.indexOf(Math.max(...arr));
It performs twice as many comparisons as necessary and will throw a RangeError on large arrays, though. I’d stick to the function.
In one line and probably faster then arr.indexOf(Math.max.apply(Math, arr)):
var a = [0, 21, 22, 7];
var indexOfMaxValue = a.reduce((iMax, x, i, arr) => x > arr[iMax] ? i : iMax, 0);
document.write("indexOfMaxValue = " + indexOfMaxValue); // prints "indexOfMaxValue = 2"
Where:
iMax - the best index so far (the index of the max element so far, on the first iteration iMax = 0 because the second argument to reduce() is 0, we can't omit the second argument to reduce() in our case)
x - the currently tested element from the array
i - the currently tested index
arr - our array ([0, 21, 22, 7])
About the reduce() method (from "JavaScript: The Definitive Guide" by David Flanagan):
reduce() takes two arguments. The first is the function that performs the reduction operation. The task of this reduction function is to somehow combine or reduce two values into a single value, and to return that reduced value.
Functions used with reduce() are different than the functions used with forEach() and map(). The familiar value, index, and array values are passed as the second, third, and fourth arguments. The first argument is the accumulated result of the reduction so far. On the first call to the function, this first argument is the initial value you passed as the
second argument to reduce(). On subsequent calls, it is the value returned by the previous invocation of the function.
When you invoke reduce() with no initial value, it uses the first element of the array as the initial value. This means that the first call to the reduction function will have the first and second array elements as its
first and second arguments.
Another solution of max using reduce:
[1,2,5,0,4].reduce((a,b,i) => a[0] < b ? [b,i] : a, [Number.MIN_VALUE,-1])
//[5,2]
This returns [5e-324, -1] if the array is empty. If you want just the index, put [1] after.
Min via (Change to > and MAX_VALUE):
[1,2,5,0,4].reduce((a,b,i) => a[0] > b ? [b,i] : a, [Number.MAX_VALUE,-1])
//[0, 3]
To complete the work of #VFDan, I benchmarked the 3 methods: the accepted one (custom loop), reduce, and find(max(arr)) on an array of 10000 floats.
Results on chromimum 85 linux (higher is better):
custom loop: 100%
reduce: 94.36%
indexOf(max): 70%
Results on firefox 80 linux (higher is better):
custom loop: 100%
reduce: 96.39%
indexOf(max): 31.16%
Conclusion:
If you need your code to run fast, don't use indexOf(max).
reduce is ok but use the custom loop if you need the best performances.
You can run this benchmark on other browser using this link:
https://jsben.ch/wkd4c
If you are utilizing underscore, you can use this nice short one-liner:
_.indexOf(arr, _.max(arr))
It will first find the value of the largest item in the array, in this case 22. Then it will return the index of where 22 is within the array, in this case 2.
Unless I'm mistaken, I'd say it's to write your own function.
function findIndexOfGreatest(array) {
var greatest;
var indexOfGreatest;
for (var i = 0; i < array.length; i++) {
if (!greatest || array[i] > greatest) {
greatest = array[i];
indexOfGreatest = i;
}
}
return indexOfGreatest;
}
var arr=[0,6,7,7,7];
var largest=[0];
//find the largest num;
for(var i=0;i<arr.length;i++){
var comp=(arr[i]-largest[0])>0;
if(comp){
largest =[];
largest.push(arr[i]);
}
}
alert(largest )//7
//find the index of 'arr'
var arrIndex=[];
for(var i=0;i<arr.length;i++){
var comp=arr[i]-largest[0]==0;
if(comp){
arrIndex.push(i);
}
}
alert(arrIndex);//[2,3,4]
EDIT: Years ago I gave an answer to this that was gross, too specific, and too complicated. So I'm editing it. I favor the functional answers above for their neat factor but not their readability; but if I were more familiar with javascript then I might like them for that, too.
Pseudo code:
Track index that contains largest value. Assume index 0 is largest initially. Compare against current index. Update index with largest value if necessary.
Code:
var mountains = [3, 1, 5, 9, 4];
function largestIndex(array){
var counter = 1;
var max = 0;
for(counter; counter < array.length; counter++){
if(array[max] < array[counter]){
max = counter;
}
}
return max;
}
console.log("index with largest value is: " +largestIndex(mountains));
// index with largest value is: 3
function findIndicesOf(haystack, needle)
{
var indices = [];
var j = 0;
for (var i = 0; i < haystack.length; ++i) {
if (haystack[i] == needle)
indices[j++] = i;
}
return indices;
}
pass array to haystack and Math.max(...array) to needle. This will give all max elements of the array, and it is more extensible (for example, you also need to find min values)
If you create a copy of the array and sort it descending, the first element of the copy will be the largest. Than you can find its index in the original array.
var sorted = [...arr].sort((a,b) => b - a)
arr.indexOf(sorted[0])
Time complexity is O(n) for the copy, O(n*log(n)) for sorting and O(n) for the indexOf.
If you need to do it faster, Ry's answer is O(n).
A minor modification revised from the "reduce" version of #traxium 's solution taking the empty array into consideration:
function indexOfMaxElement(array) {
return array.reduce((iMax, x, i, arr) =>
arr[iMax] === undefined ? i :
x > arr[iMax] ? i : iMax
, -1 // return -1 if empty
);
}
A stable version of this function looks like this:
// not defined for empty array
function max_index(elements) {
var i = 1;
var mi = 0;
while (i < elements.length) {
if (!(elements[i] < elements[mi]))
mi = i;
i += 1;
}
return mi;
}
To find the index of the greatest value in an array, copy the original array into the new array and then sort the original array in decreasing order to get the output [22, 21, 7, 0]; now find the value 22 index in the copyNumbers array using this code copyNumbers.indexOf(numbers[0]);
<script>
const numbers = [0, 21, 22, 7];
const copyNumbers = [];
copyNumbers.push(...numbers);
numbers.sort(function(a, b){
return b - a
});
const index = copyNumbers.indexOf(numbers[0]);
console.log(index);
</script>
Make this
const max = arr.reduce((m, n) => Math.max(m, n)), then the indexes of the max
get index with findIndex
var index = arr.findIndex(i => i === max)