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With given array on unique numbers which are always greater than 0 I need to find all possible unique combinations of those numbers that are equal to a certain number when summed.
For example, getNumberComponents([7, 4, 3, 2, 5, 6, 8, 1], 8) should return
[ [ 7, 1 ], [ 4, 3, 1 ], [ 3, 5 ], [ 2, 5, 1 ], [ 2, 6 ], [ 8 ] ] because sum of all numbers in every subarray equals 8.
My solution:
function getNumberComponents(numArray, number) {
const arrayLength = numArray.length;
const allVariants = [];
function findComponents(currentIndex = 0, currentVariant = []) {
while (currentIndex < arrayLength) {
const currentElement = numArray[currentIndex];
const currentSum = currentVariant.reduce((acc, cur) => acc + cur, 0);
const sumWithCurrent = currentSum + currentElement;
if (sumWithCurrent === number) {
allVariants.push([...currentVariant, currentElement]);
}
currentIndex++;
if (sumWithCurrent < number) {
findComponents(currentIndex, [...currentVariant, currentElement]);
}
}
}
findComponents();
return allVariants;
}
But I wonder if it's possible to use tail recursion for that? I have no idea how to turn my solution into tail recursion.
To make this tail recursive, you could:
Keep track of all indices that were selected to arrive at the current sum. That way you can easily replace a selected index with the successor index.
In each execution of the function get the "next" combination of indices. This could be done as follows:
If the sum has not been achieved yet, add the index the follows immediately after the most recently selected index, and adjust the sum
If the sum has achieved or exceeded, remove the most recently selected index, and then add the successor index instead, and adjust the sum
If there is no successor index, then forget about this index and replace the previous one in the list, again adjusting the sum
If there are no more entries in the list of indices, then all is done.
Instead of accumulating a sum, you could also decrease the number that you pass to recursion -- saving one variable.
Make the function return the array with all variants, so there is no need for an inner function, nor any action that follows the function call.
Here is an impementation:
function getNumberComponents(numArray, number, selectedIndices=[], allVariants=[]) {
let i = selectedIndices.at(-1)??-1;
if (number < 0) { // Sum is too large. There's no use to adding more
i = numArray.length; // Force the while-condition to be true
} else if (number == 0) { // Bingo
allVariants.push(selectedIndices.map(idx => numArray[idx]));
}
while (++i >= numArray.length) { // No more successor index available
if (selectedIndices.length == 0) return allVariants; // All done
i = selectedIndices.pop(); // Undo a previous selection
number += numArray[i]; // Remove from sum
}
selectedIndices.push(i); // Select index and recur:
return getNumberComponents(numArray, number - numArray[i], selectedIndices, allVariants);
}
console.log(getNumberComponents([7, 4, 3, 2, 5, 6, 8, 1], 8));
Here is my version of your function, but using tail recursion. This is still a complex subject for me, check if there are no mistakes
function getNumberComponents(numArray, number, currentIndex = 0, currentVariant = [], allVariants = new Set()) {
if (currentIndex >= numArray.length) {
const currentSum = currentVariant.reduce((acc, cur) => acc + cur, 0);
if (currentSum === number) {
allVariants.add(currentVariant);
}
return Array.from(allVariants);
}
const currentElement = numArray[currentIndex];
const currentSum = currentVariant.reduce((acc, cur) => acc + cur, 0);
const sumWithCurrent = currentSum + currentElement;
if (sumWithCurrent <= number) {
allVariants = new Set([...allVariants, ...getNumberComponents(numArray, number, currentIndex + 1, [...currentVariant, currentElement], allVariants), ...getNumberComponents(numArray, number, currentIndex + 1, currentVariant, new Set())]);
} else {
allVariants = new Set([...allVariants, ...getNumberComponents(numArray, number, currentIndex + 1, currentVariant, new Set())]);
}
return Array.from(allVariants);
}
console.log(getNumberComponents([7, 4, 3, 2, 5, 6, 8, 1], 8));
I am trying to get 5 closest elements to given element, including that element, in array. For example, if we have:
const arr = [1, 2, 3, 4, 7, 11, 12, 13, 15, 17]
and I want to get 5 closest elements to 11, it should return [4, 7, 11, 12, 13]. If i pass 1 it should return [1, 2, 3, 4, 7]. If I pass 15 it should return [11, 12, 13, 15, 17].
I'm not sure what you meant;
You might've meant a code to find the element and return the five nearest elements to it by place in the array;
Or you might've meant a code to find the 5 numbers closest to a number you say.
IF you meant the first case
There are two ways to do so,
A value as a parameter
Use this code:
function closestNByValue(arr, value, n) {
let ind = arr.indexOf(value);
let finalArr = [];
if (n > arr.length) {
finalArr = Array.from(arr);
} else if (ind == -1) {
finalArr = [];
} else if (ind <= n/2 - 0.5) {
finalArr = arr.slice(0, n);
} else if (ind >= (arr.length - n/2) - 0.5) {
finalArr = arr.slice(-n);
} else if (n%2 == 0) {
finalArr = arr.slice(ind-(n/2), ind+(n/2));
} else {
finalArr = arr.slice(ind-(n/2 - 0.5), ind+(n/2 + 0.5));
}
return finalArr;
}
console.log(closestNByValue([1, 2, 3, 4, 7, 11, 12, 13, 15, 17], 11, 5))
How does it do the job?
Okay first we need to find the index of the value and save it in ind (short form of 'index') and we check multiple different situations for what the ind is so we'd be able to output the best answer as finalArr.
There are two exceptions; what if there was no such value in our array? then ind = -1 and we'd return an empty array; or what if the number of elements nearby that we seek is larger than or equal to the arr.length? then we'd return all of the arr.
But if there were no exceptions, there are three different situations for the ind; first, ind is a number that makes us have all of the finalArr values from the first of arr, second, ind be a number that makes us have all of the finalArr values from the last of arr, and third, ind being a number that we have to select half from the indexes smaller than ind and half, larger.
If it is the third way, the way we select we'd be different depending on the oddity of the numbers we want to select.
And we'll have a conditional statement for each situation and return the finalArr.
An index as a parameter
function closestNByIndex(arr, ind, n) {
let finalArr = [];
if (n > arr.length) {
finalArr = Array.from(arr);
} else if (ind == -1) {
finalArr = [];
} else if (ind <= n/2 - 0.5) {
finalArr = arr.slice(0, n);
} else if (ind >= (arr.length - n/2) - 0.5) {
finalArr = arr.slice(-n);
} else if (n%2 == 0) {
finalArr = arr.slice(ind-(n/2), ind+(n/2));
} else {
finalArr = arr.slice(ind-(n/2 - 0.5), ind+(n/2 + 0.5));
}
return finalArr;
}
console.log(closestNByIndex([1, 2, 3, 4, 7, 11, 12, 13, 15, 17], 5, 5))
Similar to the first code it works, though we have the index and we don't search for it.
The point is, if you use the function with value, it'd do the nearest 5 elements of the first value that equals the entry but such confusion is not being tolerated in the second code.
IF you meant the second case
This is a code I coded:
const arr = [1, 2, 3, 4, 7, 11, 12, 13, 15, 17];
function allDiff(arr, num1, num2) {
const finalArr = [];
const x = Math.abs(num2 - num1);
for (let y = 0; y < arr.length; y++) {
if (Math.abs(arr[y] - num1) == x) {
finalArr.push(arr[y]);
}
}
return finalArr;
}
function deleteArr(arr, delet) {
for (let x = 0; x < arr.length; x++) {
if (delet.includes(arr[x])) {
delete arr[x];
}
}
return arr;
}
function closest(arr, num) {
const map = new Map()
arr2 = Array.from(arr);
let key, value;
for (let x = 0; x < arr2.length; x++) {
key = Math.abs(arr2[x] - num);
value = allDiff(arr2, num, arr2[x]);
arr2 = deleteArr(arr2, value);
map.set(key, value);
}
return map;
}
function closestN(arr, num, n) {
const map = closest(arr, num);
const mapKeys = Array.from(map.keys());
const mapKeysSorted = mapKeys.sort(function(a, b) {
return a - b
});
let finalArr = [];
let y;
for (let i = 0; i < mapKeysSorted.length; i++) {
if (n <= 0) {
break;
}
y = map.get(mapKeysSorted[i]);
if (n < y.length) {
finalArr = finalArr.concat(y.slice(0, n + 1));
break;
}
finalArr = finalArr.concat(y);
n -= y.length;
}
return finalArr;
}
console.log(closestN(arr, 11, 5));
It might be a little too long, but I have programmed it as you can give it any array (arr) with integer values, an integer (num) that you'd like it to be the base and another integer (n) for the number of the size of the output array, 5 in this case.
Explaining the code
The function closest would return a map of (the difference between the numbers, a list of the numbers in the arr that differs the number equal to their key).
The main function, closestN, calls the closest function and saves the map in the map variable.
Then it sorts the keys of the map in mapKeysSorted.
Now, a for loop loops through the mapKeySorted array and pushes new elements to the finalArr until the size of the finalArr reaches the number of elements we seek.
The main function is the closestN.
Here's a way to get to your goal:
To start, first thing to do is finding the index of the wanted number in the array. Example index of 1 in your array arr is 0. The index will help in extracting the numbers later on. The method findIndex will help us in finding the index.
Then, we need to find the position at which will start extaracting the closest numbers (in terms of position not value). As seen from the desired output you have provided, usually you want the returned array to be in the following structure:
output: [
2 nearest numbers (from N left),
the wanted number,
2 nearest numbers (from N right)
]
This can get tricky so we should make sure to deal with some edge case like when the wanted element is sitting at position 0.
Extract the numbers and return them as an array as described by your desired output. The use of slice method will come in handy here which allow us to extract the numbers just as we need.
Here's a live demo demonstrating solution:
const arr = [1, 2, 3, 4, 7, 11, 12, 13, 15, 17],
/** a function that returns an array containing the "5" (depending on "arr" length that could be less) nearest numbers (in terms of position) in "arr" array to the supplied number "n" */
findClosestNumbers = n => {
/** make sure we don't exceed the array length */
const toTake = 5 > arr.length ? arr.length : 5,
/** find the index of the wanted nulber "n", if "-1" is returned then "n" cannot be found ion the array "arr" */
idx = arr.findIndex(el => n == el),
/**
* from where we should start returning the nearest numbers (the position of the first number to extract from "arr"
* the below condition help deal with some edge cases like when "n" is the last element in "arr"
*/
startIdx = idx + toTake / 2 > arr.length ?
arr.length - 5 :
(idx - 2 >= 0 ?
idx - 2 :
0);
/** return the nearest numbers or return an empty array "[]" if the number "n" is not found on the array "arr" */
return idx == -1 ? [] : arr.slice(startIdx, startIdx + 5);
};
/** run for various scenarios */
console.log('For 1 =>', findClosestNumbers(1));
console.log('For 11 =>', findClosestNumbers(11));
console.log('For 15 =>', findClosestNumbers(15));
console.log('For 17 =>', findClosestNumbers(17));
.as-console-wrapper {
max-height: 100%!important;
}
The demo above is meant to help you understand how things could work and it is not the only way to get to your goal. Also, because I kept it as simple as possible, the above demo is wide open for improvements.
I'm trying to create a function that groups an array of numbers based on a length parameter. The length represents the max length of each sub-array. The code works as it is meant to for getting the sub arrays, but what I'd like to do is make it sort by odd and even.
function myFunctionA(myArr1, myVal) {
newArr = [];
for ( x = 0; x < myArr1.length; x += myVal) {
newArr.push(myArr1.slice(x, x + myVal));
}
return newArr;
}
Console.log(myfunction([1,2,3,4,5,6,7,8,9,10],3))
This returns [[1,2,3],[4,5,6],[7,8,9],[10]]
What I'd like to do is go through each sub array at a time until the sub arrays are the correct length and add any leftover values to a sub array/s
This would look like
[[1,3,5][2,4,6][7,9][8,10]]
Since arr 0 and arr 1 are the correct length that we have stated in the console.log statement, 7 8 9 and 10 are left over. But since the can't create a full sub array and they are odds and even, they form two sub arrays with a side of 2.
Other examples:
myfunction([1,2,3,4,5,6,7],2)
Should return [[1,3],[2,4],[5,7],[6]]
myfunction([1,2,3,4,5,6,7,8],1)
Should return [[1][2][3][4][5][6][7][8]]
You could take an array for collecting all odd and even values and then push the group if it has zero items. By having the wanted size, create a new array.
function chunkenator(array, size, fn) {
let groups = [],
result = [];
for (let value of array) {
const group = fn(value);
if (!groups[group]) groups[group] = [];
if (!groups[group].length) result.push(groups[group]);
groups[group].push(value);
if (groups[group].length === size) groups[group] = [];
}
return result;
}
console.log(chunkenator([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 3, x => x % 2));
console.log(chunkenator([1, 3, 5, 7, 8, 9, 11, 13, 15], 3, x => x % 2));
One possibility would be to first seperate the numbers into even and odd numbers and then just loop over it, pushing the numbers into a new array switching between even and odd numbers.
It's not the cleanest piece of code, but it works.
function myfunction(arr, n) {
const evenOdd = arr.reduce((acc, e) => {
const ind = +(e % 2 === 0);
acc[ind] = acc[ind] || [];
acc[ind].push(e);
return acc;
}, []);
let ind = 0, res = [[]];
while (evenOdd[0].length || evenOdd[1].length) {
for (let i = n; i--;) {
const val = evenOdd[ind].shift();
if (val) res[res.length - 1].push(val)
}
ind = (ind + 1) % 2
res.push([])
}
res.pop()
return res;
}
for (const n of [1, 2, 3]) {
console.log(n,
myfunction([1, 2, 3, 4, 5, 6, 7, 8], n)
)
}
Hello I am taking an array of integers with ranging numbers from 1 - 100 and I'm counting the duplicated numbers within it. Example, array[1,1,1,1,1,100,3,5,2,5,2,23,23,23,23,23,]. Result = 1 - 5 times, 5 - 2 times, 2 - 2 times, 23 - 5 times. I cannot see how to make this work I have tried to edit this code snippet so that it counts and returns the number of duplicates of a specific integer that is a duplicate but I could not see how to do it. Please assist Thank You.
https://repl.it/#youngmaid/JS-ALGORITHMS-Counting-Duplicates
//To count or reveal duplicates within an array. Using the array method of sort() is one way.
//Sort the following array using .sort(), which put the items in the array in numerical or alphabetical order.
//Create a new variable for the sorted array.
//Also create a new variable for an empty array.
//Create a loop using the length of the first, original array with an increment of "++".
//Create an if statement that includes adding an item comparing to the index.
//Then push the emply array in the sorted array.
//console log the new array.
let duplicateArr = [5, 3, 7, 4, 7, 5, 3, 2, 7, 3, 2];
let sortArr = duplicateArr.sort();
let newArr = [];
for(let i = 0; i < duplicateArr.length; i++) {
if(sortArr[i + 1] == sortArr[i]){
newArr.push(sortArr[i]);
}
}
console.log(newArr);
//The other way or more detailed/reusable approach is to create a function and variable hash table.
//The hash table to place all the items in the array.
//Then create another variable placing duplicates in the array.
//Then go through each item in the array through a for loop. (Using arr as the argument).
//Create a conditional if/else statement. If the item in the hash table does not exist, then insert it as a duplicate.
function duplicates(arr) {
let hashTable = [];
let dups = [];
for (var i = 0; i < arr.length; i++){
if (hashTable[arr[i].toString()] === undefined) {
hashTable[arr[i].toString()] = true;
} else {
dups.push(arr[i]);
}
}
return dups;
}
duplicates([3, 24, -3, 103, 28, 3, 1, 28, 24]);
If I understand correctly, you could achieve this via Array#reduce() as shown below:
let duplicateArr = [5, 3, 7, 4, 7, 5, 3, 2, 7, 3, 2];
/* Reduce the input duplicateArr to a map relating values to counts */
const valueCounts = duplicateArr.reduce((counts, value) => {
/* Determine the count of current value from the counts dictionary */
const valueCount = (counts[ value ] === undefined ? 0 : counts[ value ])
/* Increment count for this value in the counts dictionary */
return { ...counts, ...{ [value] : valueCount + 1 } }
}, {})
/* Remove values with count of 1 (or less) */
for(const value in valueCounts) {
if(valueCounts[value] < 2) {
delete valueCounts[value]
}
}
/* Display the values and counts */
for(const value in valueCounts) {
console.log(`${ value } occours ${ valueCounts[value] } time(s)` )
}
Reasonably basic loop approach
const data = [1, 1, 1, 1, 1, 100, 3, 5, 2, 5, 2, 23, 23, 23, 23, 23, ]
function dupCounts(arr) {
var counts = {};
arr.forEach(function(n) {
// if property counts[n] doesn't exist, create it
counts[n] = counts[n] || 0;
// now increment it
counts[n]++;
});
// iterate counts object and remove any that aren't dups
for (var key in counts) {
if (counts[key] < 2) {
delete counts[key];
}
}
return counts
}
console.log(dupCounts(data))
Here using only 1 loop.
let duplicateArr = [5, 3, 7, 4, 7, 5, 3, 2, 7, 3, 2]
let sortArr = duplicateArr.sort()
let current = 0, counter = 0
sortArr.forEach(n => {
if (current === n) {
counter++
}
else {
if (counter > 1){
console.log(current + " occurs " + counter + " times.")
}
counter = 1
current = n
}
})
if (counter > 1){
console.log(current + " occurs " + counter + " times.")
}
The cleanest way is using ES6 Map
function duplicates(arr) {
// This will be the resulting map
const resultMap = new Map();
// This will store the unique array values (to detect duplicates using indexOf)
const occurrences = [];
for (let i of arr){
if (occurrences.indexOf(i) !== -1) {
// Element has a duplicate in the array, add it to resultMap
if (resultMap.has(i)) {
// Element is already in the resultMap, increase the occurrence by 1
resultMap.set(i, resultMap.get(i) + 1);
} else {
// Element is not in resultMap, set its key to 2 (the first 2 occurrences)
resultMap.set(i, 2);
}
} else {
// Element is showing for the first time (not a duplicate yet)
occurrences.push(i);
}
}
return resultMap;
}
// To iterate on the map keys and values use this
for (const [key, value] of map) {
console.log(key + ' - ' + value + ' times');
}
You can just iterate over all of the unique values and then count how many of them exists.
here is a sample code:
let duplicateArr = [5, 3, 7, 4, 7, 5, 3, 2, 7, 3, 2];
let sortArr = duplicateArr.sort();
let newArr = {};
let duplicateValues = [];
for (let i = 0; i < duplicateArr.length; i++) {
let count = 0;
let k = 0;
while (i + k < duplicateArr.length && sortArr[i] == sortArr[i + k]) {
count++;
k++;
}
if (count > 1) {
newArr[sortArr[i]] = count;
duplicateValues.push(sortArr[i]);
}
i = i + k;
}
console.log("duplicate items with count:", newArr);
console.log("duplicate items:", duplicateValues);
Using Array.prototype.reduce() you can create a hash object variable containing as keys the numbers in the duplicateArr array variable and the values are the number of repeated times..
Code:
const duplicateArr1 = [5, 3, 7, 4, 7, 5, 3, 2, 7, 3, 2];
const duplicateArr2 = [1, 1, 1, 1, 1, 100, 3, 5, 2, 5, 2, 23, 23, 23, 23, 23];
const getStringOfDuplicated = array => {
const hash = array.reduce((a, c) => (a[c] = ++a[c] || 1, a), {});
return Object.entries(hash)
.filter(([k, v]) => v > 1)
.sort(([ak, av], [bk, bv]) => bv - av)
.map(([k, v]) => `${k} - ${v} times`)
.join(', ');
};
console.log(getStringOfDuplicated(duplicateArr1));
console.log(getStringOfDuplicated(duplicateArr2));
I have a number array [2, 1, 3, 4, 5, 1] and want to remove the smallest number in the list. But somehow my IF statement gets skipped.
I checked and by itself "numbers[i + 1]" and "numbers[i]" do work, but "numbers[i + 1] < numbers[i]" doesn't...
function removeSmallest(numbers) {
var smallestNumberKEY = 0;
for (i = 0; i <= numbers.lenths; i++) {
if (numbers[i + 1] < numbers[i]) {
smallestNumberKEY = i + 1;
}
}
numbers.splice(smallestNumberKEY, 1);
return numbers;
}
document.write(removeSmallest([2, 1, 3, 4, 5, 1]));
You have a typo in your code, array doesn't have lenths property
function removeSmallest(numbers) {
var smallestNumberKEY = 0;
for (var i = 0; i < numbers.length - 1; i++) {
if (numbers[i + 1] < numbers[i]) {
smallestNumberKEY = i + 1;
numbers.splice(smallestNumberKEY, 1);
}
}
return numbers;
}
document.write(removeSmallest([2, 1, 3, 4, 5, 1]));
But your algorithm wont work for another array, e.g [5, 3, 1, 4, 1], it will remove a value 3 too.
You can find the min value with Math.min function and then filter an array
function removeSmallest(arr) {
var min = Math.min(...arr);
return arr.filter(e => e != min);
}
You can use Array#filter instead
function removeSmallest(arr) {
var min = Math.min.apply(null, arr);
return arr.filter((e) => {return e != min});
}
console.log(removeSmallest([2, 1, 3, 4, 5, 1]))
Short one liner. If the smallest value exist multiple times it will only remove ONE. This may or may not be what you want.
const result = [6,1,3,1].sort().filter((_,i) => i) // result = [1,3,6]
It works by sorting and then creating a new array from the items where indeces are truthy(anything but 0)
another solution with splice and indexOf:
array = [2, 1, 3, 4, 5, 1];
function replace(arr){
arr = arr.slice(); //copy the array
arr.splice( arr.indexOf(Math.min.apply(null, arr)),1)
return arr;
}
document.write( replace(array) ,'<br> original array : ', array)
edit : making a copy of the array will avoid the original array from being modified
"Short" solution using Array.forEach and Array.splice methods:
function removeSmallest(numbers) {
var min = Math.min.apply(null, numbers);
numbers.forEach((v, k, arr) => v !== min || arr.splice(k,1));
return numbers;
}
console.log(removeSmallest([2, 1, 3, 4, 5, 1])); // [2, 3, 4, 5]
This is a proposal with a single loop of Array#reduce and without Math.min.
The algorithm sets in the first loop min with the value of the element and returns an empty array, because the actual element is the smallest value and the result set should not contain the smallest value.
The next loop can have
a value smaller than min, then assign a to min and return a copy of the original array until the previous element, because a new minimum is found and all other previous elements are greater than the actual value and belongs to the result array.
a value greater then min, then the actual value is pushed to the result set.
a value equal to min, then the vaue is skipped.
'use strict';
var removeSmallest = function () {
var min;
return function (r, a, i, aa) {
if (!i || a < min) {
min = a;
return aa.slice(0, i);
}
if (a > min) {
r.push(a);
}
return r;
}
}();
document.write('<pre>' + JSON.stringify([2, 1, 3, 2, 4, 5, 1].reduce(removeSmallest, []), 0, 4) + '</pre>');
I like this oneliner: list.filter(function(n) { return n != Math.min.apply( Math, list ) })
check it out here: https://jsfiddle.net/rz2n4rsd/1/
function remove_smallest(list) {
return list.filter(function(n) { return n != Math.min.apply( Math, list ) })
}
var list = [2, 1, 0, 4, 5, 1]
console.log(list) // [2, 1, 0, 4, 5, 1]
list = remove_smallest(list)
console.log(list) // [2, 1, 4, 5, 1]
list = remove_smallest(list)
console.log(list) // [2, 4, 5]
I had to do this but I needed a solution that did not mutate the input array numbers and ran in O(n) time. If that's what you're looking for, try this one:
const removeSmallest = (numbers) => {
const minValIndex = numbers.reduce((finalIndex, currentVal, currentIndex, array) => {
return array[currentIndex] <= array[finalIndex] ? currentIndex : finalIndex
}, 0)
return numbers.slice(0, minValIndex).concat(numbers.slice(minValIndex + 1))
}
function sumOfPaiars(ints){
var array = [];
var min = Math.min(...ints)
console.log(min)
for(var i=0;i<ints.length;i++){
if(ints[i]>min){
array.push(ints[i])
}
}
return array
}
If you only wish to remove a single instance of the smallest value (which was my use-case, not clear from the op).
arr.sort().shift()
Here is a piece of code that is work properly but is not accepted from codewars:
let numbers = [5, 3, 2, 1, 4];
numbers.sort(function numbers(a, b) {
return a - b;
});
const firstElement = numbers.shift();