I solved this problem by iterating through the array then find the item when the sum equals to array[i] + item returning true otherwise returning false.
My Question is => How I can return the indices of those numbers that add up to sum not just true? Using the same code below:
function hasPairsWithSum(array,sum) {
for (let i = 0; i < array.length; i++) {
if (array.find((item) => {return sum === array[i] + item}
));
return true;
};
return false;
};
console.log(hasPairsWithSum([1,2,4,4],8))
Note: Time complexity must be less than O(n ^ 2).
JavaScript O(n) Solution.
function hasPairsWithSum(array, sum) {
const map = new Map ();
for(let i = 0; i < array.length; i++) {
let currVal = array[i];
if (map.has(currVal)) {
return [map.get(currVal),i]
}
// difference value = sum - current value
let diff = sum - currVal
map.set(diff,i)
}
};
console.log(hasPairsWithSum([2,2,4,4], 8))
Please refer this code.
function hasPairsWithSum(array,sum) {
let result = [];
for (let i = 0; i < array.length; i++) {
if (array.some((item, index) => {return i === index ? false : sum === array[i] + item}))
result.push(i);
};
return result;
};
console.log(hasPairsWithSum([1,2,4,4],8))
console.log(hasPairsWithSum([3,2,4],6))
console.log(hasPairsWithSum([0,4,3,0],0))
O(n) Soln ... using math concept a+b = n then if a is present in our array then need to find b = n - a is present or not ..
def hasPairsWithSum(array,sum):
d = {}
for i in range(len(array)):
if(array[i] in d):
d[array[i]].append(i)
else:
d[array[i]] = [i]
ans = []
for i in range(len(array)):
val = sum - array[i]
if(val in d):
if(d[val][0] == i):
if(len(d[val]) > 1):
ans.append((i,d[val][1]))
break
else:
continue
else:
ans.append((i,d[val][0]))
break
return ans
print(hasPairsWithSum([4, 4, 4, 4], 8))
O(nlogn) soln ....just store the index with elements .. then sort it by their values .. next step run a loop with complexity of O(n) [concept : Two pointers]
def hasPairsWithSum(array,sum):
arr = []
for i in range(len(array)):
arr.append((array[i],i))
arr.sort()
i = 0
j = len(array)-1
ans = []
while(i<j):
tmp_sum = arr[i][0] + arr[j][0]
if(tmp_sum == sum):
ans.append((arr[i][1] , arr[j][1]))
#add your logic if you want to find all possible indexes instead of break
break
elif(tmp_sum < sum):
i = i + 1
elif(tmp_sum > sum):
j = j - 1
return ans
print(hasPairsWithSum([1,2,4,4],8))
note : if you want to find all possible soln then these approaches will not work either add you own logic in while loop or another approach is use binary search with traversal on every element and store the indexes in set (worst case this will be O(n^2) as we have to find all possible values) Eg: [4,4,4,4,4,4] , sum = 8 and you want to print all possible indexes then we end up running it upto n^2 (why? reason: total possible solns. are 5+4+3+2+1 = n*(n-1)/2 ≈ n^2)
You have to iterate over the array elements checking at every iteration for every element of the array (except the last one) all the elements at the right of it like below:
function findIndexes(array, sum) {
const result = [];
for (let i = 0; i < array.length -1; ++i) {
for (let j = i + 1; j < array.length; ++j) {
if ((array[i] + array[j]) === sum) {
result.push([i, j]);
}
}
}
return result;
}
console.log(findIndexes([1, 2, 4, 4], 8));
console.log(findIndexes([3, 2, 4], 6));
Update:
It is possible to obtain a linear O(n) complexity using an auxiliary Map structure associating an integer value as key with as a value the list containing all the indexes of the elements in the array equal to the integer key like below:
function findIndexes(array, sum) {
const map = new Map();
const result = [];
for (let i = 0; i < array.length; ++i) {
const a = array[i];
const b = sum - a;
if (map.has(b)) {
for (const index of map.get(b)) {
result.push([index, i]);
}
}
const l = map.has(a) ? map.get(a) : [];
l.push(i);
map.set(a, l);
}
return result;
}
console.log(findIndexes([1, 2, 4, 4], 8));
console.log(findIndexes([3, 2, 4], 6));
console.log(findIndexes([1, 1, 1], 2));
Related
I am trying to solve a problem , Two Sum for those who know it , I started learning JavaScript coming from Lua , and I am stuck , I don't know why the function returns "undefined" I fill like the variable is defined
var num = [2, 7, 11, 15]
function numbers(target) {
var idx = {}
num.forEach(function(n, i) {
var j = idx[target - n]
if (j) {
var res = '[ ${j} ${i} ]'
return res
}
console.log(n)
idx[n] = i
})
}
console.log(numbers(9))
output:
2
7
11
15
undefined
Problem: https://leetcode.com/problems/two-sum/
I found the solution I think , I don't think it's the best one , but it works
Code:
var nums = [2,7,11,15]
var twoSum = function(target) {
for (let i = 0; i < nums.length; i++) {
for (let j = i + 1; j < nums.length; j++) {
if (nums[i] + nums[j] == target) {
return [i, j]
}
}
}
}
console.log(twoSum(9))
You have a few issues:
Your current code is returning a string '[ ${j} ${i} ]', you need to return an array [j, i].
Using return from within .forEach() does not return from your numbers function, it instead returns from the inner callback function function(n, i) { which doesn't do much except just skips to the next item in the loop. You need to change your forEach() to a regular for loop so that when you return you return from your numbers function, and not an inner callback.
Your condition if (j) { is flawed. Consider if j is 0, which occurs when the number you're after to add to the current number n is at index 0. In this case, your if-block won't run because 0 is considered falsy. Your condition should instead be more specific by checking that it returned a value if(j !== undefined). See this answer for more details about the condition.
You should be passing nums into your numbers() function rather than relying on a global variable to exist for your function to work correctly.
Example of modified code:
function numbers(nums, target) {
const idx = {}
for(let i = 0; i < nums.length; i++) {
let n = nums[i];
let j = idx[target - n]
if (j !== undefined) {
return [j, i];
}
idx[n] = i;
}
}
console.log(numbers([2, 7, 11, 15], 9));
console.log(numbers([3,2,4], 6));
console.log(numbers([3,3], 6));
See the answers for this question for more details for solving this problem.
var twoSum = function(nums, target) {
let newArr = nums
let sum = 0
outputArr = []
for(let i = 0 ; i < nums.length ; i++){
let val1 = nums[i]
for(let j = 0 ; j < newArr.length ; j++){
if(i !==j) {
let val2 = newArr[j]
if(val1 + val2 === target) {
sum = val1 + val2
outputArr.push(i)
outputArr.push(j)
break;
}
}
}
if(sum === target)
break;
}
return outputArr
};
try this If helps and you need explanation I will help you
I want to count the unique values in a given array without altering the original array but the solution has to be within the time complexity of O(n). so far all of the solutions I've seen, have a time complexity of O(n^2) like here. I can't find the error in my solution's logic. I'm new to Data Structure & Algorithms and would like a simple solution.
MY CODE -
const countUniqueValues = (arr) =>{
if(arr.length === 0){
return console.log(arr.length);
}else if(arr.length === 1){
return console.log(arr.length);
}
const unique = [];
let i = 0;
for( let j = 1; j < arr.length; j++){
if(arr[i] !== arr[j]){
i ++;
unique.push(arr[i]);
}
}
return console.log(unique);
}
//test cases
countUniqueValues([1,1,1,1,1,2]) // 2
countUniqueValues([1,2,3,4,4,4,7,7,12,12,13]) // 7
countUniqueValues([]) // 0
countUniqueValues([-2,-1,-1,0,1]) // 4
Wrong Output -
[ 1 ]
[
2, 3, 4, 4,
4, 7, 7, 12
]
0
[ -1, -1, 0 ]
Turn the array into a Set (O(n)) and count the set's size:
const countUniqueValues = arr => new Set(arr).size;
NB - very important - the arrays must be sorted for this to work:
This should do the trick:
var prevValue = "";
const countUniqueValues = (arr) =>{
if(arr.length === 0){
return console.log(arr.length);
}else if(arr.length === 1){
return console.log(arr.length);
}
prevValue = arr[0];
let i = 1;
for( let j = 1; j < arr.length; ++j){
if(arr[j] != prevValue){
++i;
prevValue = arr[j];
}
}
console.log(i);
return i;
}
const makeUniqueAndCount = arr => {
const uniqueKeysObject = {};
arr.forEach(number => {
uniqueKeysObject[number] = true;
});
return Object.keys(uniqueKeysObject).length;
};
This solution uses objects in javascript. The keys for a javascript object are always unique. You can then use the keys method of the javascript object prototype to turn it into an array to get its length. This solution will work for an unsorted array as well.
I have an array of integers and a number k. Is it necessary to determine whether there are two numbers in the array whose sum is k?
function findPairs(nums, k) {
var s = [];
var length = nums.length;
for (var i = 0; i < length; i++) {
if (s[nums[i]] === k - nums[i]) {
console.log(nums[i], k - nums[i])
return true;
} else {
s[k - nums[i]] = nums[i];
}
}
return false;
}
var nums = [10, 15, 3, 7]
var k = 17
console.log(findPairs(nums, k))
why my code is not working?
My guess is that you had a syntax error or forgot to actually declare a function, since the code in the current version of your question appears to work as expected.
As an aside, I would suggest using an object or a Set instead of an array to store the other pair, because for large values of k, your s array may consume a lot of memory on some JavaScript engines as a result.
function findPairs(nums, k) {
var s = {};
var length = nums.length;
for (var i = 0; i < length; i++) {
if (s[nums[i]] === k - nums[i]) {
console.log(nums[i], k - nums[i]);
return true;
}
s[k - nums[i]] = nums[i];
console.log(s); // see the lookup table after each iteration
}
return false;
}
var nums = [10, 15, 3, 7];
var k = 17;
console.log(findPairs(nums, k));
Functional solution
const findPairs = (nums, k) =>
nums
.flatMap((v, i, arr) => arr.slice(i + 1).map(w => [v, w]))
.filter(pair => pair[0] + pair[1] === k);
console.log(findPairs([1, 2, 3, 4, 5, 6], 5));
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
Example:
Given nums = [3, 2, 4], target = 6,
Because nums[1] + nums[2] = 2 + 4 = 6
return [1, 2].
Solution
var twoSum = function(nums, target) {
for(let i = 0; i <= nums.length; i++){
for(let j = 0; j <= nums.length; j++){
if(nums[i] + nums[j] == target){
return [i, j]
}
}
}
};
The code above works in other cases but not this one.
Expected result [1,2]
Output [0,0]
For instance, I've tried to use a different array of numbers and a different target and it works even if you change the order of the numbers
Example:
New array: [15, 7, 11, 2], target = 9,
Output: [1, 3].
I don't understand what is wrong with the solution and I hope that someone can explain. Thanks
You can use a very simple technique.
Basically, you can check if the difference of target & the element in the current iteration, exists in the array.
Assuming same index cannot be used twice
nums = [3, 2, 4], target = 6
nums[0] = 3
target = 6
diff = 6 - 3 = 3
nums.indexOf[3] = 0 // FAILURE case because it's the same index
// move to next iteration
nums[1] = 2
target = 6
diff = 6 - 2 = 4
nums.indexOf(4) = 2 // SUCCESS '4' exists in the array nums
// break the loop
Here's the accepted answer by the leetcode.
/**
* #param {number[]} nums
* #param {number} target
* #return {number[]}
*/
var twoSum = function(nums, target) {
for (let index = 0; index < nums.length; index++) {
const diff = target - nums[index];
const diffIndex = nums.indexOf(diff);
// "diffIndex !== index" takes care of same index not being reused
if (diffIndex !== -1 && diffIndex !== index) {
return [index, diffIndex];
}
}
};
Runtime: 72 ms, faster than 93.74% of JavaScript online submissions for Two Sum.
Memory Usage: 38.5 MB, less than 90.55% of JavaScript online submissions for Two Sum.
Can anybody help me in reducing the memory usage also?
I don't understand what is wrong with the solution and I hope that
someone can explain ?
Here you're both inner and outer loop start from 0th so in the case [3,2,4] and target 6 it will return [0,0] as 3 + 3 is equal to target, so to take care of same index element not being used twice created a difference of 1 between outer and inner loop
Make outer loop to start from 0th index and inner loop with value i+1
var twoSum = function(nums, target) {
for(let i = 0; i < nums.length; i++){
for(let j = i+1; j < nums.length; j++){
if(nums[i] + nums[j] == target){
return [i, j]
}
}
}
};
console.log(twoSum([15, 7, 11, 2],9))
console.log(twoSum([3, 2, 4],6))
we can solve this problem in O(n) by using the map/object.
We can maintain a map or object which will save all values with index and then we can iterate the array and find target-nums[i] for each value and will find that value in map/object.
let's see this by example:-
nums=[2,7,11,15]
target = 9;
then map/object will be
mm={
2 : 0,
7: 1,
11: 2,
15: 3
}
then for each value, we will find diff and find that diff in map/object.
for i=0 diff= 9-2=7 and mm.has(7) is true so our answer is 2 and 7.
for their index we can use mm.get(7) and i.
return [mm.get(7), i]
var twoSum = function(nums, target) {
let mm=new Map();
for(let i=0;i<nums.length;i++){
mm.set(nums[i],i);
}
let diff=0;
let j;
for(let i=0;i<nums.length;i++){
diff=target-nums[i];
if(mm.has(diff) && i!=mm.get(diff)){
j=mm.get(diff);
if(j>i){
return [i,j];
}else{
return [j,i];
}
}
}
};
Runtime: 76 ms, faster than 88.18% of JavaScript online submissions for Two Sum.
Memory Usage: 41.4 MB, less than 13.32% of JavaScript online submissions for Two Sum.
Your solution works as expected. For nums = [3, 2 ,4] and target = 6, [0, 0] is a valid solution for the outlined problem as nums[0] + nums[0] = 3 + 3 = 6.
If you need two different indices (In my understanding this is not required by the task) you can add an additional check for inequality (nums[i] + nums[j] == target && i != j).
var twoSum = function(nums, target) {
for(let i=0; i<nums.length; i++){
for(let j=i+1; j<nums.length; j++){
if(nums[j] === target - nums[i]){
return [i, j];
}
}
}
};
Here's a simple method to solve this problem and its efficient using different type of inputs using JavaScript.
Like with input of ([3,3], 6) and its expected output will be [0,1] and input like ([3,2,4], 6) with expected output will be [2,4]
var twoSum = function (nums, target) {
for (let i = 0; i <= nums.length; i++) {
for (let j = 0; j <= nums.length; j++) {
if (i !== j) {
if (nums[i] + nums[j] === target) {
return [i, j];
}
}
}
}
};
console.log(twoSum([3, 2, 4], 6));
var twoSum = function (nums, target) {
var len = nums.length;
for (var i = 0; i < len; i++) {
for (var j = i + 1; j < len; j++) {
if (nums[i] + nums[j] == target) {
return [i,j];
}
}
}
};
var twoSum = function(nums, target) {
for(var i=0;i<nums.length;i++){
for(var j=i+1;j<nums.length;j++){
temp = nums[i]+nums[j];
if(temp == target){
return [i,j]
}
}
}
};
console.log(twoSum([15, 7, 11, 2],9))
console.log(twoSum([3, 2, 4],6))
console.log(twoSum([3,3],6))
This works perfectly and the Runtime: 72 ms lesser than 84ms
Another efficient solution way to solve this problem with an O(n) time complexity is not using nested loops. I commented the steps, so JS developers can easy understand. Here is my solution using golang:
func twoSum(intArray []int, target int) []int {
response := []int{-1, -1} // create an array as default response
if len(intArray) == 0 { // return default response if the input array is empty
return response
}
listMap := map[int]int{} // create a Map, JS => listMap = new Map()
for index, value := range intArray { // for loop to fill the map
listMap[value] = index
}
for i, value := range intArray { // for loop to verify if the subtraction is in the map
result := target - value
if j, ok := listMap[result]; ok && i != j { // this verify if a property exists on a map in golang. In the same line we verify it i == j.
response[0] = i
response[1] = j
return response
}
}
return response
}
var twoSum = function(nums, target) {
var numlen = nums.length;
for(let i=0; i<=numlen; i++){
for(let j=i+1;j<numlen;j++){
var num1 = parseInt(nums[i]);
var num2 = parseInt(nums[j]);
var num3 = num1 + num2;
if(num3 == target){
return[i,j]
}
}
}
};
I suppose this could be a better solution. Instead of nesting loops, this provides a linear solution.
(PS: indexOf is kinda a loop too with O(n) complexity)
var twoSum = function (nums, target) {
const hm = {}
nums.forEach((num, i) => {
hm[target - num] = i
})
for (let i = 0; i < nums.length; i++) {
if(hm[nums[i]] !== undefined && hm[nums[i]] !== i) {
return ([hm[nums[i]], i])
}
}
};
var twoSum = function(nums, target) {
for (var i = 0; i < nums.length; i++)
{
if( nums.indexOf(target - nums[i]) !== -1)
{
return [i , nums.indexOf(target - nums[i])]
}
}
};
For clearity console a & b in inner for loop . This will give you clear insight
var twoSum = function(nums, target) {
var arr=[];
for(var a=0;a<nums.length;a++){
for(var b=1;b<nums.length;b++){
let c=nums[a]+nums[b];
if(c== target && a != b){
arr[0]=a;
arr[1]=b;
}
}
}
return arr;
};
My solution:
Create a Set
Then loop the array for retrieve all the differences between the target number minus all the elements in array. Insert only which are positive (as is a Set, duplicates will not be included).
Then iterates array again, now only compare if array elements is in the set, if yes take the index from i store in the index array and return it.
public static Object solution(int[] a, int target){
Set<Integer> s = new HashSet<>();
ArrayList<Integer> indexes = new ArrayList<Integer>();
for (int e : a){
Integer diff = new Integer(target - e);
if(diff>0){
s.add(diff);
}
}
int i = 0;
for (int e : a){
if(s.contains(e)){
indexes.add(i);
}
i++;
}
return indexes;
}
def twoSum(self, nums: List[int], target: int) -> List[int]:
for i in range(len(nums)):
comp = target - nums[i]
if comp in nums:
j = nums.index(comp)
if(i!=j):
return [i,j]
var twoSum = function(nums, target) {
for(let i = 0; i < nums.length; i++){
for(let j = i+1; j < nums.length; j++){
if(nums[i] + nums[j] == target){
return [i, j]
}
}
}
};
console.log(twoSum([15, 7, 11, 2],9))
console.log(twoSum([3, 2, 4],6))
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
indx =[]
for i in range(len(nums)):
for j in range(i+1,len(nums)):
if nums[j]==(target-nums[i]):
indx.append(i)
indx.append(j)
return indx
twoSummation = (numsArr, estTarget) => {
for(let index = 0; index < numsArr.length; index++){
for(let n = index+1; n < numsArr.length; n++){
if(numsArr[index] + numsArr[n] == estTarget){
return [index, n]
}
}
}
}
console.log(twoSummation([2,7,11,15], 9))
console.log(twoSummation([3,2,4], 6))
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Document</title>
</head>
<body>
<script>
var twoSum = function(nums, target) {
for(var i=0;i<nums.length;i++){
for(var j=i+1;j<nums.length;j++){
temp = nums[i]+nums[j];
if(temp == target){
return [nums[i],nums[j]]
}
}
}
};
console.log(twoSum([15, 7, 11, 2],9))
console.log(twoSum([3, 2, 4],6))
console.log(twoSum([3,3],6))
</script>
</body>
</html>
Solution from my leetcode submission in Java. Top 75% in runtime, top 25% in memory. It is 0(n) however uses an external HashMap.
import java.util.HashMap;
class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer, Integer> numbers= new HashMap<Integer, Integer>();
for(int i=0; i<nums.length; i++){
int cur = nums[i];
numbers.put(cur, i);
}
for(int i=0; i<nums.length; i++){
//check map for diff of target - cur
int diff = target - nums[i];
if(numbers.containsKey(diff) && numbers.get(diff)!=i){
return new int[]{i, numbers.get(diff)};
}
}
return null;
}
}
I'm trying to solve this problem:
Given two parameters: an array a and integer i, find the largest array length where the sum of all elements is <= i.
For example, having the following array: [3, 1, 2, 1], and i = 4, all the combinations that are <= i are: [3], [1], [2], [3, 1], [1, 2], [1, 2, 1]. The largest subarray is [1, 2, 1], so the return of the function should be 3 (the array length).
What would be a an efficient approach to solve this problem?
This is my algorithm until now, but I know that I'm missing something:
function sumArray(a) {
return a.reduce((a, b) => a + b, 0)
}
function maxLength(a, i) {
let max = 0
let array = [a[0]]
a.splice(1, a.length).forEach(number => {
array.push(number)
if (sumArray(array) <= k) max = array.length
else array.splice(array.indexOf(Math.max.apply(null, array)), 1)
})
return max
}
Here's how I would do it.
First, we'll get the longest sub-array by taking only the smallest elements out of the original, because that way the sum will be smallest possible as we go along. So:
const original = [3, 1, 2, 1];
const maxsum = 4;
// To make sure we take only the smallest, let's just
// sort the array straight away. That way the smallest
// element will always be the first.
// The .slice call is to make a copy, so we don't change
// the original:
const sorted = original.slice().sort();
// Here's your sum function. Looks legit.
function sumArray(a) {
return a.reduce((a, b) => a + b, 0)
}
// Now take items from the front of the sorted array and
// put them in the new array until either the original is
// empty or the max is reached.
let subarray = [];
while (sorted.length) {
// Get the next smallest element. It's always the
// first one because of the sort.
const smallest = sorted.shift();
// If the sum of what we have plus this new element
// is bigger than the max, our work is done:
if (sumArray(subarray) + smallest > maxsum) break;
// Otherwise, add it to our sub array and continue.
subarray.push(smallest)
}
// Once that loop has run, either we ran out of elements,
// or we hit the max. Either way, our job is done.
console.log("Original array:", original);
console.log("Maximal subset:", subarray);
console.log("Maximal subset length:", subarray.length);
Finally, if you want to get fancy, you can even do this with a single .reduce call:
const original = [3, 1, 2, 1];
const maxsum = 4;
const maximalSubset = original.slice().sort().reduce((subset, current) => {
if (subset.reduce((s, c) => s + c, 0) + current <= maxsum) subset.push(current);
return subset;
}, []);
console.log("Orignal:", original);
console.log("Maximal subset:", maximalSubset);
console.log("Maximal subset length:", maximalSubset.length);
Although while shorter, that second snippet has the downside that we have to iterate the entire array before we get the result, whereas the first one will stop once the maximum is reached.
EDIT:
It turns out that the subarray needs to be a continuous piece of the original, so changing the order of the original won't work since we need to make sure the result is a continuous slice of the original.
To do that, instead, just check each subslice of the array, and keep the best one:
let original = [74,659,931,273,545,879,924,710,441,166,493,43,988,504,328,730,841,613,304,170,710,158,561,934,100,279,817,336,98,827,513,268,811,634,980,150,580,822,968,673,394,337,486,746,229,92,195,358,2,154,709,945,669,491,125,197,531,904,723,667,550];
const maxsum = 22337;
function arraySum(arr) {
return arr.reduce((p, c) => p + c, 0);
}
// Double for loop will do the trick.
let bestSoFar = [];
for (let i = 0; i < original.length; i++) {
for (let j = i+1; j < original.length; j++) {
if (j-i > bestSoFar.length && arraySum(original.slice(i, j)) < maxsum) {
bestSoFar = original.slice(i, j);
}
}
}
console.log("Longest continuous subarray is:", bestSoFar.length);
A brute force approach is likely the best solution to this problem. Start at each entry and see how far you can go before arriving at a sum > i and if it's better than the best you've seen so far save it. I provided a sample Java solution below, I haven't actually run it so one or two of my indices may be off but I think you can get the gist. Runtime is O(n^2), memory is O(n) (both occurring with, for example, #getMaxSubArray(new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, 1000000))
private int[] getMaxSubArray(int[] a, int i) {
int minIndex = 0;
int maxIndex = -1;
for (int j = 0; j < a.length; j+=1) {
if (a[j] > i) {
continue;
}
int sum = a[j];
for (int k = j + 1; k < a.length; k++) {
if (sum + a[k] > i) {
if ((k - 1) - j > maxIndex - minIndex) {
maxIndex = k - 1;
minIndex = j;
}
break;
}
sum += a[k];
}
}
if (maxIndex - minIndex < 0) {
return null;
}
int[] result = new int[maxIndex - minIndex + 1];
for (int p = minIndex; p <= maxIndex; p += 1) {
result[p - minIndex] = a[p];
}
return result;
}
This is my solution. It will return the max length of the subarray. Can you please take a look, guys?
function maxLength(a, k) {
const sortedArray = a.sort((i,j) => i - j);
let sum = 0;
let length = 0;
const subArray = [];
for (let i=0; i < sortedArray.length; i++) {
sum = sum + sortedArray[i];
if (sum <= k) {
length++;
subArray.push(sortedArray[i]);
} else {
return length;
}
}
return length;
}