[Don't comment your code here - You will be immediately disqualified]
Given an array of positive numbers and a positive number ‘k,’ find the maximum sum of any contiguous subarray of size ‘k’.
Example 1:
Input: [2, 1, 5, 1, 3, 2], k=3 Output: 9 Explanation: Subarray with maximum sum is [5, 1, 3].
Example 2:
Input: [2, 3, 4, 1, 5], k=2 Output: 7 Explanation: Subarray with maximum sum is [3, 4].
function max_sub_array_of_size_k(k, arr) {
let maxSum = 0,
windowSum = 0;
// loop through start till array length - k
for (i = 0; i < arr.length - k + 1; i++) {
windowSum = 0;
// loop through i to i + k elements
for (j = i; j < i + k; j++) {
windowSum += arr[j];
}
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}
console.log(Maximum sum of a subarray of size K: ${max_sub_array_of_size_k(3, [2, 1, 5, 1, 3, 2])})
Run solution at Code
function arrSum() {
var arr = [2, 1, 5, 1, 3, 2];
const k = 2;
var number = [];
var len = arr.length;
if (k > 0) {
if (k < len / 2) {
number = arr.slice(k - 1, k + k - 1);
var sum = number.reduce(function(a, b) {
return a + b;
}, 0);
} else {
number = arr.slice(0, k);
var sum = number.reduce(function(a, b) {
return a + b;
}, 0);
}
console.log( "Sum: " + sum);
console.log( "Number: " + [number]);
}
function maxSum(arr, k) {
let sum=0, res=Number.NEGATIVE_INFINITY;
for(let i=0; i<k; i++) {
sum+=arr[i];
}
res=sum;
for(let i=1; i+k<=arr.length; i++) {
sum-=arr[i-1];
sum+=arr[i+k-1];
res = Math.max(res, sum);
}
return res;
}
console.log(maxSum([2, 3, 4, 1, 5], 3));