maxSubarraySum.js

// Given an array of integers and a number, write a function called maxSubarraySum, which finds the maximum sum of a subarray with the length of the number passed to the function.

// Note that a subarray must consist of consecutive elements from the original array. In the first example below, [100, 200, 300] is a subarray of the original array, but [100, 300] is not.

// maxSubarraySum([100,200,300,400], 2) // 700
// maxSubarraySum([1,4,2,10,23,3,1,0,20], 4)  // 39
// maxSubarraySum([-3,4,0,-2,6,-1], 2) // 5
// maxSubarraySum([3,-2,7,-4,1,-1,4,-2,1],2) // 5
// maxSubarraySum([2,3], 3) // null
// Constraints:

// Time Complexity - O(N)

// Space Complexity - O(1)

function maxSubarraySum(arr, length) {
  if(arr.length < length){
    return null;
  }
  let maxSum = 0;
  for(let i = 0; i < length; i++){
    maxSum += arr[i];
  }
  let iterateSum =maxSum;
  for(let i = length; i < arr.length; i++){
    iterateSum += arr[i] - arr[i - length];
    if(iterateSum > maxSum){
      maxSum = iterateSum;
    }
  }
  return maxSum;
}

console.log(maxSubarraySum([100,200,300,400], 2));