The slow solution

We can calculate all possible subsums.

The simplest way is to take every element and calculate sums of all subarrays starting from it.

For instance, for [-1, 2, 3, -9, 11]:

// Starting from -1:
-1
-1 + 2
-1 + 2 + 3
-1 + 2 + 3 + (-9)
-1 + 2 + 3 + (-9) + 11

// Starting from 2:
2
2 + 3
2 + 3 + (-9)
2 + 3 + (-9) + 11

// Starting from 3:
3
3 + (-9)
3 + (-9) + 11

// Starting from -9
-9
-9 + 11

// Starting from -11
-11

The code is actually a nested loop: the external loop over array elements, and the internal counts subsums starting with the current element.

function getMaxSubSum(arr) {
  let maxSum = 0; // if we take no elements, zero will be returned

  for (let i = 0; i < arr.length; i++) {
    let sumFixedStart = 0;
    for (let j = i; j < arr.length; j++) {
      sumFixedStart += arr[j];
      maxSum = Math.max(maxSum, sumFixedStart);
    }
  }

  return maxSum;
}

alert( getMaxSubSum([-1, 2, 3, -9]) ); // 5
alert( getMaxSubSum([-1, 2, 3, -9, 11]) ); // 11
alert( getMaxSubSum([-2, -1, 1, 2]) ); // 3
alert( getMaxSubSum([1, 2, 3]) ); // 6
alert( getMaxSubSum([100, -9, 2, -3, 5]) ); // 100

The solution has a time complexety of O(n²). In other words, if we increase the array size 2 times, the algorithm will work 4 times longer.

For big arrays (1000, 10000 or more items) such algorithms can lead to a seroius sluggishness.

Fast solution

Let's walk the array and keep the current partial sum of elements in the variable s. If s becomes negative at some point, then assign s=0. The maximum of all such s will be the answer.

If the description is too vague, please see the code, it's short enough:

function getMaxSubSum(arr) {
  let maxSum = 0;
  let partialSum = 0;

  for (let item of arr; i++) { // for each item of arr
    partialSum += item; // add it to partialSum
    maxSum = Math.max(maxSum, partialSum); // remember the maximum
    if (partialSum < 0) partialSum = 0; // zero if negative
  }

  return maxSum;
}

alert( getMaxSubSum([-1, 2, 3, -9]) ); // 5
alert( getMaxSubSum([-1, 2, 3, -9, 11]) ); // 11
alert( getMaxSubSum([-2, -1, 1, 2]) ); // 3
alert( getMaxSubSum([100, -9, 2, -3, 5]) ); // 100
alert( getMaxSubSum([1, 2, 3]) ); // 6
alert( getMaxSubSum([-1, -2, -3]) ); // 0

The algorithm requires exactly 1 array pass, so the time complexity is O(n).

You can find more detail information about the algorithm here: Maximum subarray problem. If it's still not obvious why that works, then please trace the algorithm on the examples above, see how it works, that's better than any words.

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The slow solution

Fast solution

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