A maximal subarray

The input is an array of numbers, e.g. arr = [1, -2, 3, 4, -9, 6].

The task is: find the contiguous subarray of arr with the maximal sum of items.

Write the function getMaxSubSum(arr) that will return that sum.

For instance:

          getMaxSubSum([-1, 2, 3, -9]) == 5 (the sum of highlighted items)
getMaxSubSum([2, -1, 2, 3, -9]) == 6
getMaxSubSum([-1, 2, 3, -9, 11]) == 11
getMaxSubSum([-2, -1, 1, 2]) == 3
getMaxSubSum([100, -9, 2, -3, 5]) == 100
getMaxSubSum([1, 2, 3]) == 6 (take all)
        

If all items are negative, it means that we take none (the subarray is empty), so the sum is zero:

getMaxSubSum([-1, -2, -3]) = 0

Please try to think of a fast solution: O(n²) or even O(n) if you can.

Open a sandbox with tests.

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 complexity 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 serious 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) { // 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 detailed 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.

Open the solution with tests in a sandbox.