maximal-sum v1.0.0

Maximal sum of contiguous subarray of an array

It provides two functions to get the job done. Both of them calculate sum of contiguous subarray of an array where the subarray produces maximum possible sum.

For example

expect(getMaxSubSum([-1, 2, 3, -9])).toBe(5);
expect(getMaxSubSum([2, -1, 2, 3, -9])).toBe(6);
expect(getMaxSubSum([-1, 2, 3, -9, 11])).toBe(11);
expect(getMaxSubSum([-2, -1, 1, 2])).toBe(3);
expect(getMaxSubSum([100, -9, 2, -3, 5])).toBe(100);
expect(getMaxSubSum([1, 2, 3])).toBe(6);

Install

npm i maximal-sum

getMaxSubSum

const { getMaxSubSum } = require('maximal-sum');
const sum = getMaxSubSum([2, -1, 2, 3, -9]);
// > 6
// The subarray is [2, -1, 2, 3]

Use this method to have O(n) complexity. The concept is simple:

1. Let maximum sum be 0;
2. Let current sum be 0;
3. Loop over all items of the array: a. add the item to current. If it becomes < 0, then override to 0. b. maximum is the greatest between maximum and current.

It works because, we don't actually want to get the subarray itself. Just the sum. Also when comparing current with maximum, if at some point, the sum becomes greater, starting from an index !== 0, we override at 3.b.

getMaxSubSumBrute

Just for testing purpose. It produces the same result but has worst performance. The reason is, it uses nested loops to brute-force all possible subarrays and checks for the maximum of sums.

const { getMaxSubSumBrute } = require('maximal-sum');
const sum = getMaxSubSumBrute([2, -1, 2, 3, -9]);
// > 6
// The subarray is [2, -1, 2, 3]