Naive solution for this problem would be caluclate all the possible combinations:
const numbers = [1, -3, 2 - 5, 7, 6, -1, -4, 11, -23]; // O(n^3) const findMaxSubAry = numbers => { let answer = Number.MIN_VALUE; /** * Calculate all the possible values and pick the max one * All possible values should be * length = 1, 2, ,3 ... n * Pick differnet start point */ // For different lenght for (let l = 0; l < numbers.length; l++) { // O(n) // For different start for (let s = 0; s < l; s++) { // O(n) if (s + l >= numbers.length) { break; } let sum = 0; for (let i = s; i < s + l; i++) { // O(n) sum += numbers[i]; } answer = Math.max(answer, sum); } } return answer; }; console.log(findMaxSubAry(numbers)); // 19
The maximum subarray problem is one of the nicest examples of dynamic programming application.
In this lesson we cover an example of how this problem might be presented and what your chain of thought should be to tackle this problem efficiently.
/** * Maximum Contiguous subarray algorithm * * Max(i) = Max(i-1) + v(i) * Max(i-1) < 0 ? v(i) : Max(i-1) * * Combining --------- maxInc(i) = maxInc(i - 1) > 0 ? maxInc(i - 1) + val(i) : val(i) max(i) = maxInc(i) > max(i - 1) ? maxInc(i) : max(i - 1) */ function maxSumSubArray(arr) { /** * inx | val | max_inc | max * 0 0 0 * 0 -2 0 0 * 1 -3 0 0 * 2 4 4 4 ---> start = 2 * 3 -1 3 4 * 4 -2 1 4 * 5 1 2 4 * 6 5 7 7 ---> end = 6 * 7 -3 4 7 */ let val = 0, max_inc = 0, max = 0, start = 0, end = 0; for (let i = 1; i < arr.length; i++) { val = arr[i]; max_inc = Math.max(max_inc + val, val); max = Math.max(max, max_inc); if (val === max_inc) { start = i; } if (max === max_inc) { end = i; } } if (end === 0) { end = start; } console.log(start, end); return arr.slice(start, end + 1); } console.log(maxSumSubArray([-2, -3, 4, -1, -2, 1, 5, -3])); //[4, -1, -2, 1, 5] console.log(maxSumSubArray([-2,-3,-4,-1,-2])); // [-2]