Maximum Subarray solution using TypeScript

This post outlines my TypeScript solution to the "Maximum Subarray" question on LeetCode using a greedy algorithm.

Algorithm: The solution finds the subarray with the largest sum by iterating over each number in the array. For each number it will do the following:

  • Set the current sum to be whichever is larger:
    • The current number + current sum
    • The current number
  • Set the max sum to be whichever is larger:
    • The current sum
    • The current max sum

Once all the elements have been iterated over, the max sum will be returned.

Time Complexity: Because each element in the array will be visited once, the time complexity it O(n) where n represents the number of elements in the array.

Space Complexity: Because the solution will use a constant amount of space, the space complexity is O(1).

 * Maximum Subarray
 * Given an integer array nums, find the contiguous 
 * subarray (containing at least one number) 
 * which has the largest sum and return its sum.
 * Time Complexity: O(n)
 * Space Complexity: O(1)
 * maxSubArray([1, 2, 2, -2, 2]) // 5
 * maxSubArray([1, 5, -7, 4, 2, 1]) // 7
function maxSubArray(nums: number[]): number {
    let sum = nums[0]; // current sum
    let max = nums[0]; // max sum
    let ptr = 1; // pointer
    while(ptr < nums.length) {
        sum = Math.max(nums[ptr], sum + nums[ptr]);
        max = Math.max(max, sum);
        ptr += 1;

    return max;

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