Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn’t one, return 0 instead.

Example 1:
Given nums = [1, -1, 5, -2, 3], k = 3,
return 4. (because the subarray [1, -1, 5, -2] sums to 3 and is the longest)

Example 2:
Given nums = [-2, -1, 2, 1], k = 1,
return 2. (because the subarray [-1, 2] sums to 1 and is the longest)

Follow Up:
Can you do it in O(n) time?

Brute force solution (TLC :( )

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public class Solution {
    public int maxSubArrayLen(int[] nums, int k) {
        int length = 0;
        int copy = k;
        for (int i = 0; i < nums.length; i++){
            k = copy;
            for (int j = i; j < nums.length; j++){
                k -= nums[j];
                if (k == 0){
                    length = Math.max(length, j - i + 1);
                }
            }
        }
        return length;
    }
}

O(n) solution

The HashMap stores the sum of all elements before index i as key, and i as value. For each i, check not only the current sum but also (currentSum - previousSum) to see if there is any that equals k, and update max length (from Leetcode discussion board)

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public class Solution {
    public int maxSubArrayLen(int[] nums, int k) {
        int sum = 0, max = 0;
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        
        for (int i = 0; i < nums.length; i++){
            sum = sum + nums[i];
            if (sum == k){max = i + 1;}
            else if (map.containsKey(sum - k)){
                max = Math.max(max, i - map.get(sum - k));
            }
            if (!map.containsKey(sum)){
                map.put(sum, i);
            }
        }
        
        return max;
    }
}