Leetcode: 325. Maximum Size Subarray Sum Equals k

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?

Solution:
The idea of the problem is to check where there is a range from i to j, inclusive, so that its sum equals to k, and the length of the range is the maximum. 

So we can naturally think of this question as a range summary problem, and we need to calculate the prefix sum of the array first. So the sum(i, j) = presum[j] - presum[i - 1] = k

In order to achieve the O(n) time, we can leverage the same idea of the "Two Sum" problem by using a hash map. So we store the presum[i - 1] + k into the map, and check if presum[j] is in the map for each iteration. Note that we can do this in one-pass of loop iteration because for each j, i - 1 must be in the position above j. 

Code (Java):

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public class Solution {     public int maxSubArrayLen(int[] nums, int k) {         if (nums == null || nums.length == 0) {             return 0;         }                   // step 1: calculate the prefix sum for all numbers of the nums array         int n = nums.length;         int[] preSum = new int[n + 1];         int sum = 0;         for (int i = 0; i < nums.length; i++) {             sum += nums[i];             preSum[i + 1] = sum;         }                   // step 2: put the preSum + target into a map         int max = 0;         Map<Integer, Integer> map = new HashMap<>();         for (int j = 0; j < preSum.length; j++) {             if (map.containsKey(preSum[j])) {                 max = Math.max(max, j - map.get(preSum[j]));             }                           if (!map.containsKey(preSum[j] + k)) {                 map.put(preSum[j] + k, j);             }         }                             return max;     } }