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

return

*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

return

*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(

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):**

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; } }

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