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?**Understand the problem:**

The problem is equal to: find out a range from i to j, in which the sum (nums[i], ..., nums[j]) = k. What is the maximal range?

So we can first calculate the prefix sum of each number, so sum(i, j) = sum(j) - sum(i - 1) = k. Therefore, for each sum(j), we only need to check if there was a sum(i - 1) which equals to sum(j) - k. We can use a hash map to store the previous calculated sum.

**Code (Java):**

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

**Comments:**

Note the map.put(0, -1). We need to put this entry into the map before, because if the maximal range starts from 0, we need to calculate sum(j) - sum(i - 1).

Good Solution!

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