# Question

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.

Note:

The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.

Example 1:

Input: nums = [1, -1, 5, -2, 3], k = 3
Output: 4
Explanation: The subarray [1, -1, 5, -2] sums to 3 and is the longest.


Example 2:

Input: nums = [-2, -1, 2, 1], k = 1
Output: 2
Explanation: The subarray [-1, 2] sums to 1 and is the longest.


Can you do it in O(n) time?

# Solution

Use DP. Store the sum from 0 to i inclusive as map key, and i as value. For i+1, if the sum from 0 to i+1 is k, then return. Otherwise, find if sum - k is in the map. If so, then the range j+1 (where j is the value of key sum-k) to i sums to k.

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;
}
}