325. Maximum Size Subarray Sum Equals k (maximum length of a subarray that sums to k.)

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

 

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.

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