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.

Follow Up:

  • Can you do it in O(n) time?

Solution

  • Method 1: O(n^2) 
      class Solution {
      public int maxSubArrayLen(int[] nums, int k) {
          int max = 0;
          for(int i = 0; i < nums.length; i++){
              int sum = 0;
              for(int j = i; j < nums.length; j++){
                  sum += nums[j];
                  if(sum == k){
                      max = Math.max(max, j - i + 1);
                  }
              }
          }
          return max;
      }
  }
  • Method 2: O(n) 
      class Solution {
      public int maxSubArrayLen(int[] nums, int k) {
          Map<Integer, Integer> map = new HashMap<>();
          for(int i = 1; i < nums.length; i++){
              nums[i] += nums[i - 1];
          }
          int max = 0;
          map.put(0, -1);
          for(int i = 0; i < nums.length; i++){
              if(map.containsKey(nums[i] - k)){
                  max = Math.max(max, i - map.get(nums[i] - k));
              }
              if(!map.containsKey(nums[i])){
                  map.put(nums[i], i);
              }
          }
          return max;
      }
  }