325. Maximum Size Subarray Sum Equals k

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:

Given 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 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(n) time?

Solution:

This is a range sum question. Whenever we see similar problem, we shall try prefixSum.

prefixSum[i] is the sum from 0 to i.

So prefixSum[j] - prefixSum[i - 1] is the range sum from i to j.

Now we go through the input array and calculate the prefixSum at index i.

At the same time, if we check and find a previous prefixSum with value prefixSum[i] - k in a HashMap, we know the range sum from map.get(prefixSum[i] - k) to i is k.

Therefore, we need O(n) time and O(n) space to solve the problem.



Code:

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