560. Subarray Sum Equals K

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.

Example 1:

Input:nums = [1,1,1], k = 2
Output: 2

Note:

1. The length of the array is in range [1, 20,000].
2. The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].

Solution:

The idea is to use a HashMap to store the prefixSum and its frequency.

When we are at index i with prefixSum sum, we check if there exists a prefixSum (sum - k).

If so, the frequency of this (sum - k) means how many different continuous subarray we can have that end with i will sum up to k.

The time complexity is O(n) and the space complexity is also O(n).



Code:

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