LeetCode 560. Subarray Sum Equals K (Medium)

Description:

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:

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

---

Analysis:

1. Brute force is always a correct method. However, we can also make a tweak. For example, we use $sum[i,j]$ to denote the sum of elements from $a[i]$ to $a[j]$ (inclusive), if we know the $sum[0,i-1]$ and $sum[0,j]$ we can get $sum[i,j]$ by executing $sum[0,j]-sum[0,i-1]$(Here we assume $i-1<j$).
2. In this problem, we need to find the number of $sum[i,j]$ which equals to $k$. When we traverse the array, we can get $sum[0,j]$ trivially, so what we need to do is to find the number of $sum[i-1]$ that satisfies $sum[0,i-1]=sum[0,j]-k$.
3. What we need to be advised is that when $sum[0,j]$ is equal to $k$, $i=0$ and $i-1$ is equal to $-1$, which seems illegal. So under this special circumstance, we should consider $sum[0,i-1]$ as $0$.

---

Code:

class Solution {
    public int subarraySum(int[] nums, int k) {
        int rs = 0;
        Map<Integer, Integer> sumFreq = new HashMap<>(); // { sum[0, i-1]: frequency }
        sumFreq.put(0, 1);
        
        int curSum = 0; // sum[0, j] (inclusive)
        for(int j = 0; j < nums.length; j++) {
            curSum += nums[j]; 
            // Try finding sum[0, i-1] that satisfies sum[0, i-1] = sum[0, j] - k .
            if(sumFreq.containsKey(curSum - k)) { 
                rs += sumFreq.get(curSum - k);
            }
            sumFreq.put(curSum, sumFreq.getOrDefault(curSum, 0)+1); //sum[0, j] --> sum[0, i-1], map.get(sum[0, i-1])++;
        }
        return rs;
    }
}