leetcode 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].

这道题还算简单。如果用 DP[i][j] 来做（i是子串的开始index，j是子串的结束index，DP[i][j] 存储 i ~ j 的和），会 Memory Limit Exceed.

需要注意的测试用例有两个。

1.

2.  [ -1, 1, 0, -1, 1]

这道题用最简单的方法就可以AC。

public int subarraySum(int[] nums, int k) {
	int count=0;		
	for(int begin=0;begin

这道题有solutions： https://leetcode.com/problems/subarray-sum-equals-k/solution/

Approach #2 Using Cummulative sum [Accepted]

Algorithm

我们定义了一个累积数组$sum[]$, 其中$sum[i]$ 存储了 0 ~ (i−1)​th​​ index 的和。要想得到 SUM[i, j]，只需要知道 SUM[0, i - 1] 和 SUM[0, j] 就好了。SUM$[i~j]$= $sum[j]-sum[i-1]$.

Java

public class Solution {
    public int subarraySum(int[] nums, int k) {
        int count = 0;
        int[] sum = new int[nums.length + 1];
        sum[0] = 0;
        for (int i = 1; i <= nums.length; i++)
            sum[i] = sum[i - 1] + nums[i - 1];
        for (int start = 0; start < nums.length; start++) {
            for (int end = start + 1; end <= nums.length; end++) {
                if (sum[end] - sum[start] == k)
                    count++;
            }
        }
        return count;
    }
}

Complexity Analysis

• Time complexity : O(n^2)O(n​2​​). Considering every possible subarray takes O(n^2)O(n​2​​) time. Finding out the sum of any subarray takes $O(1)$ time after the initial processing of $O(n)$ for creating the cumulative sum array.

• Space complexity : $O(n)$. Cumulative sum array $sum$ of size $n+1$ is used.

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Approach #3 Without space [Accepted]

Algorithm

这个解法跟我的解法是一样的。选择一个特定的$start$ 点，遍历其之后的 $end$ 点。计算从 start ~ end 的 sum。当 $sum$ = $k$ 时，更新 $count$ 值。

Java

public class Solution {
    public int subarraySum(int[] nums, int k) {
        int count = 0;
        for (int start = 0; start < nums.length; start++) {
            int sum=0;
            for (int end = start; end < nums.length; end++) {
                sum+=nums[end];
                if (sum == k)
                    count++;
            }
        }
        return count;
    }
}

Complexity Analysis

• Time complexity : O(n^2)O(n​2​​). We need to consider every subarray possible.

• Space complexity : $O(1)$. Constant space is used.

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Approach #4 Using hashmap [Accepted]

Algorithm

we know the key to solve this problem is SUM[i, j]. So if we know SUM[0, i - 1] and SUM[0, j], then we can easily get SUM[i, j]. To achieve this, we just need to go through the array, calculate the current sum and save number of all seen PreSum to a HashMap. Time complexity O(n), Space complexity O(n).

Java

public class Solution {
    public int subarraySum(int[] nums, int k) {
        int sum = 0, count= 0;
        Map preSum = new HashMap<>();
        preSum.put(0, 1);
        
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (preSum.containsKey(sum - k)) {
                count += preSum.get(sum - k);
            }
            preSum.put(sum, preSum.getOrDefault(sum, 0) + 1);
        }
        
        return count;
    }
}

Complexity Analysis

• Time complexity : $O(n)$. The entire $nums$ array is traversed only once.

• Space complexity : $O(n)$. Hashmap $map$ can contain upto $n$ distinct entries in the worst case.