# 560. Subarray Sum Equals K

## Description

Given an array of integers nums and an integer k, return the total number of subarrays whose sum equals to k.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

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


Example 2:

Input: nums = [1,2,3], k = 3
Output: 2


Constraints:

• 1 <= nums.length <= 2 * 104
• -1000 <= nums[i] <= 1000
• -107 <= k <= 107

## Solutions

Use a map to store each sum and the count of subarrays of the sum. Put sum 0 with count 1 to the map initially.

Loop over nums and calculate the total sum sum. For each num in nums, add num to sum, and obtain the count of subarrays with sum sum - k. Add the sum to the result. Then update the map with the count of subarrays with sum sum. Finally, return the result.

• class Solution {
public int subarraySum(int[] nums, int k) {
Map<Integer, Integer> counter = new HashMap<>();
counter.put(0, 1);
int ans = 0, s = 0;
for (int num : nums) {
s += num;
ans += counter.getOrDefault(s - k, 0);
counter.put(s, counter.getOrDefault(s, 0) + 1);
}
return ans;
}
}

• class Solution {
public:
int subarraySum(vector<int>& nums, int k) {
unordered_map<int, int> counter;
counter[0] = 1;
int ans = 0, s = 0;
for (int& num : nums) {
s += num;
ans += counter[s - k];
++counter[s];
}
return ans;
}
};

• class Solution:
def subarraySum(self, nums: List[int], k: int) -> int:
dp = Counter({0: 1})
ans = s = 0
for num in nums:
s += num
ans += dp[s - k] # not matter if the same num like [1,1]
dp[s] += 1
return ans

#############

class Solution: # over time limit, not fast enough
def subarraySum(self, nums: List[int], k: int) -> int:
count = 0

# sum value from 1 to i-th element
sum_arr = [0] * (len(nums) + 1)

# cached sum
for i in range(1, len(nums) + 1):
sum_arr[i] = sum_arr[i - 1] + nums[i - 1]

for start in range(len(nums)):
for end in range(start + 1, len(nums) + 1):
if sum_arr[end] - sum_arr[start] == k:
count += 1

return count


• func subarraySum(nums []int, k int) int {
counter := map[int]int{0: 1}
ans, s := 0, 0
for _, num := range nums {
s += num
ans += counter[s-k]
counter[s]++
}
return ans
}

• function subarraySum(nums: number[], k: number): number {
let ans = 0,
s = 0;
const counter = new Map();
counter.set(0, 1);
for (const num of nums) {
s += num;
ans += counter.get(s - k) || 0;
counter.set(s, (counter.get(s) || 0) + 1);
}
return ans;
}


• use std::collections::HashMap;

impl Solution {
pub fn subarray_sum(nums: Vec<i32>, k: i32) -> i32 {
let mut res = 0;
let mut sum = 0;
let mut map = HashMap::new();
map.insert(0, 1);
nums.iter().for_each(|num| {
sum += num;
res += map.get(&(sum - k)).unwrap_or(&0);
map.insert(sum, map.get(&sum).unwrap_or(&0) + 1);
});
res
}
}