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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.
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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 } }