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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.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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
      }
  }
  
  

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