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1248. Count Number of Nice Subarrays

Description

Given an array of integers nums and an integer k. A continuous subarray is called nice if there are k odd numbers on it.

Return the number of nice sub-arrays.

 

Example 1:

Input: nums = [1,1,2,1,1], k = 3
Output: 2
Explanation: The only sub-arrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].

Example 2:

Input: nums = [2,4,6], k = 1
Output: 0
Explanation: There is no odd numbers in the array.

Example 3:

Input: nums = [2,2,2,1,2,2,1,2,2,2], k = 2
Output: 16

 

Constraints:

• 1 <= nums.length <= 50000
• 1 <= nums[i] <= 10^5
• 1 <= k <= nums.length

Solutions

Solution 1: Prefix Sum + Array or Hash Table

The problem asks for the number of subarrays that contain exactly $k$ odd numbers. We can calculate the number of odd numbers $t$ in each prefix array and record it in an array or hash table $cnt$. For each prefix array, we only need to find the number of prefix arrays with $t-k$ odd numbers, which is the number of subarrays ending with the current prefix array.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

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

• class Solution {
      public int numberOfSubarrays(int[] nums, int k) {
          int n = nums.length;
          int[] cnt = new int[n + 1];
          cnt[0] = 1;
          int ans = 0, t = 0;
          for (int v : nums) {
              t += v & 1;
              if (t - k >= 0) {
                  ans += cnt[t - k];
              }
              cnt[t]++;
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int numberOfSubarrays(vector<int>& nums, int k) {
          int n = nums.size();
          vector<int> cnt(n + 1);
          cnt[0] = 1;
          int ans = 0, t = 0;
          for (int& v : nums) {
              t += v & 1;
              if (t - k >= 0) {
                  ans += cnt[t - k];
              }
              cnt[t]++;
          }
          return ans;
      }
  };
  

• class Solution:
      def numberOfSubarrays(self, nums: List[int], k: int) -> int:
          cnt = Counter({0: 1})
          ans = t = 0
          for v in nums:
              t += v & 1
              ans += cnt[t - k]
              cnt[t] += 1
          return ans
  
  

• func numberOfSubarrays(nums []int, k int) (ans int) {
  	n := len(nums)
  	cnt := make([]int, n+1)
  	cnt[0] = 1
  	t := 0
  	for _, v := range nums {
  		t += v & 1
  		if t >= k {
  			ans += cnt[t-k]
  		}
  		cnt[t]++
  	}
  	return
  }
  

• function numberOfSubarrays(nums: number[], k: number): number {
      const n = nums.length;
      const cnt = new Array(n + 1).fill(0);
      cnt[0] = 1;
      let ans = 0;
      let t = 0;
      for (const v of nums) {
          t += v & 1;
          if (t - k >= 0) {
              ans += cnt[t - k];
          }
          cnt[t] += 1;
      }
      return ans;
  }
  
  

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