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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 $nums$.

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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

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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;
    }
}

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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;
    }
};

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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
}

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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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