# 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 $nums$.

• 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;
}