LeetCode 1248. Count Number of Nice Subarrays Solution in Java, C++, Python & More | Explanation + Code

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 are 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.

PythonJavaC++GoTypeScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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