Leetcode 1248 count number of nice subarrays

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

Topic link ：https://leetcode.com/problems/count-number-of-nice-subarrays/

The main idea of the topic ： Seeking inclusion k Number of odd subarrays

Topic analysis ： Use an array pos Record the subscript of odd numbers , Accumulate the number of even numbers on the left and right of each odd number as an array l0 and r0, Then enumerate and calculate , Right. i Odd number and number i+k-1 An odd number , The length they can form is k The subarray of is (1+l0[pos[i]]) * (1 + r0[pos[i+k-1]]) 

12ms, Time beats 81.3%

class Solution {
    public int numberOfSubarrays(int[] nums, int k) {
        int n = nums.length, sum = 0;
        int[] pos = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            if (nums[i - 1] % 2 == 1) {
                pos[++sum] = i;
            }
        }

        int[] r0 = new int[n + 1];
        int cntr0 = 0;
        for (int i = n; i >= 1; i--) {
            if (nums[i - 1] % 2 == 0) {
                cntr0++;
            } else {
                r0[i] = cntr0;
                cntr0 = 0;
            }
        }
            
        int[] l0 = new int[n + 1];
        int cntl0 = 0;
        for (int i = 1; i <= n; i++) {
            if (nums[i - 1] % 2 == 0) {
                cntl0++;
            } else {
                l0[i] = cntl0;
                cntl0 = 0;
            }
        }
        int ans = 0;
        for (int i = 1; i + k - 1 <= sum; i++) {
            if (pos[i] != 0 && i + k - 1 <= sum && pos[i + k - 1] != 0) {
                ans += (1 + l0[pos[i]]) * (1 + r0[pos[i + k - 1]]);
            }
        }
        return ans;
    }
}