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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1201_1300.s1248_count_number_of_nice_subarrays;

// #Medium #Array #Hash_Table #Math #Sliding_Window
// #2022_03_12_Time_14_ms_(69.78%)_Space_74.9_MB_(51.91%)

/**
 * 1248 - Count Number of Nice Subarrays\.
 *
 * Medium
 *
 * 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`
**/
public class Solution {
    public int numberOfSubarrays(int[] nums, int k) {
        int oddLen = 0;
        int startIndex = 0;
        int num = 0;
        int endIndex;
        int res = 0;
        boolean hasK;
        for (int i = 0; i < nums.length; i++) {
            hasK = false;
            endIndex = i;
            if (nums[i] % 2 == 1) {
                oddLen++;
            }
            while (oddLen >= k) {
                hasK = true;
                if (nums[startIndex++] % 2 == 1) {
                    oddLen--;
                }
                num++;
            }
            res += num;
            while (hasK && ++endIndex < nums.length && nums[endIndex] % 2 == 0) {
                res += num;
            }
            num = 0;
        }
        return res;
    }
}