1248. Count Number of Nice Subarrays

class Solution {
    public int numberOfSubarrays(int[] nums, int k) {
        int[] preArray = new int[nums.length + 1]; // 注意不要忘记多分配一个
        preArray[0] = 1;

        int sum = 0, res = 0;
        for (int num : nums) {
            sum += num & 1; // 前缀数组奇数的个数
            if (sum >= k) {
                res += preArray[sum - k];
            }
            preArray[sum]++;
        }

        return res;
    }
}

Reference

1248. Count Number of Nice Subarrays