LeetCode 1248: Count Number of Nice Subarrays Solution

Master LeetCode problem 1248 (Count Number of Nice Subarrays), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1248. Count Number of Nice Subarrays

Medium

Array Hash Table Math Sliding Window Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can use a sliding window approach. We will maintain two pointers - left and right, and keep expanding the window until we have exactly k odd numbers in the window. For each valid window, we can calculate the number of subarrays that can be formed with that window as a nice subarray. We can keep track of the count of such subarrays and return the final result.

Initialize left = 0, right = 0, count = 0, oddCount = 0, result = 0
Iterate over the array while moving the right pointer:
- If nums[right] is odd, increment oddCount
- While oddCount == k, increment count for each valid window and move the left pointer:
  - While nums[left] is not odd, increment count and move left
  - After exiting the inner loop, calculate the number of subarrays that can be formed with the current window
  - Increment result by the count of subarrays
  - Move the left pointer to continue the search for more valid windows
Return the final result

Time Complexity: O(N) where N is the number of elements in the array. Space Complexity: O(1)

Solution Code

class Solution {
    public int numberOfSubarrays(int[] nums, int k) {
        int left = 0, right = 0, count = 0, oddCount = 0, result = 0;
        
        for (right = 0; right < nums.length; right++) {
            if (nums[right] % 2 == 1) {
                oddCount++;
            }
            
            while (oddCount == k) {
                count = 0;
                while (oddCount == k) {
                    if (nums[left] % 2 == 0) {
                        count++;
                    } else {
                        oddCount--;
                    }
                    left++;
                }
                result += count;
            }
        }
        
        return result;
    }
}

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