2845. Count of Interesting Subarrays

Explanation:

To solve this problem, we can iterate through all possible subarrays and check if each subarray is interesting based on the given conditions. We can maintain a count of indices within each subarray that satisfy the condition and then check if this count modulo modulo is equal to k.

Here are the detailed steps:

Initialize a variable count to keep track of the total count of interesting subarrays.
Iterate through all possible subarrays using two nested loops for the start index and end index of the subarray.
For each subarray, iterate through the elements to count the number of indices that satisfy the condition nums[i] % modulo == k.
Check if this count modulo modulo is equal to k. If it is, increment the count.
Return the final count as the result.

Time Complexity: The time complexity of this approach is O(N^2), where N is the number of elements in the input array nums, as we are checking all possible subarrays.

Space Complexity: The space complexity is O(1) as we are not using any extra space that grows with the input size.

:

View Leetcode Problem

class Solution {
    public int countInterestingSubarrays(int[] nums, int modulo, int k) {
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i; j < nums.length; j++) {
                int cnt = 0;
                for (int idx = i; idx <= j; idx++) {
                    if (nums[idx] % modulo == k) {
                        cnt++;
                    }
                }
                if (cnt % modulo == k) {
                    count++;
                }
            }
        }
        return count;
    }
}

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