LeetCode 3395: Subsequences with a Unique Middle Mode I Solution

Master LeetCode problem 3395 (Subsequences with a Unique Middle Mode I), a hard 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.

3395. Subsequences with a Unique Middle Mode I

Hard

Array Hash Table Math Combinatorics

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible subsequences of size 5 of the given array nums. For each subsequence, we check if the middle element (seq[2]) is a unique mode, i.e., it appears with the maximum frequency in the subsequence. If it is a unique mode, we increment the count of valid subsequences. Finally, we return the count modulo 10^9 + 7 as the result.

Algorithm:

Initialize a variable count to store the count of subsequences with a unique middle mode.
Iterate through all possible subsequences of size 5 of the input array nums.
For each subsequence, check if the middle element (seq[2]) is a unique mode.
If the middle element is a unique mode, increment the count.
Return count modulo 10^9 + 7 as the result.

Time Complexity: O(n^5), where n is the length of the input array nums. Space Complexity: O(1)

Solution Code

class Solution {
    public int countMiddleModes(int[] nums) {
        int mod = 1000000007;
        int count = 0;
        
        for(int i = 0; i < nums.length; i++) {
            for(int j = i+1; j < nums.length; j++) {
                for(int k = j+1; k < nums.length; k++) {
                    for(int l = k+1; l < nums.length; l++) {
                        for(int m = l+1; m < nums.length; m++) {
                            int[] seq = {nums[i], nums[j], nums[k], nums[l], nums[m]};
                            Arrays.sort(seq);
                            if(seq[2] == seq[1] || seq[2] == seq[3]) continue;
                            count = (count + 1) % mod;
                        }
                    }
                }
            }
        }
        
        return count;
    }
}

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