LeetCode 3400: Maximum Number of Matching Indices After Right Shifts Solution

Master LeetCode problem 3400 (Maximum Number of Matching Indices After Right Shifts), 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.

3400. Maximum Number of Matching Indices After Right Shifts

Medium

Array Two Pointers Simulation

Problem Explanation

Explanation:

Given an array of integers nums and an integer k, we need to find the maximum number of matching indices after performing right shifts on the array nums for each value of k in the range from 0 to nums.length-1.

To solve this problem, we can iterate over each possible value of k from 0 to nums.length-1. For each value of k, we can compare the elements after right shifting the array nums by k positions with the original array nums to find the number of matching indices.

Algorithm:

Initialize a variable maxMatches to store the maximum number of matching indices.
Iterate over each value of k from 0 to nums.length-1.
Right shift the array nums by k positions.
Compare each element of the original array nums with the corresponding element in the right-shifted array.
Count the number of matching indices and update maxMatches if the count is greater than the current maximum.
Reset the array nums to its original state before the next iteration.
Return the maximum number of matching indices maxMatches.

Time Complexity: O(n^2) where n is the length of the input array nums.

Space Complexity: O(n) for storing the right-shifted array for each value of k.

Solution Code

class Solution {
    public int maxMatching(int[] nums) {
        int n = nums.length;
        int maxMatches = 0;
        
        for (int k = 0; k < n; k++) {
            int matches = 0;
            for (int i = 0; i < n; i++) {
                if (nums[i] == nums[(i + k) % n]) {
                    matches++;
                }
            }
            maxMatches = Math.max(maxMatches, matches);
        }
        
        return maxMatches;
    }
}

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