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Leetcode 2946. Matrix Similarity After Cyclic Shifts

Problem Statement

Given a 2D integer matrix grid, the task is to determine the minimum number of cells that need to be changed to make the matrix similar after any number of cyclical shifts of its rows or columns. A matrix is considered similar to another if one can be transformed into the other by replacing some of its elements with other elements present in the matrix.

Clarifying Questions

  1. Matrix Dimensions: What is the typical range of the matrix dimensions (n x m)?
  2. Element Range: What is the range of values for the elements in the matrix?
  3. Similitude Definition: Are two matrices considered similar if they can be made identical after any number of shifts and substitutions?
  4. Substitution Cost: Is there a specified cost or constraint on how many elements can be replaced, or is it simply about minimizing changes?

With typical constraint ranges in mind, let’s proceed.

Strategy

  1. Understand the Concept of Cyclic Shifts:
    • Cyclic shifts of rows mean shifting rows circularly.
    • Cyclic shifts of columns mean shifting columns circularly.
  2. Similarity Condition:
    • We need to determine the configurations after shifts that require the minimal number of element substitutions.
  3. Transformative Pattern Finding:
    • Identify patterns within rows and columns that optimize for similarity with minimal change.

Code

Below, I’ll provide the Java solution to address the problem statement:

import java.util.HashMap;
import java.util.Map;

public class MatrixSimilarity {
    
    public int minChangesToMakeSimilar(int[][] grid) {
        int rowCount = grid.length;
        int colCount = grid[0].length;
        
        // Initialize a hash map to store frequencies of each element in the matrix
        Map<Integer, Integer> freqMap = new HashMap<>();
        
        // Populate the frequency map
        for (int r = 0; r < rowCount; r++) {
            for (int c = 0; c < colCount; c++) {
                freqMap.put(grid[r][c], freqMap.getOrDefault(grid[r][c], 0) + 1);
            }
        }
        
        // Determine the maximum frequency value
        int maxFreq = freqMap.values().stream().max(Integer::compare).orElse(0);
        
        // Minimum changes needed to make the matrix similar after any shifts
        int totalElements = rowCount * colCount;
        return totalElements - maxFreq;
    }
    
    public static void main(String[] args) {
        MatrixSimilarity ms = new MatrixSimilarity();
        int[][] grid = {
            {2, 3, 1},
            {1, 3, 2},
            {3, 2, 1}
        };
        System.out.println(ms.minChangesToMakeSimilar(grid)); // Should output the minimal number of changes needed
    }
}

Time Complexity

The overall time complexity is (O(n \times m + k)), which is efficient for typical problem constraints.

The space complexity is (O(k)), which is the space needed to store frequencies of unique elements.

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