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Minimum changes required to make each path in a matrix palindrome
  • Difficulty Level : Hard
  • Last Updated : 06 Aug, 2020

Given a matrix with N rows and M columns, the task is make all possible paths from the cell (N, M) to (1, 1) palindrome by minimum changes in the cell values. 
 

Possible moves from any cell (x, y) is either move Left(x – 1, y) or move Down (x, y – 1).

Examples: 
 

Input: mat[ ][ ] = { { 1, 2, 2 }, { 1, 0, 0 } } 
Output:
Explanation: 
For each path in matrix to be Palindrome, possible matrices (after changes) are 
{ { 0, 2, 2 }, { 2, 2, 0 } } or { { 1, 2, 2 }, { 2, 2, 1 } }.
Input: mat[ ][ ] = { { 5, 3 }, { 0, 5 } } 
Output:
Explanation: 
No change required in above matrix. Each path from (N, M) to (1, 1) is already Palindrome 
 

 



Approach: 
 

  • A path is called palindromic if the value of the last cell is equal to the value of the first cell, the value of the second last cell is equal to the value of the second cell, and so on. 
     
  • So we can conclude that, to make a path palindromic, the cells at distance (Manhatten distance) x from (N, M) must be equal to the cells at distance x from (1, 1)
     
  • To minimize the number of changes, convert each cell at distance x from (1, 1) and (N, M) to the most frequent among all values present in those cells. 
     

Below is the implementation of the above approach.
 

C++




// C++ Program to implement the
// above approach
#include <bits/stdc++.h>
using namespace std;
 
#define N 7
 
// Function for counting changes
int countChanges(int matrix[][N],
                 int n, int m)
{
    // Maximum distance possible
    // is (n - 1 + m - 1)
    int dist = n + m - 1;
 
    // Stores the maximum element
    int Max_element = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            // Update the maximum
            Max_element = max(Max_element,
                              matrix[i][j]);
        }
    }
 
    // Stores frequencies of
    // values for respective
    // distances
    int freq[dist][Max_element + 1];
 
    // Initialize frequencies of
    // cells as 0
    for (int i = 0; i < dist; i++) {
        for (int j = 0; j < Max_element + 1;
             j++)
            freq[i][j] = 0;
    }
 
    // Count frequencies of cell
    // values in the matrix
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
 
            // Increment frequency of
            // value at distance i+j
            freq[i + j][matrix[i][j]]++;
        }
    }
 
    int min_changes_sum = 0;
    for (int i = 0; i < dist / 2; i++) {
 
        // Store the most frequent
        // value at i-th distance
        // from (0, 0) and (N - 1, M - 1)
        int maximum = 0;
        int total_values = 0;
 
        // Calculate max frequency
        // and total cells at distance i
        for (int j = 0; j < Max_element + 1;
             j++) {
 
            maximum
                = max(maximum,
                      freq[i][j]
                          + freq[n + m - 2
                                 - i][j]);
 
            total_values
                += freq[i][j]
                   + freq[n + m - 2 - i][j];
        }
 
        // Count changes required
        // to convert all cells
        // at i-th distance to
        // most frequent value
        min_changes_sum
            += total_values - maximum;
    }
 
    return min_changes_sum;
}
 
// Driver Code
int main()
{
    int mat[][N] = { { 7, 0, 3, 1, 8, 1, 3 },
                     { 0, 4, 0, 1, 0, 4, 0 },
                     { 3, 1, 8, 3, 1, 0, 7 } };
 
    int minChanges = countChanges(mat, 3, 7);
 
    cout << minChanges;
 
    return 0;
}

Java




// Java program to implement the
// above approach
import java.util.*;
 
class GFG{
static int N = 7;
 
// Function for counting changes
static int countChanges(int matrix[][],
                        int n, int m)
{
     
    // Maximum distance possible
    // is (n - 1 + m - 1)
    int i, j, dist = n + m - 1;
 
    // Stores the maximum element
    int Max_element = 0;
    for(i = 0; i < n; i++)
    {
        for(j = 0; j < m; j++)
        {
             
            // Update the maximum
            Max_element = Math.max(Max_element,
                                   matrix[i][j]);
        }
    }
 
    // Stores frequencies of
    // values for respective
    // distances
    int freq[][] = new int[dist][Max_element + 1];
 
    // Initialize frequencies of
    // cells as 0
    for(i = 0; i < dist; i++)
    {
        for(j = 0; j < Max_element + 1;
            j++)
            freq[i][j] = 0;
    }
 
    // Count frequencies of cell
    // values in the matrix
    for(i = 0; i < n; i++)
    {
        for(j = 0; j < m; j++)
        {
             
            // Increment frequency of
            // value at distance i+j
            freq[i + j][matrix[i][j]]++;
        }
    }
     
    int min_changes_sum = 0;
    for(i = 0; i < dist / 2; i++)
    {
         
        // Store the most frequent
        // value at i-th distance
        // from (0, 0) and (N - 1, M - 1)
        int maximum = 0;
        int total_values = 0;
 
        // Calculate max frequency
        // and total cells at distance i
        for(j = 0; j < Max_element + 1; j++)
        {
            maximum = Math.max(maximum,
                               freq[i][j] +
                               freq[n + m -
                                    2 - i][j]);
            total_values += freq[i][j] +
                            freq[n + m -
                                 2 - i][j];
        }
 
        // Count changes required
        // to convert all cells
        // at i-th distance to
        // most frequent value
        min_changes_sum += total_values -
                           maximum;
    }
    return min_changes_sum;
}
 
// Driver Code
public static void main (String []args)
{
    int mat[][] = { { 7, 0, 3, 1, 8, 1, 3 },
                    { 0, 4, 0, 1, 0, 4, 0 },
                    { 3, 1, 8, 3, 1, 0, 7 } };
 
    int minChanges = countChanges(mat, 3, 7);
 
    System.out.print(minChanges);
}
}
 
// This code is contributed by chitranayal

Python3




# Python3 program to implement the
# above approach
 
# Function for counting changes
def countChanges(matrix, n, m):
 
    # Maximum distance possible
    # is (n - 1 + m - 1)
    dist = n + m - 1
 
    # Stores the maximum element
    Max_element = 0
    for i in range(n):
        for j in range(m):
             
            # Update the maximum
            Max_element = max(Max_element,
                              matrix[i][j])
 
    # Stores frequencies of
    # values for respective
    # distances
    freq = [[0 for i in range(Max_element + 1)]
               for j in range(dist)]
 
    # Initialize frequencies of
    # cells as 0
    for i in range(dist):
        for j in range(Max_element + 1):
            freq[i][j] = 0
 
    # Count frequencies of cell
    # values in the matrix
    for i in range(n):
        for j in range(m):
 
            # Increment frequency of
            # value at distance i+j
            freq[i + j][matrix[i][j]] += 1
 
    min_changes_sum = 0
    for i in range(dist // 2):
 
        # Store the most frequent
        # value at i-th distance
        # from (0, 0) and (N - 1, M - 1)
        maximum = 0
        total_values = 0
 
        # Calculate max frequency
        # and total cells at distance i
        for j in range(Max_element + 1):
            maximum = max(maximum,
                          freq[i][j] +
                          freq[n + m - 2 - i][j])
 
            total_values += (freq[i][j] +
                             freq[n + m - 2 - i][j])
 
        # Count changes required
        # to convert all cells
        # at i-th distance to
        # most frequent value
        min_changes_sum += total_values - maximum
 
    return min_changes_sum
 
# Driver code
if __name__ == '__main__':
 
    mat = [ [ 7, 0, 3, 1, 8, 1, 3 ],
            [ 0, 4, 0, 1, 0, 4, 0 ],
            [ 3, 1, 8, 3, 1, 0, 7 ] ]
 
    minChanges = countChanges(mat, 3, 7)
 
    print(minChanges)
 
# This code is contributed by Shivam Singh

C#




// C# program to implement the
// above approach
using System;
class GFG{
   
//static int N = 7;
 
// Function for counting changes
static int countChanges(int [,]matrix,
                        int n, int m)
{
     
    // Maximum distance possible
    // is (n - 1 + m - 1)
    int i, j, dist = n + m - 1;
 
    // Stores the maximum element
    int Max_element = 0;
    for(i = 0; i < n; i++)
    {
        for(j = 0; j < m; j++)
        {
             
            // Update the maximum
            Max_element = Math.Max(Max_element,
                                   matrix[i, j]);
        }
    }
 
    // Stores frequencies of
    // values for respective
    // distances
    int [,]freq = new int[dist, Max_element + 1];
 
    // Initialize frequencies of
    // cells as 0
    for(i = 0; i < dist; i++)
    {
        for(j = 0; j < Max_element + 1;
            j++)
            freq[i, j] = 0;
    }
 
    // Count frequencies of cell
    // values in the matrix
    for(i = 0; i < n; i++)
    {
        for(j = 0; j < m; j++)
        {
             
            // Increment frequency of
            // value at distance i+j
            freq[i + j, matrix[i, j]]++;
        }
    }
     
    int min_changes_sum = 0;
    for(i = 0; i < dist / 2; i++)
    {
         
        // Store the most frequent
        // value at i-th distance
        // from (0, 0) and (N - 1, M - 1)
        int maximum = 0;
        int total_values = 0;
 
        // Calculate max frequency
        // and total cells at distance i
        for(j = 0; j < Max_element + 1; j++)
        {
            maximum = Math.Max(maximum,
                               freq[i, j] +
                               freq[n + m -
                                    2 - i, j]);
            total_values += freq[i, j] +
                            freq[n + m -
                                 2 - i, j];
        }
 
        // Count changes required
        // to convert all cells
        // at i-th distance to
        // most frequent value
        min_changes_sum += total_values -
                           maximum;
    }
    return min_changes_sum;
}
 
// Driver Code
public static void Main(String []args)
{
    int [,]mat = { { 7, 0, 3, 1, 8, 1, 3 },
                    { 0, 4, 0, 1, 0, 4, 0 },
                    { 3, 1, 8, 3, 1, 0, 7 } };
 
    int minChanges = countChanges(mat, 3, 7);
 
    Console.Write(minChanges);
}
}
 
// This code is contributed by sapnasingh4991
Output:
6



Time Complexity: O(N * M) 
Auxiliary Space: O((N + M)*maxm), where maxm is the maximum element present in the matrix.

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