Minimum steps to convert all top left to bottom right paths in Matrix as palindrome | Set 2

• Last Updated : 14 Jun, 2021

Given a matrix mat[][] with N rows and M columns. The task is to find the minimum number of changes required in the matrix such that every path from top left to bottom right is a palindromic path. In a path only right and bottom movements are allowed from one cell to another cell.

Examples:

Input: mat[][] = {{1, 2}, {3, 1}}
Output: 0
Explanation:
Every path in the matrix from top left to bottom right is palindromic.
Paths => {1, 2, 1}, {1, 3, 1}

Input: mat[][] = {{1, 2}, {3, 5}}
Output: 1
Explanation:
Only one change is required for the every path to be palindromic.
That is => mat = 1
Paths => {1, 2, 1}, {1, 3, 1}

Naive Approach: For the naive approach please refer to this post.

Efficient Approach: The idea is to discard the use of an extra space that is the use of HashMap. Follow the steps given below:

1. Distance possible from top left and bottom right are in the range 0 to N + M – 2. Hence create a 2D array of dimensions [N + M – 1].
2. Store frequency of distances in an array while considering Row number (in range 0 to N + M – 2) as distance and column number (0 to 9) as an element in the given matrix.
3. For the number of changes to be minimum, change each cell at distance X with a value that has the highest frequency among all values at distance X.
4. The minimum number of steps required is the sum of the difference of total values of frequency and the maximum value of frequency for each distance.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach#include using namespace std;#define N 7 // Function for counting minimum// number of changesint countChanges(int matrix[][N],                 int n, int m){    // Distance of elements from (0, 0)    // will is i range [0, n + m - 2]    int dist = n + m - 1;     // Store frequencies of [0, 9]    // at distance i    int freq[dist];     // Initialize frequencies as 0    for (int i = 0; i < dist; i++) {        for (int j = 0; j < 10; j++)            freq[i][j] = 0;    }     // Count frequencies of [0, 9]    for (int i = 0; i < n; i++) {         for (int j = 0; j < m; j++) {             // Increment frequency of            // value matrix[i][j]            // at distance i+j            freq[i + j][matrix[i][j]]++;        }    }     int min_changes_sum = 0;    for (int i = 0; i < dist / 2; i++) {         int maximum = 0;        int total_values = 0;         // Find value with max frequency        // and count total cells at distance i        // from front end and rear end        for (int j = 0; j < 10; j++) {             maximum = max(maximum, freq[i][j]                    + freq[n + m - 2 - i][j]);             total_values += (freq[i][j]                   + freq[n + m - 2 - i][j]);        }         // Change all values to the        // value with max frequency        min_changes_sum += (total_values                            - maximum);    }     // Return the answer    return min_changes_sum;} // Driver Codeint main(){    // Given Matrix    int mat[][N] = { { 1, 2 }, { 3, 5 } };     // Function Call    cout << countChanges(mat, 2, 2);    return 0;}

Java

 // Java program for the above approachimport java.util.*; class GFG{     static final int N = 7; // Function for counting minimum// number of changesstatic int countChanges(int matrix[][],                        int n, int m){         // Distance of elements from (0, 0)    // will is i range [0, n + m - 2]    int dist = n + m - 1;     // Store frequencies of [0, 9]    // at distance i    int [][]freq = new int[dist];     // Initialize frequencies as 0    for(int i = 0; i < dist; i++)    {        for(int j = 0; j < 10; j++)            freq[i][j] = 0;    }     // Count frequencies of [0, 9]    for(int i = 0; i < n; i++)    {        for(int j = 0; j < m; j++)        {                         // Increment frequency of            // value matrix[i][j]            // at distance i+j            freq[i + j][matrix[i][j]]++;        }    }     int min_changes_sum = 0;    for(int i = 0; i < dist / 2; i++)    {        int maximum = 0;        int total_values = 0;         // Find value with max frequency        // and count total cells at distance i        // from front end and rear end        for(int j = 0; j < 10; 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]);        }                 // Change all values to the        // value with max frequency        min_changes_sum += (total_values -                            maximum);    }     // Return the answer    return min_changes_sum;} // Driver Codepublic static void main(String[] args){         // Given matrix    int mat[][] = { { 1, 2 }, { 3, 5 } };     // Function call    System.out.print(countChanges(mat, 2, 2));}} // This code is contributed by Rohit_ranjan

Python3

 # Python3 program for the above approach # Function for counting minimum# number of changesdef countChanges(matrix, n, m):     # Distance of elements from (0, 0)    # will is i range [0, n + m - 2]    dist = n + m - 1     # Store frequencies of [0, 9]    # at distance i    # Initialize all with zero    freq = [ * 10 for i in range(dist)]     # Count frequencies of [0, 9]    for i in range(n):        for j in range(m):             # Increment frequency of            # value matrix[i][j]            # at distance i+j            freq[i + j][matrix[i][j]] += 1     min_changes_sum = 0     for i in range(dist // 2):        maximum = 0        total_values = 0         # Find value with max frequency        # and count total cells at distance i        # from front end and rear end                for j in range(10):            maximum = max(maximum, freq[i][j] +                       freq[n + m - 2 - i][j])             total_values += (freq[i][j] +                 freq[n + m - 2 - i][j])         # Change all values to the value        # with max frequency        min_changes_sum += (total_values -                            maximum)                                 # Return the answer    return min_changes_sum # Driver codeif __name__ == '__main__':     # Given matrix    mat = [ [ 1, 2 ], [ 3, 5 ] ]     # Function call    print(countChanges(mat, 2, 2)) # This code is contributed by himanshu77

C#

 // C# program for the above approachusing System; class GFG{     //static readonly int N = 7; // Function for counting minimum// number of changesstatic int countChanges(int [,]matrix,                        int n, int m){         // Distance of elements from (0, 0)    // will is i range [0, n + m - 2]    int dist = n + m - 1;     // Store frequencies of [0, 9]    // at distance i    int [,]freq = new int[dist, 10];     // Initialize frequencies as 0    for(int i = 0; i < dist; i++)    {        for(int j = 0; j < 10; j++)            freq[i, j] = 0;    }     // Count frequencies of [0, 9]    for(int i = 0; i < n; i++)    {        for(int j = 0; j < m; j++)        {                         // Increment frequency of            // value matrix[i,j]            // at distance i+j            freq[i + j, matrix[i, j]]++;        }    }     int min_changes_sum = 0;    for(int i = 0; i < dist / 2; i++)    {        int maximum = 0;        int total_values = 0;         // Find value with max frequency        // and count total cells at distance i        // from front end and rear end        for(int j = 0; j < 10; 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]);        }                 // Change all values to the        // value with max frequency        min_changes_sum += (total_values -                            maximum);    }     // Return the answer    return min_changes_sum;} // Driver Codepublic static void Main(String[] args){         // Given matrix    int [,]mat = { { 1, 2 }, { 3, 5 } };     // Function call    Console.Write(countChanges(mat, 2, 2));}} // This code is contributed by Rohit_ranjan

Javascript


Output:
1

Time Complexity: O(N*M)
Auxiliary Space: O(N*M)

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