# Minimum steps to convert all paths in matrix from top left to bottom right as palindromic paths | Set 2

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: M = 2, N = 2, mat[M][N] = {{0, 0}, {0, 1}}
Output:
Explanation:
Change matrix from 0 to 1. The two paths from (0, 0) to (1, 1) become palindromic.

Input: M = 3, N = 7, mat[M][N] = {{1, 2, 3, 4, 5, 6, 7}, {2, 2, 3, 3, 4, 3, 2}, {1, 2, 3, 2, 5, 6, 4}}
Output: 10

Time Complexity: O(N^3)
Auxiliary Space: O(N)

Efficient Approach:
The following observations have to be made:

• A unique diagonal exists for every value (i+j) where i is the row index and j is the column index.
• The task reduces to selecting two diagonals, which are at the same distance from cell (0, 0) and cell (M – 1, N – 1) respectively and making all their elements equal to a single number, which is repeated most number of times in both the chosen diagonals.
• If the number of elements between cell (0, 0) and (M – 1, N – 1) are odd, then a common diagonal equidistant from both the cells exists. Elements of that diagonal need not be modified as they do not affect the palindromic arrangement of a path.
• If the numbers of cells between (0, 0) and (M – 1, N – 1) are even, no such common diagonal exists.

Follow the below steps to solve the problem:

1. Create a 2D array frequency_diagonal that stores the frequency of all numbers in each chosen diagonal.
2. Each diagonal can be uniquely represented as the sum of (i, j).
3. Initialise a count variable that stores the count of the total number of cells where the values have to be replaced.
4. Iterate over the mat[][] and increment the frequency of current element in the diagonal ((i + j) value) where it belongs to.
5. Initialize a variable number_of_elements to 1, which stores the number of elements in each of the currently chosen pair of diagonals.
6. Initialize start = 0 and end = M + N – 2 and repeat the steps below until start < end
• Find the frequency of the number which appears maximum times in the two selected diagonals that are equidistant from (0, 0) and (M-1, N-1).
• Let the frequency in the above step be X. Add the value (total number of elements in the two diagonals – X) to count the minimum number of changes.
• Increment start by 1 and decrement end by 1 and if the number of elements in the current diagonal is less than the maximum possible elements in any diagonal of the matrix, then increment number_of_elements by 1.
7. Print the value of total count after the above steps.

Below is the implementation of the above approach:

## C++

 `// C++ program of the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate the minimum ` `// number of replacements ` `int` `MinReplacements( ` `    ``int` `M, ``int` `N, ``int` `mat[]) ` `{ ` ` `  `    ``// 2D array to store frequency ` `    ``// of all the numners in ` `    ``// each diaginal ` ` `  `    ``int` `frequency_diagonal; ` ` `  `    ``// Initialise all the elements ` `    ``// of 2D array with 0 ` `    ``memset``(frequency_diagonal, 0, ` `           ``sizeof``(frequency_diagonal)); ` ` `  `    ``// Initialise answer as 0 ` `    ``int` `answer = 0; ` ` `  `    ``// Iterate over the matrix ` `    ``for` `(``int` `i = 0; i < M; i++) { ` `        ``for` `(``int` `j = 0; j < N; j++) { ` ` `  `            ``// Update the frequency of ` `            ``// number mat[i][j] ` `            ``// for the diagonal ` `            ``// identified by (i+j) ` `            ``frequency_diagonal[i + j] ` `                              ``[mat[i][j]]++; ` `        ``} ` `    ``} ` ` `  `    ``// Initialize start as 0 ` `    ``// which indicates the ` `    ``// first diagonal ` `    ``int` `start = 0; ` ` `  `    ``// Initialize end as ` `    ``// M + N - 2 which indicates ` `    ``// the last diagonal ` `    ``int` `end = M + N - 2; ` ` `  `    ``// Number of elements in ` `    ``// the current diagonal ` `    ``int` `no_of_elemnts = 1; ` ` `  `    ``// Maximum possible number ` `    ``// of elements in a diagonal ` `    ``// can be minimum of (number of ` `    ``// rows and number of columns) ` `    ``int` `max_elements = min(M, N); ` ` `  `    ``while` `(start < end) { ` ` `  `        ``// The frequecny of number ` `        ``// which occurs for the ` `        ``// maximum number of times ` `        ``// in the two selected ` `        ``// diagonals ` `        ``int` `X = INT_MIN; ` ` `  `        ``for` `(``int` `i = 0; i <= 10000; i++) { ` `            ``X = max( ` `                ``X, ` `                ``frequency_diagonal[start][i] ` `                    ``+ frequency_diagonal[end][i]); ` `        ``} ` ` `  `        ``answer = answer + (2 * (no_of_elemnts)) - X; ` ` `  `        ``// Increment start ` `        ``start++; ` `        ``// Decrement end ` `        ``end--; ` ` `  `        ``// Increment current number ` `        ``// of elements until it reaches ` `        ``// the maximum possible value ` `        ``if` `(no_of_elemnts < max_elements) ` `            ``no_of_elemnts++; ` `    ``} ` ` `  `    ``// return the final answer ` `    ``return` `answer; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Number of rows ` `    ``int` `M = 3; ` ` `  `    ``// Number of columns ` `    ``int` `N = 7; ` ` `  `    ``int` `mat ` `        ``= { { 1, 2, 3, 4, 5, 6, 7 }, ` `            ``{ 2, 2, 3, 3, 4, 3, 2 }, ` `            ``{ 1, 2, 3, 2, 5, 6, 4 } }; ` ` `  `    ``cout << MinReplacements(M, N, mat) ` `         ``<< endl; ` `} `

## Java

 `// Java program of the above approach ` `class` `GFG{ ` ` `  `// Function to calculate the minimum ` `// number of replacements ` `static` `int` `MinReplacements(``int` `M, ``int` `N, ` `                           ``int` `mat[][]) ` `{ ` `     `  `    ``// 2D array to store frequency ` `    ``// of all the numners in ` `    ``// each diaginal ` `    ``int` `[][]frequency_diagonal = ``new` `int``[``100``][``10005``]; ` ` `  `    ``// Initialise answer as 0 ` `    ``int` `answer = ``0``; ` ` `  `    ``// Iterate over the matrix ` `    ``for``(``int` `i = ``0``; i < M; i++) ` `    ``{ ` `        ``for``(``int` `j = ``0``; j < N; j++)  ` `        ``{ ` ` `  `            ``// Update the frequency of ` `            ``// number mat[i][j] ` `            ``// for the diagonal ` `            ``// identified by (i+j) ` `            ``frequency_diagonal[i + j][mat[i][j]]++; ` `        ``} ` `    ``} ` ` `  `    ``// Initialize start as 0 ` `    ``// which indicates the ` `    ``// first diagonal ` `    ``int` `start = ``0``; ` ` `  `    ``// Initialize end as ` `    ``// M + N - 2 which indicates ` `    ``// the last diagonal ` `    ``int` `end = M + N - ``2``; ` ` `  `    ``// Number of elements in ` `    ``// the current diagonal ` `    ``int` `no_of_elemnts = ``1``; ` ` `  `    ``// Maximum possible number ` `    ``// of elements in a diagonal ` `    ``// can be minimum of (number of ` `    ``// rows and number of columns) ` `    ``int` `max_elements = Math.min(M, N); ` ` `  `    ``while` `(start < end) ` `    ``{ ` ` `  `        ``// The frequecny of number ` `        ``// which occurs for the ` `        ``// maximum number of times ` `        ``// in the two selected ` `        ``// diagonals ` `        ``int` `X = Integer.MIN_VALUE; ` ` `  `        ``for``(``int` `i = ``0``; i <= ``10000``; i++) ` `        ``{ ` `            ``X = Math.max(X, ` `                ``frequency_diagonal[start][i] + ` `                ``frequency_diagonal[end][i]); ` `        ``} ` ` `  `        ``answer = answer + (``2` `* (no_of_elemnts)) - X; ` ` `  `        ``// Increment start ` `        ``start++; ` `         `  `        ``// Decrement end ` `        ``end--; ` ` `  `        ``// Increment current number ` `        ``// of elements until it reaches ` `        ``// the maximum possible value ` `        ``if` `(no_of_elemnts < max_elements) ` `            ``no_of_elemnts++; ` `    ``} ` ` `  `    ``// return the final answer ` `    ``return` `answer; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `     `  `    ``// Number of rows ` `    ``int` `M = ``3``; ` ` `  `    ``// Number of columns ` `    ``int` `N = ``7``; ` ` `  `    ``int` `mat[][] = { { ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7` `}, ` `                    ``{ ``2``, ``2``, ``3``, ``3``, ``4``, ``3``, ``2` `}, ` `                    ``{ ``1``, ``2``, ``3``, ``2``, ``5``, ``6``, ``4` `} }; ` ` `  `    ``System.out.print(MinReplacements(M, N, mat) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 program of the above approach ` `import` `sys ` ` `  `# Function to calculate the minimum ` `# number of replacements ` `def` `MinReplacements(M, N, mat): ` ` `  `    ``# 2D array to store frequency ` `    ``# of all the numners in ` `    ``# each diaginal ` `    ``frequency_diagonal ``=` `[[``0` `for` `x ``in` `range``(``10005``)] ` `                             ``for` `y ``in` `range` `(``100``)]; ` ` `  `    ``# Initialise answer as 0 ` `    ``answer ``=` `0` ` `  `    ``# Iterate over the matrix ` `    ``for` `i ``in` `range``(M): ` `        ``for` `j ``in` `range``(N): ` ` `  `            ``# Update the frequency of ` `            ``# number mat[i][j] ` `            ``# for the diagonal ` `            ``# identified by (i+j) ` `            ``frequency_diagonal[i ``+` `j][mat[i][j]] ``+``=` `1` ` `  `    ``# Initialize start as 0 ` `    ``# which indicates the ` `    ``# first diagonal ` `    ``start ``=` `0` ` `  `    ``# Initialize end as ` `    ``# M + N - 2 which indicates ` `    ``# the last diagonal ` `    ``end ``=` `M ``+` `N ``-` `2` ` `  `    ``# Number of elements in ` `    ``# the current diagonal ` `    ``no_of_elemnts ``=` `1` ` `  `    ``# Maximum possible number ` `    ``# of elements in a diagonal ` `    ``# can be minimum of (number of ` `    ``# rows and number of columns) ` `    ``max_elements ``=` `min``(M, N) ` ` `  `    ``while` `(start < end): ` ` `  `        ``# The frequecny of number ` `        ``# which occurs for the ` `        ``# maximum number of times ` `        ``# in the two selected ` `        ``# diagonals ` `        ``X ``=` `-``sys.maxsize ``-` `1` ` `  `        ``for` `i ``in` `range``(``10001``): ` `            ``X ``=` `max``(X, ` `                    ``frequency_diagonal[start][i] ``+` `                    ``frequency_diagonal[end][i]) ` `         `  `        ``answer ``=` `answer ``+` `(``2` `*` `(no_of_elemnts)) ``-` `X ` ` `  `        ``# Increment start ` `        ``start ``+``=` `1` `         `  `        ``# Decrement end ` `        ``end ``-``=` `1` ` `  `        ``# Increment current number ` `        ``# of elements until it reaches ` `        ``# the maximum possible value ` `        ``if` `(no_of_elemnts < max_elements): ` `            ``no_of_elemnts ``+``=` `1` ` `  `    ``# Return the final answer ` `    ``return` `answer ` ` `  `# Driver Code ` ` `  `# Number of rows ` `M ``=` `3` ` `  `# Number of columns ` `N ``=` `7` ` `  `mat ``=` `[ [ ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7` `], ` `        ``[ ``2``, ``2``, ``3``, ``3``, ``4``, ``3``, ``2` `], ` `        ``[ ``1``, ``2``, ``3``, ``2``, ``5``, ``6``, ``4` `] ] ` ` `  `print``(MinReplacements(M, N, mat)) ` ` `  `# This code is contributed by chitranayal `

## C#

 `// C# program of the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to calculate the minimum ` `// number of replacements ` `static` `int` `MinReplacements(``int` `M, ``int` `N, ` `                           ``int` `[,]mat) ` `{ ` `     `  `    ``// 2D array to store frequency ` `    ``// of all the numners in ` `    ``// each diaginal ` `    ``int` `[,]frequency_diagonal = ``new` `int``[100, 10005]; ` ` `  `    ``// Initialise answer as 0 ` `    ``int` `answer = 0; ` ` `  `    ``// Iterate over the matrix ` `    ``for``(``int` `i = 0; i < M; i++) ` `    ``{ ` `        ``for``(``int` `j = 0; j < N; j++)  ` `        ``{ ` ` `  `            ``// Update the frequency of ` `            ``// number mat[i,j] ` `            ``// for the diagonal ` `            ``// identified by (i+j) ` `            ``frequency_diagonal[i + j, mat[i, j]]++; ` `        ``} ` `    ``} ` ` `  `    ``// Initialize start as 0 ` `    ``// which indicates the ` `    ``// first diagonal ` `    ``int` `start = 0; ` ` `  `    ``// Initialize end as ` `    ``// M + N - 2 which indicates ` `    ``// the last diagonal ` `    ``int` `end = M + N - 2; ` ` `  `    ``// Number of elements in ` `    ``// the current diagonal ` `    ``int` `no_of_elemnts = 1; ` ` `  `    ``// Maximum possible number ` `    ``// of elements in a diagonal ` `    ``// can be minimum of (number of ` `    ``// rows and number of columns) ` `    ``int` `max_elements = Math.Min(M, N); ` ` `  `    ``while` `(start < end) ` `    ``{ ` ` `  `        ``// The frequecny of number ` `        ``// which occurs for the ` `        ``// maximum number of times ` `        ``// in the two selected ` `        ``// diagonals ` `        ``int` `X = ``int``.MinValue; ` ` `  `        ``for``(``int` `i = 0; i <= 10000; i++) ` `        ``{ ` `            ``X = Math.Max(X, ` `                ``frequency_diagonal[start, i] + ` `                ``frequency_diagonal[end, i]); ` `        ``} ` ` `  `        ``answer = answer + (2 * (no_of_elemnts)) - X; ` ` `  `        ``// Increment start ` `        ``start++; ` `         `  `        ``// Decrement end ` `        ``end--; ` ` `  `        ``// Increment current number ` `        ``// of elements until it reaches ` `        ``// the maximum possible value ` `        ``if` `(no_of_elemnts < max_elements) ` `            ``no_of_elemnts++; ` `    ``} ` ` `  `    ``// Return the readonly answer ` `    ``return` `answer; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `     `  `    ``// Number of rows ` `    ``int` `M = 3; ` ` `  `    ``// Number of columns ` `    ``int` `N = 7; ` ` `  `    ``int` `[,]mat = { { 1, 2, 3, 4, 5, 6, 7 }, ` `                   ``{ 2, 2, 3, 3, 4, 3, 2 }, ` `                   ``{ 1, 2, 3, 2, 5, 6, 4 } }; ` ` `  `    ``Console.Write(MinReplacements(M, N, mat) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by amal kumar choubey  `

Output:

```10
```

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

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