Minimum steps to convert all paths in matrix from top left to bottom right as palindromic paths

• Last Updated : 03 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:
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:
Explanation:
Only one change is required for the every path to be palindromic.
That is => mat = 1
Paths => {1, 2, 1}, {1, 3, 1}

Simple Approach:

The key observation in the problem is that elements at the same distance from the front end or rear end are equal. Therefore, find all the elements at equal distance from (0, 0) and (N-1, M-1) and then make all of them equal in a minimum number of changes. Maintain a count variable to get the total number of changes. Below is the illustration of the approach:

• The distance possible from the top left and bottom right is 0 to N + M – 2.
• Maintain two-pointers one at the top left that is the distance at 0 and another at N + M – 2.
• Iterate over the matrix and for all distance maintain an hash-map of the elements of the matrix at the current distance.
• Update the matrix elements with the minimum number of changes required.
• Finally, increment the left distance by 1 and decrement the right distance by 1.

Below is the implementation of the above approach:

C++

 // C++ implementation to find the// minimum number of changes required// such that every path from top left// to the bottom right// are palindromic paths #include using namespace std;#define M 3#define N 3 // Function to find the minimum number// of the changes required for the// every path to be palindromicint minchanges(int mat[N][M]){    // count variable for    // maintaining total changes.    int count = 0;     // left and right variables for    // keeping distance values    // from cell(0, 0) and    // (N-1, M-1) respectively.    int left = 0, right = N + M - 2;     while (left < right) {         unordered_map mp;        int totalsize = 0;         // Iterating over the matrix        for (int i = 0; i < N; i++) {            for (int j = 0; j < M; j++) {                if (i + j == left) {                    mp[mat[i][j]]++;                    totalsize++;                }                else if (i + j == right) {                    mp[mat[i][j]]++;                    totalsize++;                }            }        }         // Finding minimum number        // of changes required.        unordered_map::iterator itr = mp.begin();        int changes = 0;        for (; itr != mp.end(); itr++)            changes = max(changes, itr->second);         // Minimum no. of changes will        // be the the minimum no.        // of different values and        // we will assume to        // make them equals to value        // with maximum frequency element        count += totalsize - changes;         // Moving ahead with        // greater distance        left++;        right--;    }    return count;} // Drive Codeint main(){    int mat[][M]        = { { 1, 4, 1 }, { 2, 5, 3 }, { 1, 3, 1 } };     // Function Call    cout << minchanges(mat);    return 0;}

Java

 // Java implementation to find the// minimum number of changes required// such that every path from top left// to the bottom right are palindromic// pathsimport java.io.*;import java.util.*; class GFG {     static final int M = 3;    static final int N = 3;     // Function to find the minimum number    // of the changes required for the    // every path to be palindromic    static int minchanges(int[][] mat)    {         // count variable for        // maintaining total changes.        int count = 0;         // left and right variables for        // keeping distance values        // from cell(0, 0) and        // (N-1, M-1) respectively.        int left = 0, right = N + M - 2;         while (left < right) {            Map mp = new HashMap<>();             int totalsize = 0;             // Iterating over the matrix            for (int i = 0; i < N; i++) {                for (int j = 0; j < M; j++) {                    if (i + j == left) {                        mp.put(mat[i][j],                               mp.getOrDefault(mat[i][j], 0)                                   + 1);                        totalsize++;                    }                    else if (i + j == right) {                        mp.put(mat[i][j],                               mp.getOrDefault(mat[i][j], 0)                                   + 1);                        totalsize++;                    }                }            }             // Finding minimum number            // of changes required.            int changes = 0;            for (Map.Entry itr :                 mp.entrySet())                changes = Math.max(changes, itr.getValue());             // Minimum no. of changes will            // be the the minimum no.            // of different values and            // we will assume to            // make them equals to value            // with maximum frequency element            count += totalsize - changes;             // Moving ahead with            // greater distance            left++;            right--;        }        return count;    }     // Driver Code    public static void main(String[] args)    {        int mat[][]            = { { 1, 4, 1 }, { 2, 5, 3 }, { 1, 3, 1 } };            // Function Call        System.out.println(minchanges(mat));    }} // This code is contributed by offbeat

Python3

 # Python3 implementation to find the# minimum number of changes required# such that every path from top left# to the bottom right# are palindromic pathsM = 3N = 3 # Function to find the minimum number# of the changes required for the# every path to be palindromic  def minchanges(mat):         # count variable for    # maintaining total changes.    count = 0     # left and right variables for    # keeping distance values    # from cell(0, 0) and    # (N-1, M-1) respectively.    left = 0    right = N + M - 2     while (left < right):        mp = {}        totalsize = 0         # Iterating over the matrix        for i in range(N):            for j in range(M):                if (i + j == left):                    mp[mat[i][j]] =                    mp.get(mat[i][j], 0) + 1                    totalsize += 1                elif (i + j == right):                    mp[mat[i][j]] =                    mp.get(mat[i][j], 0) + 1                    totalsize += 1         # Finding minimum number        # of changes required.        changes = 0        for itr in mp:            changes = max(changes, mp[itr])         # Minimum no. of changes will        # be the the minimum no.        # of different values and        # we will assume to        # make them equals to value        # with maximum frequency element        count += totalsize - changes         # Moving ahead with        # greater distance        left += 1        right -= 1    return count  # Driver Codeif __name__ == '__main__':    mat = [[1, 4, 1],           [2, 5, 3],           [1, 3, 1]]         # Function Call    print(minchanges(mat)) # This code is contributed by Mohit Kumar 29

C#

 // C# implementation to find the// minimum number of changes required// such that every path from top left// to the bottom right are palindromic// pathsusing System;using System.Collections;using System.Collections.Generic; class GFG {     static int M = 3;    static int N = 3;     // Function to find the minimum number    // of the changes required for the    // every path to be palindromic    static int minchanges(int[, ] mat)    {         // count variable for        // maintaining total changes.        int count = 0;         // left and right variables for        // keeping distance values        // from cell(0, 0) and        // (N-1, M-1) respectively.        int left = 0, right = N + M - 2;         while (left < right) {            Dictionary mp                = new Dictionary();            int totalsize = 0;             // Iterating over the matrix            for (int i = 0; i < N; i++) {                for (int j = 0; j < M; j++) {                    if (i + j == left) {                        if (mp.ContainsKey(mat[i, j])) {                            mp[mat[i, j]]++;                        }                        else {                            mp[mat[i, j]] = 1;                        }                        totalsize++;                    }                    else if (i + j == right) {                        if (mp.ContainsKey(mat[i, j])) {                            mp[mat[i, j]]++;                        }                        else {                            mp[mat[i, j]] = 1;                        }                        totalsize++;                    }                }            }             // Finding minimum number            // of changes required.            int changes = 0;            foreach(KeyValuePair itr in mp)            {                changes = Math.Max(changes, itr.Value);            }             // Minimum no. of changes will            // be the the minimum no.            // of different values and            // we will assume to            // make them equals to value            // with maximum frequency element            count += totalsize - changes;             // Moving ahead with            // greater distance            left++;            right--;        }        return count;    }     // Driver Code    public static void Main(string[] args)    {        int[, ] mat            = { { 1, 4, 1 }, { 2, 5, 3 }, { 1, 3, 1 } };         // Function Call        Console.Write(minchanges(mat));    }} // This code is contributed by rutvik_56

Javascript


Output:
2

Performance Analysis:

• Time Complexity: O(N3)
• Auxiliary Space: O(N)

Efficient Approach:

Algorithm:

• We will use a hashmap to count the frequency of every element which are the same distance from the top and bottom.
• We will use a 2D map for this where the key will be index and the value will be another map which will count frequency
• After counting frequency we will iterate l=0 to r=m+n-1 while l<r and each time we will count the sum and find the max frequency value ≥f
• We will replace (sum-f) elements with the element whose frequency is maximum and store result+=(sum-f)
• Print result

Below is the implementation of the above approach:

C++

 // C++ Program to count minimum change// required to convert all the paths// pallindromic from top left to// right bottom cell.#include using namespace std; // Function to find the minimum number// of the changes required for the// every path to be palindromicint minchanges(vector >& a){    int res = 0; // use to store final result     // Row and column    int N = a.size(), M = a.size();     // mp_key -> (i+j) , mp_value -> nw_map    // nw_map_key -> elements_of_matrix    // nw_map_value -> frequency of elements     // 2-D map    map > mp;     for (int i = 0; i < N; i++) {        for (int j = 0; j < M; j++) {                         // calculating position            int ind = i + j;                         // increase the frequency of a[i][j]            // at position ind            mp[ind][a[i][j]]++;        }    }         // Define left and right limit    int r = M + N - 2, l = 0;    while (l < r) {                 // s-> count total number of elements        // at index l and r        // mx-> store maximum frequency of any element        int s = 0, mx = 0;         // store all elements frequency at index l        for (auto x : mp[r]) {            mp[l][x.first] += x.second;        }         // Count total elements and mx->max_frequency        for (auto x : mp[l]) {            s += x.second;            mx = max(x.second, mx);        }         // We will replace (s-mx) elements with        // the element whose frequency is mx        res += (s - mx);        l++;        r--;    }       // return res    return res;} // Driver Codeint main(){    // Function Call    vector > mat        = { { 1, 4, 1 }, { 2, 5, 3 }, { 1, 3, 1 } };    cout << "Total number of changes requires "         << minchanges(mat) << "\n";     // Function Call    vector > mat1        = { { 1, 4 }, { 2, 5 }, { 1, 3 }, { 2, 5 } };    cout << "Total number of changes requires "         << minchanges(mat1) << "\n";       return 0;} // This code is contributed by ajaykr00kj
Output
Total number of changes requires 2
Total number of changes requires 3

Time Complexity: O(m*n)

Space Complexity: O(m*n)

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