将矩阵中所有路径从左上角到右下角转换为回文路径的最小步骤

给定一个矩阵mat[][]有 N 行和 M 列。任务是找到矩阵中所需的最小更改次数，使得从左上角到右下角的每条路径都是回文路径。在路径中，只允许从一个单元格到另一个单元格的向右和底部移动。
例子：

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][1] = 1
Paths => {1, 2, 1}, {1, 3, 1}

编程需要懂一点英语

简单的方法：

问题中的关键观察是与前端或后端相同距离的元素是相等的。因此，找到与 (0, 0) 和 (N-1, M-1) 等距的所有元素，然后以最少的更改次数使所有元素相等。维护一个计数变量以获取更改的总数。下面是该方法的说明：

从左上角和右下角可能的距离是 0 到 N + M – 2。
保持两个指针，一个在左上角，即 0 处的距离，另一个在 N + M – 2 处。
迭代矩阵，并为所有距离维护当前距离的矩阵元素的哈希图。
使用所需的最少更改次数更新矩阵元素。
最后，将左侧距离增加 1，将右侧距离减少 1。

下面是上述方法的实现：

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 palindromic
int 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 Code
int 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
// paths
import 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 paths
M = 3
N = 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 Code
if __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
// paths
using 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

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 palindromic
int minchanges(vector >& a)
{
    int res = 0; // use to store final result
 
    // Row and column
    int N = a.size(), M = a[0].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 Code
int 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

输出：

性能分析：

时间复杂度： O(N ³ )
辅助空间： O(N)

有效的方法：

算法：

我们将使用哈希图来计算与顶部和底部距离相同的每个元素的频率。
我们将为此使用 2D 地图，其中键将是索引，值将是另一个将计算频率的地图
在计算频率后，我们将迭代 l=0 到 r=m+n-1 而 l
我们将(sum-f) 元素替换为频率最大的元素并存储结果+=(sum-f)
打印结果

下面是上述方法的实现：

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 palindromic
int minchanges(vector >& a)
{
    int res = 0; // use to store final result
 
    // Row and column
    int N = a.size(), M = a[0].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 Code
int 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

输出

Total number of changes requires 2
Total number of changes requires 3

时间复杂度： O(m*n)

空间复杂度： O(m*n)

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