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}
朴素的方法：关于朴素的方法，请参考这篇文章。
高效的方法：这个想法是丢弃使用 HashMap 的额外空间。请按照以下步骤操作：
- 从左上角和右下角可能的距离在0 到 N + M – 2的范围内。因此，创建一个维度为[N + M – 1][10]的二维数组。
- 将距离的频率存储在数组中，同时将行号(范围 0 到 N + M – 2)视为距离，将列号(0 到 9)视为给定矩阵中的元素。
- 用于改变的次数为最小，更改在距离X的每个细胞与具有距离X的所有值中的最高频率的值。
- 所需的最小步数是每个距离的频率总值与频率最大值的差值之和。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
#define N 7
// Function for counting minimum
// number of changes
int 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][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 = 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 Code
int 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 approach
import java.util.*;
class GFG{
static final int N = 7;
// Function for counting minimum
// number of changes
static 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 Code
public 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_ranjanPython3
# Python3 program for the above approach
# Function for counting minimum
# number of changes
def 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 = [[0] * 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 code
if __name__ == '__main__':
# Given matrix
mat = [ [ 1, 2 ], [ 3, 5 ] ]
# Function call
print(countChanges(mat, 2, 2))
# This code is contributed by himanshu77C#
// C# program for the above approach
using System;
class GFG{
//static readonly int N = 7;
// Function for counting minimum
// number of changes
static 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 Code
public 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输出：
1
时间复杂度： O(N*M)
辅助空间： O(N*M)
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