给定一个具有N行和M列的矩阵，任务是通过单元格值的最小变化来制作从单元格(N, M)到(1, 1)回文的所有可能路径。
Possible moves from any cell (x, y) is either move Left(x – 1, y) or move Down (x, y – 1).
例子：
Input: mat[ ][ ] = { { 1, 2, 2 }, { 1, 0, 0 } }
Output: 3
Explanation:
For each path in matrix to be Palindrome, possible matrices (after changes) are
{ { 0, 2, 2 }, { 2, 2, 0 } } or { { 1, 2, 2 }, { 2, 2, 1 } }.
Input: mat[ ][ ] = { { 5, 3 }, { 0, 5 } }
Output: 0
Explanation:
No change required in above matrix. Each path from (N, M) to (1, 1) is already Palindrome
方法：
- 如果最后一个单元格的值等于第一个单元格的值，倒数第二个单元格的值等于第二个单元格的值，以此类推，则该路径称为回文。
- 因此我们可以得出结论，要使路径回文，距离 (Manhatten distance) x与(N, M)的单元格必须等于距离x与(1, 1)的单元格。
- 为了最大限度地减少变化次数，请将距离(1, 1)和(N, M) 的距离为x 的每个单元格转换为这些单元格中存在的所有值中最常见的值。
下面是上述方法的实现。
C++
// C++ Program to implement the
// above approach
#include
using namespace std;
#define N 7
// Function for counting changes
int countChanges(int matrix[][N],
int n, int m)
{
// Maximum distance possible
// is (n - 1 + m - 1)
int dist = n + m - 1;
// Stores the maximum element
int Max_element = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
// Update the maximum
Max_element = max(Max_element,
matrix[i][j]);
}
}
// Stores frequencies of
// values for respective
// distances
int freq[dist][Max_element + 1];
// Initialize frequencies of
// cells as 0
for (int i = 0; i < dist; i++) {
for (int j = 0; j < Max_element + 1;
j++)
freq[i][j] = 0;
}
// Count frequencies of cell
// values in the matrix
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
// Increment frequency of
// value at distance i+j
freq[i + j][matrix[i][j]]++;
}
}
int min_changes_sum = 0;
for (int i = 0; i < dist / 2; i++) {
// Store the most frequent
// value at i-th distance
// from (0, 0) and (N - 1, M - 1)
int maximum = 0;
int total_values = 0;
// Calculate max frequency
// and total cells at distance i
for (int j = 0; j < Max_element + 1;
j++) {
maximum
= max(maximum,
freq[i][j]
+ freq[n + m - 2
- i][j]);
total_values
+= freq[i][j]
+ freq[n + m - 2 - i][j];
}
// Count changes required
// to convert all cells
// at i-th distance to
// most frequent value
min_changes_sum
+= total_values - maximum;
}
return min_changes_sum;
}
// Driver Code
int main()
{
int mat[][N] = { { 7, 0, 3, 1, 8, 1, 3 },
{ 0, 4, 0, 1, 0, 4, 0 },
{ 3, 1, 8, 3, 1, 0, 7 } };
int minChanges = countChanges(mat, 3, 7);
cout << minChanges;
return 0;
} Java
// Java program to implement the
// above approach
import java.util.*;
class GFG{
static int N = 7;
// Function for counting changes
static int countChanges(int matrix[][],
int n, int m)
{
// Maximum distance possible
// is (n - 1 + m - 1)
int i, j, dist = n + m - 1;
// Stores the maximum element
int Max_element = 0;
for(i = 0; i < n; i++)
{
for(j = 0; j < m; j++)
{
// Update the maximum
Max_element = Math.max(Max_element,
matrix[i][j]);
}
}
// Stores frequencies of
// values for respective
// distances
int freq[][] = new int[dist][Max_element + 1];
// Initialize frequencies of
// cells as 0
for(i = 0; i < dist; i++)
{
for(j = 0; j < Max_element + 1;
j++)
freq[i][j] = 0;
}
// Count frequencies of cell
// values in the matrix
for(i = 0; i < n; i++)
{
for(j = 0; j < m; j++)
{
// Increment frequency of
// value at distance i+j
freq[i + j][matrix[i][j]]++;
}
}
int min_changes_sum = 0;
for(i = 0; i < dist / 2; i++)
{
// Store the most frequent
// value at i-th distance
// from (0, 0) and (N - 1, M - 1)
int maximum = 0;
int total_values = 0;
// Calculate max frequency
// and total cells at distance i
for(j = 0; j < Max_element + 1; 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];
}
// Count changes required
// to convert all cells
// at i-th distance to
// most frequent value
min_changes_sum += total_values -
maximum;
}
return min_changes_sum;
}
// Driver Code
public static void main (String []args)
{
int mat[][] = { { 7, 0, 3, 1, 8, 1, 3 },
{ 0, 4, 0, 1, 0, 4, 0 },
{ 3, 1, 8, 3, 1, 0, 7 } };
int minChanges = countChanges(mat, 3, 7);
System.out.print(minChanges);
}
}
// This code is contributed by chitranayalPython3
# Python3 program to implement the
# above approach
# Function for counting changes
def countChanges(matrix, n, m):
# Maximum distance possible
# is (n - 1 + m - 1)
dist = n + m - 1
# Stores the maximum element
Max_element = 0
for i in range(n):
for j in range(m):
# Update the maximum
Max_element = max(Max_element,
matrix[i][j])
# Stores frequencies of
# values for respective
# distances
freq = [[0 for i in range(Max_element + 1)]
for j in range(dist)]
# Initialize frequencies of
# cells as 0
for i in range(dist):
for j in range(Max_element + 1):
freq[i][j] = 0
# Count frequencies of cell
# values in the matrix
for i in range(n):
for j in range(m):
# Increment frequency of
# value at distance i+j
freq[i + j][matrix[i][j]] += 1
min_changes_sum = 0
for i in range(dist // 2):
# Store the most frequent
# value at i-th distance
# from (0, 0) and (N - 1, M - 1)
maximum = 0
total_values = 0
# Calculate max frequency
# and total cells at distance i
for j in range(Max_element + 1):
maximum = max(maximum,
freq[i][j] +
freq[n + m - 2 - i][j])
total_values += (freq[i][j] +
freq[n + m - 2 - i][j])
# Count changes required
# to convert all cells
# at i-th distance to
# most frequent value
min_changes_sum += total_values - maximum
return min_changes_sum
# Driver code
if __name__ == '__main__':
mat = [ [ 7, 0, 3, 1, 8, 1, 3 ],
[ 0, 4, 0, 1, 0, 4, 0 ],
[ 3, 1, 8, 3, 1, 0, 7 ] ]
minChanges = countChanges(mat, 3, 7)
print(minChanges)
# This code is contributed by Shivam SinghC#
// C# program to implement the
// above approach
using System;
class GFG{
//static int N = 7;
// Function for counting changes
static int countChanges(int [,]matrix,
int n, int m)
{
// Maximum distance possible
// is (n - 1 + m - 1)
int i, j, dist = n + m - 1;
// Stores the maximum element
int Max_element = 0;
for(i = 0; i < n; i++)
{
for(j = 0; j < m; j++)
{
// Update the maximum
Max_element = Math.Max(Max_element,
matrix[i, j]);
}
}
// Stores frequencies of
// values for respective
// distances
int [,]freq = new int[dist, Max_element + 1];
// Initialize frequencies of
// cells as 0
for(i = 0; i < dist; i++)
{
for(j = 0; j < Max_element + 1;
j++)
freq[i, j] = 0;
}
// Count frequencies of cell
// values in the matrix
for(i = 0; i < n; i++)
{
for(j = 0; j < m; j++)
{
// Increment frequency of
// value at distance i+j
freq[i + j, matrix[i, j]]++;
}
}
int min_changes_sum = 0;
for(i = 0; i < dist / 2; i++)
{
// Store the most frequent
// value at i-th distance
// from (0, 0) and (N - 1, M - 1)
int maximum = 0;
int total_values = 0;
// Calculate max frequency
// and total cells at distance i
for(j = 0; j < Max_element + 1; 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];
}
// Count changes required
// to convert all cells
// at i-th distance to
// most frequent value
min_changes_sum += total_values -
maximum;
}
return min_changes_sum;
}
// Driver Code
public static void Main(String []args)
{
int [,]mat = { { 7, 0, 3, 1, 8, 1, 3 },
{ 0, 4, 0, 1, 0, 4, 0 },
{ 3, 1, 8, 3, 1, 0, 7 } };
int minChanges = countChanges(mat, 3, 7);
Console.Write(minChanges);
}
}
// This code is contributed by sapnasingh4991Javascript
输出：
6时间复杂度： O(N * M)
辅助空间： O((N + M)*maxm)，其中 maxm 是矩阵中存在的最大元素。
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