给定一个 N x N 矩阵，确定最大 K，使得 K x K 是一个具有所有相等元素的子矩阵，即该子矩阵中的所有元素必须相同。
约束：
1 <= N <= 1000
0 <= A i , j <= 10 9
例子：
Input : a[][] = {{2, 3, 3},
{2, 3, 3},
{2, 2, 2}}
Output : 2
Explanation: A 2x2 matrix is formed from index
A0,1 to A1,2
Input : a[][] = {{9, 9, 9, 8},
{9, 9, 9, 6},
{9, 9, 9, 3},
{2, 2, 2, 2}
Output : 3
Explanation : A 3x3 matrix is formed from index
A0,0 to A2,2方法一(天真的方法)
我们可以很容易地找到为O的所有子矩阵正方形(N 3)和时间检查每个子矩阵是否包含相等的元件或不为O(n 2)时间这使得运行算法为O(N 5)的总时间。
方法二(动态规划)
对于每个单元格 (i, j)，我们存储 K 的最大值，使得 K x K 是一个子矩阵，其中所有元素都相等，并且 (i, j) 的位置是最右下角的元素。
DP i,j取决于 {DP i-1, j , DP i, j-1 , DP i-1, j-1 }
If Ai, j is equal to {Ai-1, j, Ai, j-1, Ai-1, j-1},
all the three values:
DPi, j = min(DPi-1, j, DPi, j-1, DPi-1, j-1) + 1
Else
DPi, j = 1 // Matrix Size 1
The answer would be the maximum of all DPi, j's下面是上述步骤的实现。
C++
// C++ program to find maximum K such that K x K
// is a submatrix with equal elements.
#include
#define Row 6
#define Col 6
using namespace std;
// Returns size of the largest square sub-matrix
// with all same elements.
int largestKSubmatrix(int a[][Col])
{
int dp[Row][Col];
memset(dp, sizeof(dp), 0);
int result = 0;
for (int i = 0 ; i < Row ; i++)
{
for (int j = 0 ; j < Col ; j++)
{
// If elements is at top row or first
// column, it wont form a square
// matrix's bottom-right
if (i == 0 || j == 0)
dp[i][j] = 1;
else
{
// Check if adjacent elements are equal
if (a[i][j] == a[i-1][j] &&
a[i][j] == a[i][j-1] &&
a[i][j] == a[i-1][j-1] )
dp[i][j] = min(min(dp[i-1][j], dp[i][j-1]),
dp[i-1][j-1] ) + 1;
// If not equal, then it will form a 1x1
// submatrix
else dp[i][j] = 1;
}
// Update result at each (i,j)
result = max(result, dp[i][j]);
}
}
return result;
}
// Driven Program
int main()
{
int a[Row][Col] = { 2, 2, 3, 3, 4, 4,
5, 5, 7, 7, 7, 4,
1, 2, 7, 7, 7, 4,
4, 4, 7, 7, 7, 4,
5, 5, 5, 1, 2, 7,
8, 7, 9, 4, 4, 4
};
cout << largestKSubmatrix(a) << endl;
return 0;
} Java
// Java program to find maximum
// K such that K x K is a
// submatrix with equal elements.
class GFG
{
static int Row = 6, Col = 6;
// Returns size of the largest
// square sub-matrix with
// all same elements.
static int largestKSubmatrix(int [][]a)
{
int [][]dp = new int [Row][Col];
int result = 0;
for (int i = 0 ;
i < Row ; i++)
{
for (int j = 0 ;
j < Col ; j++)
{
// If elements is at top
// row or first column,
// it wont form a square
// matrix's bottom-right
if (i == 0 || j == 0)
dp[i][j] = 1;
else
{
// Check if adjacent
// elements are equal
if (a[i][j] == a[i - 1][j] &&
a[i][j] == a[i][j - 1] &&
a[i][j] == a[i - 1][j - 1])
{
dp[i][j] = (dp[i - 1][j] > dp[i][j - 1] &&
dp[i - 1][j] > dp[i - 1][j - 1] + 1) ?
dp[i - 1][j] :
(dp[i][j - 1] > dp[i - 1][j] &&
dp[i][j - 1] > dp[i - 1][j - 1] + 1) ?
dp[i][j - 1] :
dp[i - 1][j - 1] + 1;
}
// If not equal, then it
// will form a 1x1 submatrix
else dp[i][j] = 1;
}
// Update result at each (i,j)
result = result > dp[i][j] ?
result : dp[i][j];
}
}
return result;
}
// Driver code
public static void main(String[] args)
{
int [][]a = {{2, 2, 3, 3, 4, 4},
{5, 5, 7, 7, 7, 4},
{1, 2, 7, 7, 7, 4},
{4, 4, 7, 7, 7, 4},
{5, 5, 5, 1, 2, 7},
{8, 7, 9, 4, 4, 4}};
System.out.println(largestKSubmatrix(a));
}
}
// This code is contributed
// by ChitraNayalC#
// C# program to find maximum
// K such that K x K is a
// submatrix with equal elements.
using System;
class GFG
{
static int Row = 6, Col = 6;
// Returns size of the
// largest square sub-matrix
// with all same elements.
static int largestKSubmatrix(int[,] a)
{
int[,] dp = new int [Row, Col];
int result = 0;
for (int i = 0 ; i < Row ; i++)
{
for (int j = 0 ;
j < Col ; j++)
{
// If elements is at top
// row or first column,
// it wont form a square
// matrix's bottom-right
if (i == 0 || j == 0)
dp[i, j] = 1;
else
{
// Check if adjacent
// elements are equal
if (a[i, j] == a[i - 1, j] &&
a[i, j] == a[i, j - 1] &&
a[i, j] == a[i - 1, j - 1])
{
dp[i, j] = (dp[i - 1, j] > dp[i, j - 1] &&
dp[i - 1, j] > dp[i - 1, j - 1] + 1) ?
dp[i - 1, j] :
(dp[i, j - 1] > dp[i - 1, j] &&
dp[i, j - 1] > dp[i - 1, j - 1] + 1) ?
dp[i, j - 1] :
dp[i - 1, j - 1] + 1;
}
// If not equal, then
// it will form a 1x1
// submatrix
else dp[i, j] = 1;
}
// Update result at each (i,j)
result = result > dp[i, j] ?
result : dp[i, j];
}
}
return result;
}
// Driver Code
public static void Main()
{
int[,] a = {{2, 2, 3, 3, 4, 4},
{5, 5, 7, 7, 7, 4},
{1, 2, 7, 7, 7, 4},
{4, 4, 7, 7, 7, 4},
{5, 5, 5, 1, 2, 7},
{8, 7, 9, 4, 4, 4}};
Console.Write(largestKSubmatrix(a));
}
}
// This code is contributed
// by ChitraNayalPython 3
# Python 3 program to find
# maximum K such that K x K
# is a submatrix with equal
# elements.
Row = 6
Col = 6
# Returns size of the
# largest square sub-matrix
# with all same elements.
def largestKSubmatrix(a):
dp = [[0 for x in range(Row)]
for y in range(Col)]
result = 0
for i in range(Row ):
for j in range(Col):
# If elements is at top
# row or first column,
# it wont form a square
# matrix's bottom-right
if (i == 0 or j == 0):
dp[i][j] = 1
else:
# Check if adjacent
# elements are equal
if (a[i][j] == a[i - 1][j] and
a[i][j] == a[i][j - 1] and
a[i][j] == a[i - 1][j - 1]):
dp[i][j] = min(min(dp[i - 1][j],
dp[i][j - 1]),
dp[i - 1][j - 1] ) + 1
# If not equal, then
# it will form a 1x1
# submatrix
else:
dp[i][j] = 1
# Update result at each (i,j)
result = max(result, dp[i][j])
return result
# Driver Code
a = [[ 2, 2, 3, 3, 4, 4],
[ 5, 5, 7, 7, 7, 4],
[ 1, 2, 7, 7, 7, 4],
[ 4, 4, 7, 7, 7, 4],
[ 5, 5, 5, 1, 2, 7],
[ 8, 7, 9, 4, 4, 4]];
print(largestKSubmatrix(a))
# This code is contributed
# by ChitraNayalPHP
Javascript
输出：
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