Finding the maximum square sub-matrix with all equal elements

• Difficulty Level : Medium
• Last Updated : 26 Aug, 2021

Given a N x N matrix, determine the maximum K such that K x K is a submatrix with all equal elements i.e., all the elements in this submatrix must be same.
Constraints:
1 <= N <= 1000
0 <= Ai , j <= 109
Examples:

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

Method I ( Naive approach )
We can easily find all the square submatrices in O(n3) time and check whether each submatrix contains equal elements or not in O(n2) time Which makes the total running time of the algorithm as O(n5).
Method II ( Dynamic Programming )
For each cell (i, j), we store the largest value of K such that K x K is a submatrix with all equal elements and position of (i, j) being the bottom-right most element.
And DPi,j depends upon {DPi-1, j, DPi, j-1, DPi-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

Below is the implementation of above steps.

C++

 // C++ program to find maximum K such that K x K// is a submatrix with equal elements.#include#define Row 6#define Col 6using 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 Programint 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 ChitraNayal

C#

 // 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 ChitraNayal

Python 3

 # Python 3 program to find # maximum K such that K x K# is a submatrix with equal # elements.Row = 6Col = 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 Codea = [[ 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 ChitraNayal



Javascript



Output:

3

Time Complexity : O(Row * Col)
Auxiliary Space : O(Row * Col)
This article is contributed by Shubham Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.