Given an integer matrix mat [ ][ ] dimensions, the task is to find the largest possible square matrix from the given matrix with maximum AND value.
AND value of a matrix is defined as the value obtained after performing bitwise AND operation on all elements of the matrix.
Examples:
Input: mat [ ][ ] = {{2, 3, 3}, {2, 3, 3}, {2, 2, 2}}
Output: 4
Explanation:
Given square submatrix has AND value 2.
The submatrix
{{3, 3}
{3, 3}}
of size 4 has maximum AND value 3. All other square submatrices of size 4 have AND value 2.Input: mat [ ][ ] =
{{9, 9, 9, 8},
{9, 9, 9, 6},
{9, 9, 9, 3},
{2, 2, 2, 2}}
Output: 9
Explanation:
The submatrix of size 9
{{9, 9, 9},
{9, 9, 9},
{9, 9, 9}}
have maximum AND value 9.
Naive Approach:
Generate all square submatrices from the given matrix. Initialize a variable answer to store the maximum & value for submatrices and another variable count to store the number of elements in the submatrix. Print the maximum value of count corresponding to maximum AND value answer obtained from all square submatrices.
Efficient Approach:
Follow the steps below to optimize the above solution:
- To maximize the & value, we need to find a submatrix that consists only of the maximum element in the matrix. This is because the maximum possible AND value in the matrix is the maximum element present in the matrix.
- Find the maximum possible value present in the matrix.
- Use Dynamic programming approach to get maximum size submatrix filled by the maximum matrix element only.
- Create an auxiliary dp[][] such that dp[i][j] stores the largest possible square submatrix mat[i][j] can be a part of such that the AND value of that submatrix is equal to mat[i][j].
- The recurrence relation is as follows:
If mat[i][j] is equal to {mat[i-1][j], mat[i][j-1], mat[i-1][j-1]} then consider all the three values as a square submatrix and update DP[i][j] as:
DP[i][j] = min(DP[i-1][j], DP[i][j-1], DP[i-1][j-1]) + 1
Otherwise,
DP[i][j] = 1
The answer would be the maximum of all DP [i][j]
- Finally, iterate over the dp[][] matrix and find the largest dp[i][j] for every mat[i][j] equal to the maximum element in the array.
Below is implementation of above approach:
C++
// C++ program to find // the length of longest // possible square submatrix // with maximum AND value // from the given matrix #include <bits/stdc++.h> using namespace std;
// Function to calculate and // return the length of // square submatrix with // maximum AND value int MAX_value(vector<vector<int> > arr)
{ // Extract dimensions
int row = arr.size();
int col = arr[0].size();
// Auxiliary array
int dp[row][col];
// Initialize auxiliary array
memset(dp, sizeof(dp), 0);
// c: Stores the maximum
// value in the matrix
// p: Stores the number
// of elements in the
// submatrix having
// maximum AND value
int i = 0, j = 0;
int c = arr[0][0], p = 0;
int d = row;
// Iterate over the matrix
// to fill the auxiliary
// matrix
for (i = 0; i < d; i++) {
for (j = 0; j < d; j++) {
// Find the max element in the
// matrix side by side
if (c < arr[i][j]) {
c = arr[i][j];
}
// Fill first row and
// column with 1's
if (i == 0 || j == 0) {
dp[i][j] = 1;
}
else {
// For every cell, check if
// the elements at the left,
// top and top left cells
// from the current cell
// are equal or not
if (arr[i - 1][j - 1] == arr[i][j]
&& arr[i - 1][j] == arr[i][j]
&& arr[i][j - 1] == arr[i][j]) {
// Store the minimum possible
// submatrix size these
// elements are part of
dp[i][j]
= min(dp[i - 1][j - 1],
min(dp[i - 1][j],
dp[i][j - 1]))
+ 1;
}
else {
// Store 1 otherwise
dp[i][j] = 1;
}
}
}
}
for (i = 0; i < d; i++) {
for (j = 0; j < d; j++) {
// checking maximum value
if (arr[i][j] == c) {
// If the maximum AND
// value occurs more
// than once
if (p < dp[i][j]) {
// Update the maximum
// size of submatrix
p = dp[i][j];
}
}
}
}
// final output
return p * p;
} // Driver Program int main()
{ vector<vector<int> > arr
= { { 9, 9, 3, 3, 4, 4 },
{ 9, 9, 7, 7, 7, 4 },
{ 1, 2, 7, 7, 7, 4 },
{ 4, 4, 7, 7, 7, 4 },
{ 5, 5, 1, 1, 2, 7 },
{ 2, 7, 1, 1, 4, 4 } };
cout << MAX_value(arr) << endl;
return 0;
} |
Java
// Java program to find the length// of longest possible square // submatrix with maximum AND // value from the given matriximport java.util.*;
class GFG{
// Function to calculate and return// the length of square submatrix // with maximum AND valuestatic int MAX_value(int [][]arr)
{ // Extract dimensions
int row = arr.length;
int col = arr[0].length;
// Auxiliary array
int [][]dp = new int[row][col];
// c: Stores the maximum
// value in the matrix
// p: Stores the number
// of elements in the
// submatrix having
// maximum AND value
int i = 0, j = 0;
int c = arr[0][0], p = 0;
int d = row;
// Iterate over the matrix
// to fill the auxiliary
// matrix
for(i = 0; i < d; i++)
{
for(j = 0; j < d; j++)
{
// Find the max element in
// the matrix side by side
if (c < arr[i][j])
{
c = arr[i][j];
}
// Fill first row and
// column with 1's
if (i == 0 || j == 0)
{
dp[i][j] = 1;
}
else
{
// For every cell, check if the
// elements at the left, top and
// top left cells from the current
// cell are equal or not
if (arr[i - 1][j - 1] == arr[i][j] &&
arr[i - 1][j] == arr[i][j] &&
arr[i][j - 1] == arr[i][j])
{
// Store the minimum possible
// submatrix size these
// elements are part of
dp[i][j] = Math.min(dp[i - 1][j - 1],
Math.min(dp[i - 1][j],
dp[i][j - 1])) + 1;
}
else
{
// Store 1 otherwise
dp[i][j] = 1;
}
}
}
}
for(i = 0; i < d; i++)
{
for(j = 0; j < d; j++)
{
// Checking maximum value
if (arr[i][j] == c)
{
// If the maximum AND
// value occurs more
// than once
if (p < dp[i][j])
{
// Update the maximum
// size of submatrix
p = dp[i][j];
}
}
}
}
// Final output
return p * p;
}// Driver codepublic static void main(String[] args)
{ int [][]arr = { { 9, 9, 3, 3, 4, 4 },
{ 9, 9, 7, 7, 7, 4 },
{ 1, 2, 7, 7, 7, 4 },
{ 4, 4, 7, 7, 7, 4 },
{ 5, 5, 1, 1, 2, 7 },
{ 2, 7, 1, 1, 4, 4 } };
System.out.print(MAX_value(arr) + "\n");
}}// This code contributed by amal kumar choubey |
Python3
# Python3 program to find the length# of longest possible square submatrix # with maximum AND value from the given# matrix # Function to calculate and return the # length of square submatrix with # maximum AND value def MAX_value(arr):
# Extract dimensions
row = len(arr)
col = len(arr)
# Auxiliary array
# Initialize auxiliary array
dp = [[0 for i in range(col)]
for j in range(row)]
# c: Stores the maximum
# value in the matrix
# p: Stores the number
# of elements in the
# submatrix having
# maximum AND value
i, j = 0, 0
c, p = arr[0][0], 0
d = row
# Iterate over the matrix
# to fill the auxiliary
# matrix
for i in range(d):
for j in range(d):
# Find the max element in the
# matrix side by side
if (c < arr[i][j]):
c = arr[i][j]
# Fill first row and
# column with 1's
if (i == 0 or j == 0):
dp[i][j] = 1
else:
# For every cell, check if
# the elements at the left,
# top and top left cells
# from the current cell
# are equal or not
if (arr[i - 1][j - 1] == arr[i][j] and
arr[i - 1][j] == arr[i][j] and
arr[i][j - 1] == arr[i][j]):
# Store the minimum possible
# submatrix size these
# elements are part of
dp[i][j] = min(dp[i - 1][j - 1],
min(dp[i - 1][j],
dp[i][j - 1])) + 1
else:
# Store 1 otherwise
dp[i][j] = 1
for i in range(d):
for j in range(d):
# Checking maximum value
if (arr[i][j] == c):
# If the maximum AND
# value occurs more
# than once
if (p < dp[i][j]):
# Update the maximum
# size of submatrix
p = dp[i][j]
# Final output
return p * p
# Driver Code arr = [ [ 9, 9, 3, 3, 4, 4 ],
[ 9, 9, 7, 7, 7, 4 ],
[ 1, 2, 7, 7, 7, 4 ],
[ 4, 4, 7, 7, 7, 4 ],
[ 5, 5, 1, 1, 2, 7 ],
[ 2, 7, 1, 1, 4, 4 ]]
print(MAX_value(arr))
# This code is contributed by divyeshrabadiya07 |
C#
// C# program to find the length// of longest possible square // submatrix with maximum AND // value from the given matrixusing System;
class GFG{
// Function to calculate and return// the length of square submatrix // with maximum AND valuestatic int MAX_value(int [,]arr)
{ // Extract dimensions
int row = arr.GetLength(0);
int col = arr.GetLength(1);
// Auxiliary array
int [,]dp = new int[row, col];
// c: Stores the maximum
// value in the matrix
// p: Stores the number
// of elements in the
// submatrix having
// maximum AND value
int i = 0, j = 0;
int c = arr[0, 0], p = 0;
int d = row;
// Iterate over the matrix
// to fill the auxiliary
// matrix
for(i = 0; i < d; i++)
{
for(j = 0; j < d; j++)
{
// Find the max element in
// the matrix side by side
if (c < arr[i, j])
{
c = arr[i, j];
}
// Fill first row and
// column with 1's
if (i == 0 || j == 0)
{
dp[i, j] = 1;
}
else
{
// For every cell, check if the
// elements at the left, top and
// top left cells from the current
// cell are equal or not
if (arr[i - 1, j - 1] == arr[i, j] &&
arr[i - 1, j] == arr[i, j] &&
arr[i, j - 1] == arr[i, j])
{
// Store the minimum possible
// submatrix size these
// elements are part of
dp[i, j] = Math.Min(dp[i - 1, j - 1],
Math.Min(dp[i - 1, j],
dp[i, j - 1])) + 1;
}
else
{
// Store 1 otherwise
dp[i, j] = 1;
}
}
}
}
for(i = 0; i < d; i++)
{
for(j = 0; j < d; j++)
{
// Checking maximum value
if (arr[i, j] == c)
{
// If the maximum AND
// value occurs more
// than once
if (p < dp[i, j])
{
// Update the maximum
// size of submatrix
p = dp[i, j];
}
}
}
}
// Final output
return p * p;
}// Driver codepublic static void Main(String[] args)
{ int [,]arr = { { 9, 9, 3, 3, 4, 4 },
{ 9, 9, 7, 7, 7, 4 },
{ 1, 2, 7, 7, 7, 4 },
{ 4, 4, 7, 7, 7, 4 },
{ 5, 5, 1, 1, 2, 7 },
{ 2, 7, 1, 1, 4, 4 } };
Console.Write(MAX_value(arr) + "\n");
}}// This code is contributed by gauravrajput1 |
Javascript
<script>// Javascript program to find the length// of longest possible square // submatrix with maximum AND // value from the given matrix // Function to calculate and return
// the length of square submatrix
// with maximum AND value
function MAX_value(arr) {
// Extract dimensions
var row = arr.length;
var col = arr[0].length;
// Auxiliary array
var dp = Array(row);
for(var i = 0;i<row;i++)
dp[i] = Array(col).fill(0);
// c: Stores the maximum
// value in the matrix
// p: Stores the number
// of elements in the
// submatrix having
// maximum AND value
var i = 0, j = 0;
var c = arr[0][0], p = 0;
var d = row;
// Iterate over the matrix
// to fill the auxiliary
// matrix
for (i = 0; i < d; i++) {
for (j = 0; j < d; j++) {
// Find the max element in
// the matrix side by side
if (c < arr[i][j]) {
c = arr[i][j];
}
// Fill first row and
// column with 1's
if (i == 0 || j == 0) {
dp[i][j] = 1;
} else {
// For every cell, check if the
// elements at the left, top and
// top left cells from the current
// cell are equal or not
if (arr[i - 1][j - 1] == arr[i][j] &&
arr[i - 1][j] == arr[i][j] &&
arr[i][j - 1] == arr[i][j]) {
// Store the minimum possible
// submatrix size these
// elements are part of
dp[i][j] = Math.min(dp[i - 1][j - 1],
Math.min(dp[i - 1][j],
dp[i][j - 1])) + 1;
} else {
// Store 1 otherwise
dp[i][j] = 1;
}
}
}
}
for (i = 0; i < d; i++) {
for (j = 0; j < d; j++) {
// Checking maximum value
if (arr[i][j] == c) {
// If the maximum AND
// value occurs more
// than once
if (p < dp[i][j]) {
// Update the maximum
// size of submatrix
p = dp[i][j];
}
}
}
}
// Final output
return p * p;
}
// Driver code
var arr = [ [ 9, 9, 3, 3, 4, 4 ],
[ 9, 9, 7, 7, 7, 4 ],
[ 1, 2, 7, 7, 7, 4 ],
[ 4, 4, 7, 7, 7, 4 ],
[ 5, 5, 1, 1, 2, 7 ],
[ 2, 7, 1, 1, 4, 4 ] ];
document.write(MAX_value(arr) + "\n");
// This code contributed by umadevi9616</script> |
Output:
4
Time Complexity: O(N*N), as we are using nested loops to traverse N*N times.
Auxiliary Space: O(N*N), as we are using extra space for matrix.