Given a matrix mat[][], the task is to check if the given matrix is strictly increasing or not. A matrix is said to be strictly increasing if all of its rows as well as all of its columns are strictly increasing.
Examples:
Input: mat[][] = {{2, 10}, {11, 20}}
Output: Yes
All the rows and columns are strictly increasing.
Input: mat[][] = {{2, 1}, {11, 20}}
Output: No
First row doesn’t satisfy the required condition.
Approach: Linearly traverse for every element and check if there are increasing row-wise and column-wise or not. The two conditions are (a[i][j] > a[i – 1][j]) and (a[i][j] > a[i][j – 1]) . If any of the two conditions fail, then the matrix is not strictly increasing.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
#define N 2 #define M 2 // Function that returns true if the matrix // is strictly increasing bool isMatrixInc( int a[N][M])
{ // Check if the matrix
// is strictly increasing
for ( int i = 0; i < N; i++) {
for ( int j = 0; j < M; j++) {
// Out of bound condition
if (i - 1 >= 0) {
if (a[i][j] <= a[i - 1][j])
return false ;
}
// Out of bound condition
if (j - 1 >= 0) {
if (a[i][j] <= a[i][j - 1])
return false ;
}
}
}
return true ;
} // Driver code int main()
{ int a[N][M] = { { 2, 10 },
{ 11, 20 } };
if (isMatrixInc(a))
cout << "Yes" ;
else
cout << "No" ;
return 0;
} |
// Java implementation of the approach import java.io.*;
class GFG
{ static int N = 2 ;
static int M = 2 ;
// Function that returns true if the matrix // is strictly increasing static boolean isMatrixInc( int a[][])
{ // Check if the matrix
// is strictly increasing
for ( int i = 0 ; i < N; i++)
{
for ( int j = 0 ; j < M; j++)
{
// Out of bound condition
if (i - 1 >= 0 )
{
if (a[i][j] <= a[i - 1 ][j])
return false ;
}
// Out of bound condition
if (j - 1 >= 0 )
{
if (a[i][j] <= a[i][j - 1 ])
return false ;
}
}
}
return true ;
} // Driver code public static void main (String[] args)
{ int a[][] = { { 2 , 10 },
{ 11 , 20 } };
if (isMatrixInc(a))
System.out.print( "Yes" );
else
System.out.print( "No" );
} } // This code is contributed by anuj_67.. |
# Python3 implementation of the approach N, M = 2 , 2
# Function that returns true if the matrix # is strictly increasing def isMatrixInc(a) :
# Check if the matrix
# is strictly increasing
for i in range (N) :
for j in range (M) :
# Out of bound condition
if (i - 1 > = 0 ) :
if (a[i][j] < = a[i - 1 ][j]) :
return False ;
# Out of bound condition
if (j - 1 > = 0 ) :
if (a[i][j] < = a[i][j - 1 ]) :
return False ;
return True ;
# Driver code if __name__ = = "__main__" :
a = [ [ 2 , 10 ],
[ 11 , 20 ] ];
if (isMatrixInc(a)) :
print ( "Yes" );
else :
print ( "No" );
# This code is contributed by AnkitRai01 |
// C# implementation of the approach using System;
class GFG
{ static int N = 2;
static int M = 2;
// Function that returns true if the matrix // is strictly increasing static Boolean isMatrixInc( int [,]a)
{ // Check if the matrix
// is strictly increasing
for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < M; j++)
{
// Out of bound condition
if (i - 1 >= 0)
{
if (a[i,j] <= a[i - 1,j])
return false ;
}
// Out of bound condition
if (j - 1 >= 0)
{
if (a[i,j] <= a[i,j - 1])
return false ;
}
}
}
return true ;
} // Driver code public static void Main (String[] args)
{ int [,]a = { { 2, 10 },
{ 11, 20 } };
if (isMatrixInc(a))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
} } // This code is contributed by Princi Singh |
<script> // Java Script implementation of the approach let N = 2; let M = 2; // Function that returns true if the matrix // is strictly increasing function isMatrixInc(a)
{ // Check if the matrix
// is strictly increasing
for (let i = 0; i < N; i++)
{
for (let j = 0; j < M; j++)
{
// Out of bound condition
if (i - 1 >= 0)
{
if (a[i][j] <= a[i - 1][j])
return false ;
}
// Out of bound condition
if (j - 1 >= 0)
{
if (a[i][j] <= a[i][j - 1])
return false ;
}
}
}
return true ;
} // Driver code let a = [[2, 10 ],
[ 11, 20 ]];
if (isMatrixInc(a))
document.write( "Yes" );
else
document.write( "No" );
// This code is contributed by sravan kumar </script> |
Yes
Time Complexity: O(N*M)
Auxiliary Space: O(1)