Check if the Matrix satisfies the given conditions
Last Updated :
07 Oct, 2022
Given a matrix mat[][], the task is to check whether the matrix is valid or not. A valid matrix is the matrix that satisfies the given conditions:
- For every row, it must contain a single distinct character.
- No two consecutive rows have a character in common.
Examples:
Input: mat[][] = {
{0, 0, 0},
{1, 1, 1},
{0, 0, 2}}
Output: No
The last row doesn’t consist of a single distinct character.
Input: mat[][] = {
{8, 8, 8, 8, 8},
{4, 4, 4, 4, 4},
{6, 6, 6, 6, 6},
{5, 5, 5, 5, 5},
{8, 8, 8, 8, 8}}
Output: Yes
Approach: First of all check the first row if it contains same characters or not. If it contains same characters then iterate the second row and compare the characters of the current row with the first element of the previous row, if the two elements are equal or the characters of the current row are different, return false. Repeat the above process for all the consecutive row.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
const int M = 5, N = 5;
bool isPerfect( char board[M][N + 1])
{
char m;
for ( int i = 0; i < M; i++) {
string s = board[i];
if (i > 0 && s[0] == m)
return false ;
for ( int j = 0; j < N; j++) {
if (j == 0)
m = s[0];
else {
if (m != s[j])
return false ;
}
}
}
return true ;
}
int main()
{
char board[M][N + 1] = { "88888" ,
"44444" ,
"66666" ,
"55555" ,
"88888" };
if (isPerfect(board))
cout << "Yes" ;
else
cout << "No" ;
return 0;
}
|
Java
class GFG
{
static final int M = 5 , N = 5 ;
static boolean isPerfect(String board[])
{
char m = 0 ;
for ( int i = 0 ; i < M; i++)
{
String s = board[i];
if (i > 0 && s.charAt( 0 ) == m)
return false ;
for ( int j = 0 ; j < N; j++)
{
if (j == 0 )
m = s.charAt( 0 );
else
{
if (m != s.charAt(j))
return false ;
}
}
}
return true ;
}
public static void main (String[] args)
{
String board[] = { "88888" , "44444" ,
"66666" , "55555" ,
"88888" };
if (isPerfect(board))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Python3
M = 5
N = 5
def isPerfect(board):
m = ''
for i in range (M):
s = board[i]
if (i > 0 and s[ 0 ] = = m):
return False
for j in range (N):
if (j = = 0 ):
m = s[ 0 ]
else :
if (m ! = s[j]):
return False
return True
board = [ "88888" ,
"44444" ,
"66666" ,
"55555" ,
"88888" ]
if (isPerfect(board)):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG
{
static readonly int M = 5, N = 5;
static bool isPerfect(String []board)
{
char m = '0' ;
for ( int i = 0; i < M; i++)
{
String s = board[i];
if (i > 0 && s[0] == m)
return false ;
for ( int j = 0; j < N; j++)
{
if (j == 0)
m = s[0];
else
{
if (m != s[j])
return false ;
}
}
}
return true ;
}
public static void Main (String[] args)
{
String []board = { "88888" , "44444" ,
"66666" , "55555" ,
"88888" };
if (isPerfect(board))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
|
Javascript
<script>
var M = 5, N = 5;
function isPerfect(board)
{
var m;
for ( var i = 0; i < M; i++)
{
var s = board[i];
if (i > 0 && s[0] == m)
return false ;
for ( var j = 0; j < N; j++)
{
if (j == 0)
m = s[0];
else
{
if (m != s[j])
return false ;
}
}
}
return true ;
}
var board = [ "88888" , "44444" ,
"66666" , "55555" ,
"88888" ];
if (isPerfect(board))
document.write( "Yes" );
else
document.write( "No" );
</script>
|
Time complexity: O(M*N)
Auxiliary space: O(N) because using extra space for string s
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