Given a string S, the task is to check if we can make the string lexicographically smaller by reversing any substring of the given string.
Examples:
Input: S = “striver”
Output: Yes
Reverse “rive” to get “stevirr” which is lexicographically smaller.
Input: S = “rxz”
Output: No
Approach: Iterate in the string and check if for any index s[i] > s[i + 1]. If there exists at least one such index, then it is possible else not.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function that returns true if s // can be made lexicographically smaller // by reversing a sub-string in s bool check(string &s)
{ int n = s.size();
// Traverse in the string
for ( int i = 0; i < n - 1; i++) {
// Check if s[i+1] < s[i]
if (s[i] > s[i + 1])
return true ;
}
// Not possible
return false ;
} // Driver code int main()
{ string s = "geeksforgeeks" ;
if (check(s))
cout << "Yes" ;
else
cout << "No" ;
return 0;
} |
// Java implementation of the approach class GFG
{ // Function that returns true if s // can be made lexicographically smaller // by reversing a sub-string in s static boolean check(String s)
{ int n = s.length();
// Traverse in the string
for ( int i = 0 ; i < n - 1 ; i++)
{
// Check if s[i+1] < s[i]
if (s.charAt(i) > s.charAt(i + 1 ))
return true ;
}
// Not possible
return false ;
} // Driver code public static void main(String args[])
{ String s = "geeksforgeeks" ;
if (check(s))
System.out.println( "Yes" );
else
System.out.println( "No" );
} } // This code is contributed by Arnab Kundu |
# Python 3 implementation of the approach # Function that returns true if s # can be made lexicographically smaller # by reversing a sub-string in s def check(s):
n = len (s)
# Traverse in the string
for i in range (n - 1 ):
# Check if s[i+1] < s[i]
if (s[i] > s[i + 1 ]):
return True
# Not possible
return False
# Driver code if __name__ = = '__main__' :
s = "geeksforgeeks"
if (check(s)):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by # Surendra_Gangwar |
// C# implementation of the approach using System;
class GFG
{ // Function that returns true if s // can be made lexicographically smaller // by reversing a sub-string in s static bool check(String s)
{ int n = s.Length;
// Traverse in the string
for ( int i = 0; i < n - 1; i++)
{
// Check if s[i+1] < s[i]
if (s[i] > s[i + 1])
return true ;
}
// Not possible
return false ;
} // Driver code public static void Main(String []args)
{ String s = "geeksforgeeks" ;
if (check(s))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
} } // This code has been contributed by 29AjayKumar |
<?php // PHP implementation of the approach // Function that returns true if s // can be made lexicographically smaller // by reversing a sub-string in s function check( $s )
{ $n = strlen ( $s );
// Traverse in the string
for ( $i = 0; $i < $n - 1; $i ++)
{
// Check if $s[$i+1] < $s[$i]
if ( $s [ $i ] > $s [ $i + 1])
return true;
}
// Not possible
return false;
} // Driver code
$s = "geeksforgeeks" ;
if (check( $s ))
echo "Yes" ;
else
echo "No" ;
// This code is contributed by jit_t ?> |
<script> // Javascript implementation of the approach
// Function that returns true if s
// can be made lexicographically smaller
// by reversing a sub-string in s
function check(s)
{
let n = s.length;
// Traverse in the string
for (let i = 0; i < n - 1; i++)
{
// Check if s[i+1] < s[i]
if (s[i] > s[i + 1])
return true ;
}
// Not possible
return false ;
}
let s = "geeksforgeeks" ;
if (check(s))
document.write( "Yes" );
else
document.write( "No" );
</script> |
Yes
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1), as constant extra space is required.