Check if string can be made lexicographically smaller by reversing any substring

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++

// 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

// 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

Python3

# 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#

// 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
// 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
?>

Javascript

<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>

Output:

Yes

Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1), as constant extra space is required.