Reverse the given string in the range [L, R]

Given a string str, and two integers L and R, the task is to reverse the string in the range [L, R] i.e. str[L…R].
Examples:

Input: str = “geeksforgeeks”, L = 5, R = 7 
Output: geeksrofgeeks 
Reverse the characters in the range str[5…7] = “geeksforgeeks” 
and the new string will be “geeksrofgeeks”

Input: str = “ijklmn”, L = 1, R = 2 
Output: ikjlmn 

Approach:

1. If the range is invalid i.e. either L < 0 or R ? len or L > R then print the original string.
2. If the range is valid then keep swapping the characters str[L] and str[R] while L < R and update L = L + 1 and R = R – 1 after every swap operation. Print the updated string in the end.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the string after
// reversing characters in the range [L, R]
string reverse(string str, int len, int l, int r)
{

    // Invalid range
    if (l < 0 || r >= len || l > r)
        return str;

    // While there are characters to swap
    while (l < r) {

        // Swap(str[l], str[r])
        char c = str[l];
        str[l] = str[r];
        str[r] = c;

        l++;
        r--;
    }

    return str;
}

// Driver code
int main()
{
    string str = "geeksforgeeks";
    int len = str.length();
    int l = 5, r = 7;

    cout << reverse(str, len, l, r);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

    // Function to return the string after
    // reversing characters in the range [L, R]
    static String reverse(char[] str, int len,
                               int l, int r)
    {

        // Invalid range
        if (l < 0 || r >= len || l > r)
            return "Invalid range!";

        // While there are characters to swap
        while (l < r)
        {

            // Swap(str[l], str[r])
            char c = str[l];
            str[l] = str[r];
            str[r] = c;

            l++;
            r--;
        }
        String string = new String(str);
        return string;
    }

    // Driver code
    public static void main (String[] args)
    {
        String str = "geeksforgeeks";
        int len = str.length();
        int l = 5, r = 7;

        System.out.println(reverse(str.toCharArray(),
                                         len, l, r));
    }
}

// This code is contributed by Ashutosh450

Python3

# Python3 implementation of the approach

# Function to return the string after
# reversing characters in the range [L, R]
def reverse(string, length, l, r) :

    # Invalid range
    if (l < 0 or r >= length or l > r) :
        return string;

    string = list(string)

    # While there are characters to swap
    while (l < r) :

        # Swap(str[l], str[r])
        c = string[l];
        string[l] = string[r];
        string[r] = c;

        l += 1;
        r -= 1;

    return "".join(string);

# Driver code
if __name__ == "__main__" :

    string = "geeksforgeeks";
    length = len(string);
    l = 5; r = 7;

    print(reverse(string, length, l, r));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the string after
    // reversing characters in the range [L, R]
    static String reverse(char[] str, int len,
                          int l, int r)
    {

        // Invalid range
        if (l < 0 || r >= len || l > r)
            return "Invalid range!";

        // While there are characters to swap
        while (l < r)
        {

            // Swap(str[l], str[r])
            char c = str[l];
            str[l] = str[r];
            str[r] = c;

            l++;
            r--;
        }
        return String.Join("",str);
    }

    // Driver code
    public static void Main (String[] args)
    {
        String str = "geeksforgeeks";
        int len = str.Length;
        int l = 5, r = 7;

        Console.WriteLine(reverse(str.ToCharArray(),
                                        len, l, r));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
    // Javascript implementation of the approach

    // Function to return the string after
    // reversing characters in the range [L, R]
    function reverse(str, len, l, r)
    {

        // Invalid range
        if (l < 0 || r >= len || l > r)
            return "Invalid range!";

        // While there are characters to swap
        while (l < r)
        {

            // Swap(str[l], str[r])
            let c = str[l];
            str[l] = str[r];
            str[r] = c;

            l++;
            r--;
        }
        return str.join("");
    }

    let str = "geeksforgeeks";
    let len = str.length;
    let l = 5, r = 7;

    document.write(reverse(str.split(''), len, l, r));

// This code is contributed by divyeshrabadiya07.
</script>

Output:

geeksrofgeeks

Time complexity: O(N), where N = (r – l)/2
Auxiliary space: O(1)