Reverse substrings of given string according to specified array indices

Given string str of length N and an array arr[] of integers, for array element arr[i](1-based indexing), reverse the substring in indices [arr[i], N - arr[i] + 1]. The task is to print the string after every reversal.

Examples:

Input: str = "GeeksforGeeks", arr[] = {2}
Output: GkeeGrofskees
Explanation:
For first element of the array is 2:
Reverse the substring (2, 12). Now the updated string is "GkeeGrofskees".

Input: str = "abcdef", arr[] = {1, 2, 3}
Output: fbdcea
Explanation:
For first element of the array is 1:
Reverse the substring (1, 6). Now the updated string is "fedcba".
For second element of the array is 2:
Reverse the substring (2, 5). Now the updated string is "fbcdea".
For third element of the array is 3:
Reverse the substring (3, 4). Now the updated string is "fbdcea".

Naive Approach: The simplest approach is to traverse the given array and for each array element arr[i] reverse the substring {s[arr[i]], ... s[N - arr[i] + 1]} and print the resultant string obtained after very update. 

Time Complexity: O(N * K), where N is the length of the string and K is the maximum length of the substring reversed.
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by keeping the track of the number of times any character at an index has been reversed. If the count of reversal is even, then the character will come back to its original place, so there will be no change and if the count of reversal is odd, then the character has to be swapped. Below are the steps:

• Initialize an array count[] to store the number of reversals at any index of the string.
• Traverse the given array arr[] and increment the count of indices count[arr[i]] by 1.
• Now, traverse the array count[] over the range [1, N/2] using the variable i and do the following:
  • Update the element at the current index as the sum of the current and previous index.
  • Now, if current element count[i] is odd, then swap str[i] and str[N - i + 1].
• Print the string after the above steps.

Below is the implementation of the above approach:

// C++ program for the above approach

#include <iostream>
using namespace std;

// Function to perform the reversal
// operation on the given string
string modifyString(int A[], string str,
                    int K)
{
    // Size of string
    int N = str.size();

    // Stores the count of indices
    int count[N + 1] = { 0 };

    // Count the positions where
    // reversals will begin
    for (int i = 0; i < K; i++) {
        count[A[i]]++;
    }

    for (int i = 1; i <= N / 2; i++) {

        // Store the count of reversals
        // beginning at position i
        count[i] = count[i] + count[i - 1];

        // Check if the count[i] is
        // odd the swap the character
        if (count[i] & 1) {
            swap(str[i - 1], str[N - i]);
        }
    }

    // Return the updated string
    return str;
}

// Driver Code
int main()
{
    // Given string str
    string str = "abcdef";

    // Given array of reversing index
    int arr[] = { 1, 2, 3 };

    int K = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    cout << modifyString(arr, str, K);

    return 0;
}

// Java program for the above approach
import java.util.*;

class GFG{

// Function to perform the reversal
// operation on the given String
static String modifyString(int A[], String str,
                           int K)
{
    
    // Size of String
    int N = str.length();

    // Stores the count of indices
    int count[] = new int[N + 1];

    // Count the positions where
    // reversals will begin
    for(int i = 0; i < K; i++)
    {
        count[A[i]]++;
    }

    for(int i = 1; i <= N / 2; i++)
    {
        
        // Store the count of reversals
        // beginning at position i
        count[i] = count[i] + count[i - 1];

        // Check if the count[i] is
        // odd the swap the character
        if ((count[i] & 1) > 0)
        {
            str = swap(str, i - 1, N - i);
        }
    }

    // Return the updated String
    return str;
}

// Swap char of a string
static String swap(String str, int i, int j) 
{ 
    char ch[] = str.toCharArray(); 
    char temp = ch[i]; 
    ch[i] = ch[j]; 
    ch[j] = temp; 
    return String.valueOf(ch); 
}

// Driver Code
public static void main(String[] args)
{
    
    // Given String str
    String str = "abcdef";

    // Given array of reversing index
    int arr[] = { 1, 2, 3 };

    int K = arr.length;

    // Function Call
    System.out.print(modifyString(arr, str, K));
}
}

// This code is contributed by Amit Katiyar

# Python3 program for the above approach

# Function to perform the reversal
# operation on the given string
def modifyString(A, str, K):
    
    # Size of string
    N = len(str)

    # Stores the count of indices
    count = [0] * (N + 1)

    # Count the positions where
    # reversals will begin
    for i in range(K):
        count[A[i]] += 1

    for i in range(1, N // 2 + 1):
        
        # Store the count of reversals
        # beginning at position i
        count[i] = count[i] + count[i - 1]

        # Check if the count[i] is
        # odd the swap the character
        if (count[i] & 1):
            str[i - 1], str[N - i] = str[N - i], str[i - 1]

    # Return the updated string
    return "".join(str)

# Driver Code
if __name__ == '__main__':
    
    # Given str
    str1 = "abcdef"
    str = [i for i in str1]

    # Given array of reversing index
    arr = [ 1, 2, 3 ]

    K = len(arr)
    
    # Function Call
    print(modifyString(arr, str, K))

# This code is contributed by mohit kumar 29

// C# program for the above approach
using System;

class GFG{

// Function to perform the reversal
// operation on the given String
static String modifyString(int []A, String str,
                           int K)
{
    
    // Size of String
    int N = str.Length;

    // Stores the count of indices
    int []count = new int[N + 1];

    // Count the positions where
    // reversals will begin
    for(int i = 0; i < K; i++)
    {
        count[A[i]]++;
    }

    for(int i = 1; i <= N / 2; i++)
    {
        
        // Store the count of reversals
        // beginning at position i
        count[i] = count[i] + count[i - 1];

        // Check if the count[i] is
        // odd the swap the character
        if ((count[i] & 1) > 0)
        {
            str = swap(str, i - 1, N - i);
        }
    }

    // Return the updated String
    return str;
}

// Swap char of a string
static String swap(String str, int i, int j) 
{ 
    char []ch = str.ToCharArray(); 
    char temp = ch[i]; 
    ch[i] = ch[j]; 
    ch[j] = temp;  
    return String.Join("", ch); 
}

// Driver Code
public static void Main(String[] args)
{
    
    // Given String str
    String str = "abcdef";

    // Given array of reversing index
    int []arr = { 1, 2, 3 };

    int K = arr.Length;

    // Function Call
    Console.Write(modifyString(arr, str, K));
}
}

// This code is contributed by Amit Katiyar 

<script>

// JavaScript program for the above approach


// Function to perform the reversal
// operation on the given String
function modifyString(A, str, K) {

    // Size of String
    str = str.split("")
    let N = str.length;

    // Stores the count of indices
    let count = new Array(N + 1).fill(0);

    // Count the positions where
    // reversals will begin
    for (let i = 0; i < K; i++) {
        count[A[i]] += 1;
    }

    for (let i = 1; i <= N / 2; i++) {

        // Store the count of reversals
        // beginning at position i
        count[i] = count[i] + count[i - 1];

        // Check if the count[i] is
        // odd the swap the character
        if (count[i] & 1) {
            let temp = str[i - 1];
            str[i - 1] = str[N - i];
            str[N - i] = temp
        }
    }

    // Return the updated String
    return str.join("");
}



// Driver Code

// Given String str
let str = "abcdef";

// Given array of reversing index
let arr = [1, 2, 3];

let K = arr.length;

// Function Call
document.write(modifyString(arr, str, K));

// This code is contributed by gfgking

</script>

fbdcea

Time Complexity: O(N + K), where N is the length of the string and K is the maximum length of the substring reversed. 
Auxiliary Space: O(N)