Reverse the substrings of the given String according to the given Array of indices

Given a string S and an array of indices A[], the task is to reverse the substrings of the given string according to the given Array of indices.
Note: A[i] ≤ length(S), for all i.
Examples:

Input: S = “abcdef”, A[] = {2, 5} 
Output: baedcf 
Explanation:

Input: S = “abcdefghij”, A[] = {2, 5} 
Output: baedcjihgf 

Approach: The idea is to use the concept of reversing the substrings of the given string. 

• Sort the Array of Indices.
• Extract the substring formed for each index in the given array as follows: 
  • For the first index in the array A, the substring formed will be from index 0 to A[0] (exclusive) of the given string, i.e. [0, A[0])
  • For all other index in the array A (except for last), the substring formed will be from index A[i] to A[i+1] (exclusive) of the given string, i.e. [A[i], A[i+1])
  • For the last index in the array A, the substring formed will be from index A[i] to L (inclusive) where L is the length of the string, i.e. [A[i], L]
• Reverse each substring found in the given string

Below is the implementation of the above approach.

C++

// C++ implementation to reverse
// the substrings of the given String
// according to the given Array of indices

#include <bits/stdc++.h>
using namespace std;

// Function to reverse a string
void reverseStr(string& str,
                int l, int h)
{
    int n = h - l;

    // Swap character starting
    // from two corners
    for (int i = 0; i < n / 2; i++) {
        swap(str[i + l], str[n - i - 1 + l]);
    }
}

// Function to reverse the string
// with the given array of indices
void reverseString(string& s, int A[], int n)
{

    // Reverse the string from 0 to A[0]
    reverseStr(s, 0, A[0]);

    // Reverse the string for A[i] to A[i+1]
    for (int i = 1; i < n; i++)
        reverseStr(s, A[i - 1], A[i]);

    // Reverse String for A[n-1] to length
    reverseStr(s, A[n - 1], s.length());
}

// Driver Code
int main()
{
    string s = "abcdefgh";
    int A[] = { 2, 4, 6 };
    int n = sizeof(A) / sizeof(A[0]);

    reverseString(s, A, n);
    cout << s;

    return 0;
}

Java

// Java implementation to reverse
// the subStrings of the given String
// according to the given Array of indices
class GFG
{

static String s;

// Function to reverse a String
static void reverseStr(int l, int h)
{
    int n = h - l;

    // Swap character starting
    // from two corners
    for (int i = 0; i < n / 2; i++)
    {
        s = swap(i + l, n - i - 1 + l);
    }
}

// Function to reverse the String
// with the given array of indices
static void reverseString(int A[], int n)
{

    // Reverse the String from 0 to A[0]
    reverseStr(0, A[0]);

    // Reverse the String for A[i] to A[i+1]
    for (int i = 1; i < n; i++)
        reverseStr(A[i - 1], A[i]);

    // Reverse String for A[n-1] to length
    reverseStr(A[n - 1], s.length());
}
static String swap(int i, int j)
{
    char ch[] = s.toCharArray();
    char temp = ch[i];
    ch[i] = ch[j];
    ch[j] = temp;
    return String.valueOf(ch);
}

// Driver Code
public static void main(String[] args)
{
    s = "abcdefgh";
    int A[] = { 2, 4, 6 };
    int n = A.length;

    reverseString(A, n);
    System.out.print(s);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to reverse
# the substrings of the given String
# according to the given Array of indices

# Function to reverse a string
def reverseStr(str, l, h):
    n = h - l

    # Swap character starting
    # from two corners
    for i in range(n//2):
        str[i + l], str[n - i - 1 + l] = str[n - i - 1 + l], str[i + l]

# Function to reverse the string
# with the given array of indices
def reverseString(s, A, n):

    # Reverse the from 0 to A[0]
    reverseStr(s, 0, A[0])

    # Reverse the for A[i] to A[i+1]
    for i in range(1, n):
        reverseStr(s, A[i - 1], A[i])

    # Reverse String for A[n-1] to length
    reverseStr(s, A[n - 1], len(s))

# Driver Code
s = "abcdefgh"
s = [i for i in s]
A = [2, 4, 6]
n = len(A)

reverseString(s, A, n)
print("".join(s))

# This code is contributed by mohit kumar 29

C#

// C# implementation to reverse
// the subStrings of the given String
// according to the given Array of indices
using System;

class GFG
{

static String s;

// Function to reverse a String
static void reverseStr(int l, int h)
{
    int n = h - l;

    // Swap character starting
    // from two corners
    for (int i = 0; i < n / 2; i++)
    {
        s = swap(i + l, n - i - 1 + l);
    }
}

// Function to reverse the String
// with the given array of indices
static void reverseString(int []A, int n)
{

    // Reverse the String from 0 to A[0]
    reverseStr(0, A[0]);

    // Reverse the String for A[i] to A[i+1]
    for (int i = 1; i < n; i++)
        reverseStr(A[i - 1], A[i]);

    // Reverse String for A[n-1] to length
    reverseStr(A[n - 1], s.Length);
}

static String swap(int i, int j)
{
    char []ch = s.ToCharArray();
    char temp = ch[i];
    ch[i] = ch[j];
    ch[j] = temp;
    return String.Join("",ch);
}

// Driver Code
public static void Main(String[] args)
{
    s = "abcdefgh";
    int []A = { 2, 4, 6 };
    int n = A.Length;

    reverseString(A, n);
    Console.Write(s);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript implementation to reverse
// the substrings of the given String
// according to the given Array of indices

// Function to reverse a string
function reverseStr(str, l, h)
{
    var n = h - l;

    // Swap character starting
    // from two corners
    for (var i = 0; i < n / 2; i++) {
        [str[i + l], str[n - i - 1 + l]] =
        [str[n - i - 1 + l], str[i + l]];
    }
    return str;
}

// Function to reverse the string
// with the given array of indices
function reverseString(s, A, n)
{

    // Reverse the string from 0 to A[0]
    s = reverseStr(s, 0, A[0]);

    // Reverse the string for A[i] to A[i+1]
    for (var i = 1; i < n; i++)
        s = reverseStr(s, A[i - 1], A[i]);

    // Reverse String for A[n-1] to length
    s = reverseStr(s, A[n - 1], s.length);
    return s;
}

// Driver Code
var s = "abcdefgh";
var A = [2, 4, 6];
var n = A.length;
s = reverseString(s.split(''), A, n);
document.write( s.join(''));

</script>

Output:

badcfehg

Time Complexity: O(l * n), where l is the length of the given string and n is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.