Skip to content
Related Articles

Related Articles

Improve Article

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

  • Difficulty Level : Basic
  • Last Updated : 31 May, 2021

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

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :