Sort an array of strings having characters at odd and even indices sorted in decreasing and increasing order respectively

Given an array arr[] consisting of N strings, the task is to first sort the characters at odd and even indices of the string in decreasing and increasing order respectively, for each string of the array and then, sort the modified array.

Examples:

Input: arr[] = {"ball", "bat", "boy"}
Output: albl atb byo
Explanation:
S[0]="ball" is converted to "albl"
S[1]="bat" is converted to "atb"
S[2]="boy" is converted to "byo"
Sorted sequence of the modified array is: albl atb byo.

Input: arr[] = {"geeks", "gfg", "hello", "world"}
Output: dwlro eohll esekg fgg

Approach: The given problem can be solved by traversing the given array arr[] of strings and for each string firstly sort the string and then rearrange each string such that characters at even indices are in ascending order and characters at odd indices are in descending order. In the end, sort the new array of strings in ascending order. Follow the steps below to solve the problem:

• Traverse the given array of strings arr[] and for each string perform the following steps:
  • Sort the string arr[i] in alphabetical order.
  • Initialize a variable, say temp as "" that stores the resultant string.
  • Traverse the string arr[i] for the half-length of the string arr[i] and store the i^th index from the start and the end of the string to the string temp.
  • If the size of the string is odd, then add the middle character of the string arr[i] to the end of the variable temp.
  • After the above steps, update the string at the i^th index of the array arr[i] to the string temp.
• After completing the above steps, sort the new array of strings in ascending order and print the array of strings arr[].

Below is the implementation of the above approach.

// C++ program for the above approach

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

// Function to sort even indices in
// ascending order and odd indices in
// descending order and sort the array
// of strings in ascending order
void sortString(string S[], int N)
{
    // Traverse array of strings
    for (int i = 0; i < N; i++) {

        // Sort string in ascending order
        sort(S[i].begin(), S[i].end());

        // Length of string
        int n = S[i].size();

        string temp = "";

        // Traverse the string
        for (int j = 0; j < n / 2; j++) {
            temp += S[i][j];
            temp += S[i][n - j - 1];
        }

        // If length of string is odd
        if (n & 1)
            temp += S[i][n / 2];

        S[i] = temp;
    }

    // Sort array of strings
    sort(S, S + N);

    // Print array of strings
    for (int i = 0; i < N; i++) {
        cout << S[i] << " ";
    }
}

// Driver Code
int main()
{
    string arr[] = { "ball", "bat", "boy" };
    int N = sizeof(arr) / sizeof(arr[0]);
    sortString(arr, N);

    return 0;
}

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

class GFG{

static String sortString(String inputString) 
{ 
    
    // Convert input string to char array 
    char tempArray[] = inputString.toCharArray(); 
      
    // Sort tempArray 
    Arrays.sort(tempArray); 
      
    // Return new sorted string 
    return new String(tempArray); 
} 

// Function to sort even indices in
// ascending order and odd indices in
// descending order and sort the array
// of Strings in ascending order
static void sortString(String S[], int N)
{
    
    // Traverse array of Strings
    for(int i = 0; i < N; i++)
    {
        
        // Sort String in ascending order
        S[i] = sortString(S[i]);

        // Length of String
        int n = S[i].length();

        String temp = "";

        // Traverse the String
        for(int j = 0; j < n / 2; j++)
        {
            temp += S[i].charAt(j);
            temp += S[i].charAt(n - j - 1);
        }

        // If length of String is odd
        if (n %2== 1)
            temp += S[i].charAt(n / 2);

        S[i] = temp;
    }

    // Sort array of Strings
    Arrays.sort(S);

    // Print array of Strings
    for(int i = 0; i < N; i++)
    {
        System.out.print(S[i] + " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    String arr[] = { "ball", "bat", "boy" };
    int N = arr.length;
    
    sortString(arr, N);
}
}

// This code is contributed by Amit Katiyar

# Python3 program for the above approach

# Function to sort even indices in
# ascending order and odd indices in
# descending order and sort the array
# of strings in ascending order
def sortString(S, N):
    
    # Traverse array of strings
    for i in range(N):
        
        # Sort string in ascending order
        S[i] = ''.join(sorted(S[i]))

        # Length of string
        n = len(S[i])

        temp = ""

        # Traverse the string
        for j in range(n // 2):
            temp += S[i][j]
            temp += S[i][n - j - 1]

        # If length of string is odd
        if (n & 1):
            temp += S[i][n // 2]

        S[i] = temp

    # Sort array of strings
    S.sort()

    # Print array of strings
    for i in range(N):
        print(S[i], end = " ")

# Driver Code
if __name__ == '__main__':
    
    arr = ["ball", "bat", "boy"]
    N = len(arr)
    
    sortString(arr, N)

# This code is contributed by ipg2016107

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

public class GFG{

static String sortString(String inputString) 
{ 
    
    // Convert input string to char array 
    char []tempArray = inputString.ToCharArray(); 
      
    // Sort tempArray 
    Array.Sort(tempArray); 
      
    // Return new sorted string 
    return new String(tempArray); 
} 

// Function to sort even indices in
// ascending order and odd indices in
// descending order and sort the array
// of Strings in ascending order
static void sortString(String []S, int N)
{
    
    // Traverse array of Strings
    for(int i = 0; i < N; i++)
    {
        
        // Sort String in ascending order
        S[i] = sortString(S[i]);

        // Length of String
        int n = S[i].Length;

        String temp = "";

        // Traverse the String
        for(int j = 0; j < n / 2; j++)
        {
            temp += S[i][j];
            temp += S[i][n - j - 1];
        }

        // If length of String is odd
        if (n %2== 1)
            temp += S[i][n / 2];

        S[i] = temp;
    }

    // Sort array of Strings
    Array.Sort(S);

    // Print array of Strings
    for(int i = 0; i < N; i++)
    {
        Console.Write(S[i] + " ");
    }
}

// Driver Code
public static void Main(String[] args)
{
    String []arr = { "ball", "bat", "boy" };
    int N = arr.Length;
    
    sortString(arr, N);
}
}

// This code is contributed by Amit Katiyar 

<script>

// JavaScript program for the above approach

// Function to sort even indices in
// ascending order and odd indices in
// descending order and sort the array
// of strings in ascending order
function sortString(S, N)
{

    // Traverse array of strings
    for (let i = 0; i < N; i++) {

        // Sort string in ascending order
        S[i] = S[i].split("").sort().join("");

        // Length of string
        let n = S[i].length;

        let temp = "";

        // Traverse the string
        for (let j = 0; j < Math.floor(n / 2); j++) {
            temp += S[i][j];
            temp += S[i][n - j - 1];
        }

        // If length of string is odd
        if (n & 1)
            temp += S[i][(Math.floor(n / 2))];

        S[i] = temp;
    }

    // Sort array of strings
    S.sort();

    // Print array of strings
    for (let i = 0; i < N; i++) {
        document.write(S[i]," ");
    }
}

// Driver Code
let arr = [ "ball", "bat", "boy" ];
let N = arr.length;
sortString(arr, N);

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

albl atb byo

Time Complexity: O(N*log N)
Auxiliary Space: O(1)

subhammahato348

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Article Tags :

• Strings
• Sorting
• Mathematical
• DSA

Practice Tags :

• Mathematical
• Sorting
• Strings