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 ith 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 ith 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)