Reorder the position of the words in alphabetical order

Given an array arr[] of strings, the task is to reorder the strings lexicographically and print their positions in the original list.

Examples:

Input: arr[] = {“zxc”, “efg”, “jkl”} 
Output: 2 3 1 
The sorted list will be {“efg”, “jkl”, “zxc”} and their 
original positions were 2, 3 and 1 respectively.

Input: arr[] = {“live”, “place”, “travel”, “word”, “sky”} 
Output: 1 2 5 3 4  

Approach: Assign all the words with an integer number equal to their position in the array. Then they sort the list of words lexicographically and their positions are altered, and therefore, their positions are printed starting from the first word in the sorted list.

Below is the implementation of the above approach: 

C++

// CPP implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to print the ordering of words
void reArrange(string words[], int n)
{

    // Creating list of words and assigning
    // them index numbers
    map<string, int> mp;
    for (int i = 0; i < n; i++)
        mp[words[i]] = i + 1;

    // Sort the list of words
    // lexicographically
    sort(words, words + n);

    // Print the ordering
    for (int i = 0; i < n; i++)
        cout << mp[words[i]] << " ";
}

// Driver Code
int main()
{
    string words[] = { "live", "place", "travel", "word", "sky" };
    int n = sizeof(words) / sizeof(words[0]);
    reArrange(words, n);
}

// This code is contributed by
// Surendra_Gangwar

Java

// Java implementation of the approach
import java.util.*;
class GFG {

    // Function to print the ordering of words
    static void reArrange(String words[], int n)
    {

        // Creating list of words and assigning
        // them index numbers
        HashMap<String, Integer> freq = new HashMap<>();
        for (int i = 0; i < n; i++) {
            freq.put(words[i], (i + 1));
        }

        // Sort the list of words
        // lexicographically
        Arrays.sort(words);

        // Print the ordering
        for (int i = 0; i < n; i++)
            System.out.print(freq.get(words[i]) + " ");
    }

    // Driver Code
    public static void main(String[] args)
    {
        String words[] = { "live", "place", "travel", "word", "sky" };
        int n = words.length;
        reArrange(words, n);
    }
}

Python3

# Python3 implementation of the approach

# Function to print the ordering of words
def reArrange(words, n):
    # Creating list of words and assigning
    # them index numbers
    mp = {}
    for i in range(n):
        mp[words[i]] = i + 1

    # Sort the list of words
    # lexicographically
    words.sort();

    # Print the ordering
    for i in range(n):
        print(mp[words[i]], end = " ")

# Driver Code

words = [ "live", "place", "travel", "word", "sky" ]
n = len(words)
reArrange(words, n);

# This code is contributed by
# Rajnis09

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG {

    // Function to print the ordering of words
    static void reArrange(String[] words, int n)
    {

        // Creating list of words and assigning
        // them index numbers
        Dictionary<String, int> freq = new Dictionary<String, int>();
        for (int i = 0; i < n; i++) {
            freq.Add(words[i], (i + 1));
        }

        // Sort the list of words
        // lexicographically
        Array.Sort(words);

        // Print the ordering
        for (int i = 0; i < n; i++)
            Console.Write(freq[words[i]] + " ");
    }

    // Driver Code
    public static void Main(String[] args)
    {
        String[] words = { "live", "place", "travel", "word", "sky" };
        int n = words.Length;
        reArrange(words, n);
    }
}

// This code contributed by Rajput-Ji

PHP

<?php
// PHP implementation of the approach

    // Function to print the ordering of words
    function reArrange($words, $n)
    {

        // Creating list of words and assigning
        // them index numbers
        $freq = array();
        for ($i = 0; $i < $n; $i++)
        {
            $freq[$words[$i]] = ($i + 1) ;
        }

        // Sort the list of words
        // lexicographically
        sort($words);

        // Print the ordering
        for ($i = 0; $i < $n; $i++)
            echo $freq[$words[$i]], " " ;
    }

    // Driver Code
    $words = array( "live", "place", "travel", "word", "sky" );
    $n = count($words);
    reArrange($words, $n);

    // This code is contributed by Ryuga
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to print the ordering of words
function reArrange(words, n)
{

    // Creating list of words and assigning
    // them index numbers
    var mp = new Map();
    for (var i = 0; i < n; i++)
        mp.set(words[i], i + 1);

    // Sort the list of words
    // lexicographically
    words.sort();

    // Print the ordering
    for (var i = 0; i < n; i++)
    {
        document.write(mp.get(words[i])+" ");
    }
}

// Driver Code
var words = ["live", "place", "travel", "word", "sky"];
var n = words.length;
reArrange(words, n);


</script>

Output

1 2 5 3 4 

Time Complexity: O(n * log n)

Auxiliary Space: O(n)

Approach:

The approach is to create a vector of integers representing the original indices of the strings, sort the vector of indices based on the strings using a lambda function that compares the corresponding strings, and then print the sorted vector of indices.

Implementation of the above approach:

C++

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

bool compare(string a, string b) { return a < b; }

void reArrange(string words[], int n)
{
    // Create a vector of indices
    vector<int> indices(n);
    for (int i = 0; i < n; i++) {
        indices[i] = i + 1;
    }

    // Sort the vector of indices based on the strings
    sort(indices.begin(), indices.end(), [&](int a, int b) {
        return compare(words[a - 1], words[b - 1]);
    });

    // Print the sorted vector of indices
    for (auto index : indices) {
        cout << index << " ";
    }
}

int main()
{
    string words[]
        = { "live", "place", "travel", "word", "sky" };
    int n = sizeof(words) / sizeof(words[0]);
    reArrange(words, n);
    return 0;
}

Java

import java.util.*;

class GFG {
    public static boolean compare(String a, String b)
    {
        return a.compareTo(b) < 0;
    }

    public static void reArrange(String[] words)
    {

        // Create an array of indices
        Integer[] indices = new Integer[words.length];
        for (int i = 0; i < words.length; i++) {
            indices[i] = i + 1;
        }

        // Sort the array of indices based on the strings
        Arrays.sort(
            indices,
            (a, b)
                -> compare(words[a - 1], words[b - 1]) ? -1
                                                       : 1);

        // Print the sorted array of indices
        for (Integer index : indices) {
            System.out.print(index + " ");
        }
    }

    public static void main(String[] args)
    {
        String[] words
            = { "live", "place", "travel", "word", "sky" };
        reArrange(words);
    }
}

Python3

def reArrange(words):
    n = len(words)

    # Create a list of indices
    indices = [i+1 for i in range(n)]

    # Sort the list of indices based on the strings
    indices.sort(key=lambda x: words[x-1])

    # Print the sorted list of indices
    for index in indices:
        print(index, end=' ')

words = ["live", "place", "travel", "word", "sky"]
reArrange(words)

C#

using System;
using System.Linq;

public class GFG {

    static bool Compare(string a, string b)
    {
        return a.CompareTo(b) < 0;
    }

    static void reArrange(string[] words, int n)
    {

        // Create an array of indices
        int[] indices = Enumerable.Range(1, n).ToArray();

        // Sort the array of indices based on the strings
        Array.Sort(indices, (a, b) = > Compare(words[a - 1],
                                               words[b - 1])
                                         ? -1
                                         : 1);

        // Print the sorted array of indices
        Console.WriteLine(string.Join(" ", indices));
    }

    public static void Main()
    {
        string[] words
            = { "live", "place", "travel", "word", "sky" };
        int n = words.Length;
        reArrange(words, n);
    }
}
// This code is contributed by prasad264

Javascript

function compare(a, b) {
  return a.localeCompare(b) < 0;
}

function reArrange(words) {
  // Create an array of indices
  let indices = Array.from({ length: words.length }, (_, i) => i + 1);

  // Sort the array of indices based on the strings
  indices.sort((a, b) => (compare(words[a - 1], words[b - 1]) ? -1 : 1));

  // Print the sorted array of indices
  let result = "";
  for (let index of indices) {
    result += index + " ";
  }
  console.log(result.trim());
}

let words = ["live", "place", "travel", "word", "sky"];
reArrange(words);

Output

1 2 5 3 4 

Time Complexity: O(N log N)

Auxiliary Space: O(1)