Sort the Array of Strings on the basis of given substring range

Last Updated : 15 Jul, 2025

Given two positive integers I and X and an array of strings arr[], the task is to sort the given array of strings on the basis of substrings starting from index I of size X.

Examples:

Input: I = 2, X = 2, arr[] = { "baqwer", "zacaeaz", "aaqzzaa", "aacaap", "abbatyo", "bbbacztr", "bbbdaaa" }
Output: abbatyo bbbacztr bbbdaaa aacaap zacaeaz baqwer aaqzzaa
Explanation:
All sub-strings starting from index I = 2 and of size x = 2 are {"qw", "ca", "qz", "ca", "ba", "ba", "bd"}.
Sorting them in lexicographical increasing order gives {"ba", "ba", "bd", "ca", "ca", "qw", "qz" }, then print the corresponding original string in this order.

Input: I = 1, X = 3, arr[] = { "submit", "source", "skills", "epidemic", "ample", "apple" }
Output: skills ample source epidemic apple submit

Approach: The idea is to create a substring of all the strings in the given array from index I of size X and keep the count of pair of a substring with the corresponding string in a map of pairs. After inserting in the map of pairs. After inserting, traverse the map and print the string.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to sort the given array
// of strings based on substring
void sortArray(vector<string> s,
               int l, int x)
{
    // Map of pairs to sort vector
    // of strings
    map<pair<string, string>, int> mp;

    for (int i = 0; i < s.size(); i++) {

        // Create substring from index
        // 'l' and of size 'X'
        string part = s[i].substr(l, x);

        // Insert in Map
        mp[{ part, s[i] }] += 1;
    }

    // Print the sorted vector of strings
    for (auto it = mp.begin();
         it != mp.end(); ++it) {

        // Traverse the number of time
        // a string is present
        for (int j = 0; j < it->second; j++) {

            // Print the string
            cout << it->first.second << ' ';
        }
    }
}

// Driver Code
int main()
{
    // Given array of strings
    vector<string> arr;
    arr = { "baqwer", "zacaeaz", "aaqzzaa",
            "aacaap", "abbatyo", "bbbacztr",
            "bbbdaaa" };

    // Given I and X
    int I = 2, X = 2;

    // Function Call
    sortArray(arr, I, X);
    return 0;
}

Java

import java.util.*;

public class Main {
    // Function to sort the given array
    // of strings based on substring
    static void sortArray(String[] s, int l, int x) {
        // Map of pairs to sort array of strings
        Map<String, Integer> mp = new TreeMap<>();

        for (int i = 0; i < s.length; i++) {
            // Create substring from index 'l' and of size 'x'
            String part = s[i].substring(l, l + x);

            // Insert in Map
            if (mp.containsKey(part + s[i])) {
                mp.put(part + s[i], mp.get(part + s[i]) + 1);
            } else {
                mp.put(part + s[i], 1);
            }
        }

        // Print the sorted array of strings
        for (String key : mp.keySet()) {
            // Traverse the number of time a string is present
            for (int j = 0; j < mp.get(key); j++) {
                // Print the string
                System.out.print(key.substring(x) + " ");
            }
        }
    }

    // Driver Code
    public static void main(String[] args) {
        // Given array of strings
        String[] arr = { "baqwer", "zacaeaz", "aaqzzaa",
                         "aacaap", "abbatyo", "bbbacztr",
                         "bbbdaaa" };

        // Given I and X
        int I = 2, X = 2;

        // Function Call
        sortArray(arr, I, X);
    }
}

Python3

# Python program for the above approach

# Function to sort the given array
# of strings based on substring
def sortArray(s, l, x):
    
    # Map of pairs to sort vector
    # of strings
    mp = {}

    for i in range(len(s)):
        
        # Create substring from index
        # 'l' and of size 'X'
        part = s[i][l:l+x]

        # Insert in Map
        if (part, s[i]) in mp:
            mp[(part, s[i])] += 1
        else:
            mp[(part, s[i])] = 1

    # Print the sorted vector of strings
    for key, value in sorted(mp.items()):
        
        # Traverse the number of time
        # a string is present
        for j in range(value):
            
            # Print the string
            print(key[1], end=" ")

# Driver Code
if __name__ == "__main__":
    
    # Given array of strings
    arr = ["baqwer", "zacaeaz", "aaqzzaa",
           "aacaap", "abbatyo", "bbbacztr",
           "bbbdaaa"]

    # Given I and X
    I = 2
    X = 2

    # Function Call
    sortArray(arr, I, X)

JavaScript

// JavaScript code for the above approach
const sortArray = (s, l, x) => {
    // Map of pairs to sort array of strings
    let mp = new Map();

    for (let i = 0; i < s.length; i++) {
        // Create substring from index 'l' and of size 'x'
        let part = s[i].substring(l, l + x);

        // Insert in Map
        if (!mp.has(part + s[i])) {
            mp.set(part + s[i], 1);
        } else {
            mp.set(part + s[i], mp.get(part + s[i]) + 1);
        }
    }

    // Print the sorted array of strings
    let sortedArray = Array.from(mp.keys()).sort();
    for (let i = 0; i < sortedArray.length; i++) {
        // Traverse the number of time a string is present
        for (let j = 0; j < mp.get(sortedArray[i]); j++) {
            // Print the string
            console.log(sortedArray[i].substring(x));
        }
    }
};

// Given array of strings
let arr = [    "baqwer",    "zacaeaz",    "aaqzzaa",    "aacaap",    "abbatyo",    "bbbacztr",    "bbbdaaa",];

// Given I and X
let I = 2,
    X = 2;

// Function Call
sortArray(arr, I, X);

// C# code for the approach

using System;
using System.Collections.Generic;
using System.Linq;

class GFG {
    // Function to sort the given array
    // of strings based on substring
    static void SortArray(List<string> s, int l, int x)
    {
        // Map of pairs to sort vector
        // of strings
        Dictionary<Tuple<string, string>, int> mp = new Dictionary<Tuple<string, string>, int>();

        for (int i = 0; i < s.Count; i++)
        {
            // Create substring from index 'l' and of size 'x'
            string part = s[i].Substring(l, x);

            // Insert in Map
            var key = new Tuple<string, string>(part, s[i]);
            if (mp.ContainsKey(key))
            {
                mp[key]++;
            }
            else
            {
                mp[key] = 1;
            }
        }

        // Print the sorted vector of strings
        foreach (var item in mp.OrderBy(kvp => kvp.Key))
        {
            // Traverse the number of times a string is present
            for (int j = 0; j < item.Value; j++)
            {
                // Print the string
                Console.Write(item.Key.Item2 + " ");
            }
        }
    }

    // Driver Code
    static void Main(string[] args)
    {
        // Given array of strings
        List<string> arr = new List<string> {
            "baqwer",
            "zacaeaz",
            "aaqzzaa",
            "aacaap",
            "abbatyo",
            "bbbacztr",
            "bbbdaaa"
        };

        // Given I and X
        int I = 2, X = 2;

        // Function Call
        SortArray(arr, I, X);
    }
}

Output:

abbatyo bbbacztr bbbdaaa aacaap zacaeaz baqwer aaqzzaa

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

Chauhanvishesh99

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