Sort the Array of Strings on the basis of given substring range
Last Updated :
29 Mar, 2023
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;
}
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);
}
}
# 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 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)
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