Given n strings, we need to sort those strings such that every string is a substring of all strings after it. If not possible to sort, then print the same.
Examples:
Input : {"d", "zddsaaz", "ds", "ddsaa", "dds"}
Output :
d
ds
dds
ddsaa
zddsaaz
Input : {"neveropen", "ee", "neveropen", "forneveropen", "ee"}
Output :
ee
ee
neveropen
forneveropen
neveropen
- Observation 1:
- If A substring of B
- Then length of A <= length of B
- Observation 2:
- If (A substring of B ) and (B substring of C)
- Then A substring of C
Solution:
Based on the two above observations, the solutions is as follows
- Sort all the string from the shorter to the longer
- Validate that each string is a substring of the following string
If we validate that each string is a substring of the following string then based on observation 2, each string is a substring of all the strings come after it.
Implementation:
C++
// C++ code to sort substrings#include <bits/stdc++.h>using namespace std;bool compare(string s1, string s2){ return s1.length() < s2.length();}void substringSort(string arr[], int n){ // sort the given array from shorter string to longer sort(arr, arr + n, compare); // validate that each // string is a substring of // the following one for (int i = 0; i < n - 1; i++) { if (arr[i + 1].find(arr[i]) == string::npos) { // the array can't be sorted cout << "Cannot be sorted"; return; } } // The array is valid and sorted // print the strings in order for (int i = 0; i < n-1; i++) { cout << arr[i] << endl; }}int main(){ // Test 1 string arr1[] = { "d", "zddsaaz", "ds", "ddsaa", "dds" }; substringSort(arr1, sizeof(arr1) / sizeof(arr1[0])); // Test 2 string arr2[] = { "for", "rof" }; substringSort(arr2, sizeof(arr2) / sizeof(arr2[0])); return 0;} |
Java
// Java code to sort substringsimport java.util.Arrays;import java.util.Comparator;public class Demo { public static void substringSort(String[] arr, int n) { // sort the given array from shorter string to longer Arrays.sort(arr, new Comparator<String>() { public int compare(String s1, String s2) { return Integer.compare(s1.length(), s2.length()); } }); // validate that each string is a substring of // the following one' for (int i = 0; i < n - 1; i++) { if (!arr[i + 1].contains(arr[i])) { // the array can't be sorted System.out.println("Cannot be sorted"); return; } } // The array is valid and sorted // print the strings in order for (int i = 0; i < n - 1; i++) { System.out.println(arr[i]); } } public static void main(String[] args) { // Test 1 String[] arr1 = { "d", "zddsaaz", "ds", "ddsaa", "dds" }; substringSort(arr1, arr1.length); // Test 2 String[] arr2 = { "for", "rof" }; substringSort(arr2, arr2.length); }} |
Python3
# Python code to sort substringsimport functoolsdef substringSort(arr, n): # sort the given array from shorter string to longer arr.sort(key=len) # validate that each string is a substring of # the following one for i in range(n - 1): if not arr[i + 1].find(arr[i]) != -1: # the array cannot be sorted print("Cannot be sorted") return # the array is valid and sorted # print the strings in order for i in range(n - 1): print(arr[i])# Test 1arr1 = ["d", "zddsaaz", "ds", "ddsaa", "dds"]substringSort(arr1, len(arr1))# Test 2arr2 = ["for", "rof"]substringSort(arr2, len(arr2)) |
C#
using System;using System.Linq;public class Demo{ public static void SubstringSort(string[] arr, int n) { // sort the given array from shorter string to longer Array.Sort(arr, (s1, s2) => s1.Length.CompareTo(s2.Length)); // validate that each string is a substring of // the following one' for (int i = 0; i < n - 1; i++) { if (!arr[i + 1].Contains(arr[i])) { // the array can't be sorted Console.WriteLine("Cannot be sorted"); return; } } // The array is valid and sorted // print the strings in order for (int i = 0; i < n - 1; i++) { Console.WriteLine(arr[i]); } } public static void Main(string[] args) { // Test 1 string[] arr1 = { "d", "zddsaaz", "ds", "ddsaa", "dds" }; SubstringSort(arr1, arr1.Length); // Test 2 string[] arr2 = { "for", "rof" }; SubstringSort(arr2, arr2.Length); }} |
Javascript
// JavaScript code to sort substringsfunction substringSort(arr, n) { // sort the given array from shorter string to longer arr.sort(function(s1, s2) { return s1.length - s2.length; }); // validate that each string is a substring of // the following one for (let i = 0; i < n - 1; i++) { if (!arr[i + 1].includes(arr[i])) { // the array can't be sorted console.log("Cannot be sorted"); return; } } // The array is valid and sorted // print the strings in order for (let i = 0; i < n - 1; i++) { console.log(arr[i]); }}// Test 1let arr1 = ["d", "zddsaaz", "ds", "ddsaa", "dds"];substringSort(arr1, arr1.length);// Test 2let arr2 = ["for", "rof"];substringSort(arr2, arr2.length); |
Output
d ds dds ddsaa Cannot be sorted
Time Complexity: O(n log n), where n is the number of strings.
Space Complexity: O(1) as no extra space has been taken.
Alternative approach:
For better time complexity, we can use counting sort only if the maximum length of the strings is specified.
Suppose that ‘maxLen’ is the maximum length of the input strings. In this case, the solution is as follows:
- Create array of length maxLen
- Sort the input strings such that the string with length 1 is in the first place in the array
- If there is two or more string has the same length they must be equal otherwise the strings cannot be sorted
- Validate that each string is a substring of the next longer string
Implementation:
C++
// C++ Program to sort substrings#include <bits/stdc++.h>using namespace std;void substringSort(string arr[], int n, int maxLen){ int count[maxLen]; string sortedArr[maxLen]; for(int i = 0; i < maxLen; i++) { count[i] = 0; sortedArr[i] = ""; } // sort the input array for (int i = 0; i < n; i++) { string s = arr[i]; int len = s.length(); if (count[len - 1] == 0) { sortedArr[len - 1] = s; count[len - 1] = 1; } else if (sortedArr[len - 1] == s) { // repeated length should // be the same string count[len - 1]++; } else { // two different strings with the same // length input array cannot be sorted cout << "Cannot be sorted"; return; } } // validate that each string is a // substring of the following one int index = 0; // get first element while (count[index] == 0) index++; int prev = index; string prevString = sortedArr[prev]; index++; for (; index < maxLen; index++) { if (count[index] != 0) { string current = sortedArr[index]; if (current.find(prevString) != string::npos) { prev = index; prevString = current; } else { cout << "Cannot be sorted"; return; } } } // The array is valid and sorted // print the strings in order for (int i = 0; i < maxLen; i++) { string s = sortedArr[i]; for (int j = 0; j < count[i]; j++) { cout << s << endl; } }}// Driver codeint main(){ int maxLen = 100; // Test 1 string arr1[] = { "d", "zddsaaz", "ds", "ddsaa", "dds", "dds" }; substringSort(arr1, sizeof(arr1)/sizeof(arr1[0]), maxLen); // Test 2 string arr2[] = { "for", "rof" }; substringSort(arr2, sizeof(arr2)/sizeof(arr2[0]), maxLen); return 0;}// This code is contributed by rameshtravel07. |
Java
// Alternative code to sort substringsimport java.util.Arrays;public class Demo { public static void substringSort(String[] arr, int n, int maxLen) { int count[] = new int[maxLen]; String[] sortedArr = new String[maxLen]; Arrays.fill(count, 0); Arrays.fill(sortedArr, ""); // sort the input array for (int i = 0; i < n; i++) { String s = arr[i]; int len = s.length(); if (count[len - 1] == 0) { sortedArr[len - 1] = s; count[len - 1] = 1; } else if (sortedArr[len - 1].equals(s)) { // repeated length should be the same string count[len - 1]++; } else { // two different strings with the same // length input array cannot be sorted System.out.println("Cannot be sorted"); return; } } // validate that each string is a substring // of the following one int index = 0; // get first element while (count[index] == 0) index++; int prev = index; String prevString = sortedArr[prev]; index++; for (; index < maxLen; index++) { if (count[index] != 0) { String current = sortedArr[index]; if (current.contains(prevString)) { prev = index; prevString = current; } else { System.out.println("Cannot be sorted"); return; } } } // The array is valid and sorted // print the strings in order for (int i = 0; i < maxLen; i++) { String s = sortedArr[i]; for (int j = 0; j < count[i]; j++) { System.out.println(s); } } } public static void main(String[] args) { int maxLen = 100; // Test 1 String[] arr1 = { "d", "zddsaaz", "ds", "ddsaa", "dds", "dds" }; substringSort(arr1, arr1.length, maxLen); // Test 2 String[] arr2 = { "for", "rof" }; substringSort(arr2, arr2.length, maxLen); }} |
Python3
# Alternative code to sort substringsdef substringSort(arr, n, maxLen): count = [0]*(maxLen) sortedArr = [""]*(maxLen) # sort the input array for i in range(n): s = arr[i] Len = len(s) if (count[Len - 1] == 0) : sortedArr[Len - 1] = s count[Len - 1] = 1 elif (sortedArr[Len - 1] == s): # repeated length should be the same string count[Len - 1]+=1 else: # two different strings with the same # length input array cannot be sorted print("Cannot be sorted") return # validate that each string is a substring # of the following one index = 0 # get first element while (count[index] == 0): index+=1 prev = index prevString = sortedArr[prev] index+=1 while index < maxLen: if (count[index] != 0) : current = sortedArr[index] if (prevString in current): prev = index prevString = current else: print("Cannot be sorted") return index+=1 # The array is valid and sorted # print the strings in order for i in range(maxLen): s = sortedArr[i] for j in range(count[i]): print(s)maxLen = 100 # Test 1arr1 = [ "d", "zddsaaz", "ds", "ddsaa", "dds", "dds" ]substringSort(arr1, len(arr1), maxLen)# Test 2arr2 = [ "for", "rof" ]substringSort(arr2, len(arr2), maxLen)# This code is contributed by mukesh07. |
C#
// C# Program to sort substringsusing System;class GFG{ public static void substringSort(String[] arr, int n, int maxLen) { int []count = new int[maxLen]; String[] sortedArr = new String[maxLen]; for(int i = 0; i < maxLen; i++) { count[i] = 0; sortedArr[i] = ""; } // sort the input array for (int i = 0; i < n; i++) { String s = arr[i]; int len = s.Length; if (count[len - 1] == 0) { sortedArr[len - 1] = s; count[len - 1] = 1; } else if (ReferenceEquals(s,sortedArr[len - 1])) { // repeated length should // be the same string count[len - 1]++; } else { // two different strings with the same // length input array cannot be sorted Console.WriteLine("Cannot be sorted"); return; } } // validate that each string is a // substring of the following one int index = 0; // get first element while (count[index] == 0) index++; int prev = index; String prevString = sortedArr[prev]; index++; for (; index < maxLen; index++) { if (count[index] != 0) { String current = sortedArr[index]; if (current.Contains(prevString)) { prev = index; prevString = current; } else { Console.WriteLine("Cannot be sorted"); return; } } } // The array is valid and sorted // print the strings in order for (int i = 0; i < maxLen; i++) { String s = sortedArr[i]; for (int j = 0; j < count[i]; j++) { Console.WriteLine(s); } } } public static void Main(String[] args) { int maxLen = 100; // Test 1 String[] arr1 = { "d", "zddsaaz", "ds", "ddsaa", "dds", "dds" }; substringSort(arr1, arr1.Length, maxLen); // Test 2 String[] arr2 = { "for", "rof" }; substringSort(arr2, arr2.Length, maxLen); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script> // Alternative code to sort substrings function substringSort(arr, n, maxLen) { let count = new Array(maxLen); let sortedArr = new Array(maxLen); count.fill(0); for(let i = 0; i < maxLen; i++) { sortedArr[i] = ""; } // sort the input array for (let i = 0; i < n; i++) { let s = arr[i]; let len = s.length; if (count[len - 1] == 0) { sortedArr[len - 1] = s; count[len - 1] = 1; } else if (sortedArr[len - 1] == s) { // repeated length should be the same string count[len - 1]++; } else { // two different strings with the same // length input array cannot be sorted document.write("Cannot be sorted" + "</br>"); return; } } // validate that each string is a substring // of the following one let index = 0; // get first element while (count[index] == 0) index++; let prev = index; let prevString = sortedArr[prev]; index++; for (; index < maxLen; index++) { if (count[index] != 0) { let current = sortedArr[index]; if (current.includes(prevString)) { prev = index; prevString = current; } else { document.write("Cannot be sorted"); return; } } } // The array is valid and sorted // print the strings in order for (let i = 0; i < maxLen; i++) { let s = sortedArr[i]; for (let j = 0; j < count[i]; j++) { document.write(s + "</br>"); } } } let maxLen = 100; // Test 1 let arr1 = [ "d", "zddsaaz", "ds", "ddsaa", "dds", "dds" ]; substringSort(arr1, arr1.length, maxLen); // Test 2 let arr2 = [ "for", "rof" ]; substringSort(arr2, arr2.length, maxLen);// This code is contributed by decode2207.</script> |
Output
d ds dds dds ddsaa zddsaaz Cannot be sorted
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