Given a string S of length N and integer K, find the smallest length string which contains the string S as a sub string exactly K times.
Examples:
Input: S = “abba”, K = 3
Output: abbabbabba
Explanation: The string “abba” occurs K times in the string abbabbabba, i.e. {abbabbabba, abbabbabba, abbabbabba}Input: S = “neveropen”, K = 3
Output: “neveropenforneveropen”
Approach: To optimize the above approach, find the Longest Proper Prefix which is also a suffix for the given string S, and then generate a substring of S excluding the longest common prefix and add this substring to the answer exactly K – 1 times to the original string. Follow the below steps to solve the problem:
- Find the length of the longest proper prefix using KMP algorithm.
- Append substring S.substring(N-lps[N-1]) to S, exactly K-1 times.
- Print the answer.
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// KMP algorithmint* kmp(string& s){ int n = s.size(); int* lps = new int[n]; lps[0] = 0; int i = 1, len = 0; while (i < n) { if (s[i] == s[len]) { len++; lps[i] = len; i++; } else { if (len != 0) { len = lps[len - 1]; } else { lps[i] = 0; i++; } } } return lps;}// Function to return the required stringstring findString(string& s, int k){ int n = s.length(); // Finding the longest proper prefix // which is also suffix int* lps = kmp(s); // ans string string ans = ""; string suff = s.substr(0, n - lps[n - 1]); for (int i = 0; i < k - 1; ++i) { // Update ans appending the // substring K - 1 times ans += suff; } // Append the original string ans += s; // Returning min length string // which contain exactly k // substring of given string return ans;}// Driver Codeint main(){ int k = 3; string s = "neveropen"; cout << findString(s, k) << endl;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{ // KMP algorithmstatic int[] kmp(String s){ int n = s.length(); int[] lps = new int[n]; lps[0] = 0; int i = 1, len = 0; while (i < n) { if (s.charAt(i) == s.charAt(len)) { len++; lps[i] = len; i++; } else { if (len != 0) { len = lps[len - 1]; } else { lps[i] = 0; i++; } } } return lps;}// Function to return the required stringstatic String findString(String s, int k){ int n = s.length(); // Finding the longest proper prefix // which is also suffix int[] lps = kmp(s); // ans string String ans = ""; String suff = s.substring(0, n - lps[n - 1]); for(int i = 0; i < k - 1; ++i) { // Update ans appending the // substring K - 1 times ans += suff; } // Append the original string ans += s; // Returning min length string // which contain exactly k // substring of given string return ans;}// Driver codepublic static void main (String[] args){ int k = 3; String s = "neveropen"; System.out.println(findString(s, k));}}// This code is contributed by offbeat |
Python3
# Python3 program to implement# the above approach# KMP algorithmdef kmp(s): n = len(s) lps = [None] * n lps[0] = 0 i, Len = 1, 0 while (i < n): if (s[i] == s[Len]): Len += 1 lps[i] = Len i += 1 else: if (Len != 0): Len = lps[Len - 1] else: lps[i] = 0 i += 1 return lps # Function to return the required stringdef findString(s, k): n = len(s) # Finding the longest proper prefix # which is also suffix lps = kmp(s) # ans string ans = "" suff = s[0: n - lps[n - 1] : 1] for i in range(k - 1): # Update ans appending the # substring K - 1 times ans += suff # Append the original string ans += s # Returning min length string # which contain exactly k # substring of given string return ans# Driver codek = 3 s = "neveropen" print(findString(s, k))# This code is contributed by divyeshrabadiya07 |
C#
// C# program to implement// the above approachusing System;class GFG{ // KMP algorithmstatic int[] kmp(string s){ int n = s.Length; int[] lps = new int[n]; lps[0] = 0; int i = 1, len = 0; while (i < n) { if (s[i] == s[len]) { len++; lps[i] = len; i++; } else { if (len != 0) { len = lps[len - 1]; } else { lps[i] = 0; i++; } } } return lps;}// Function to return the required stringstatic string findString(string s, int k){ int n = s.Length; // Finding the longest proper prefix // which is also suffix int[] lps = kmp(s); // ans string string ans = ""; string suff = s.Substring(0, n - lps[n - 1]); for(int i = 0; i < k - 1; ++i) { // Update ans appending the // substring K - 1 times ans += suff; } // Append the original string ans += s; // Returning min length string // which contain exactly k // substring of given string return ans;}// Driver codepublic static void Main (string[] args){ int k = 3; string s = "neveropen"; Console.Write(findString(s, k));}}// This code is contributed by rutvik_56 |
Javascript
<script>// JavaScript Program to implement// the above approach// KMP algorithmfunction kmp(s){ var n = s.length; var lps = new Array(n); lps[0] = 0; var i = 1; var len = 0; while (i < n) { if (s[i] == s[len]) { len++; lps[i] = len; i++; } else { if (len != 0) { len = lps[len - 1]; } else { lps[i] = 0; i++; } } } return lps;}// Function to return the required stringfunction findString(s, k){ var n = s.length; // Finding the longest proper prefix // which is also suffix var lps = kmp(s); // ans string var ans = ""; var suff = s.substr(0, n - lps[n - 1]); for (var i = 0; i < k - 1; ++i) { // Update ans appending the // substring K - 1 times ans += suff; } // Append the original string ans += s; // Returning min length string // which contain exactly k // substring of given string return ans;}// Driver Codevar k = 3;var s = "neveropen";document.write(findString(s, k));//This code is contributed by phasing17</script> |
Output
neveropenforneveropen
Time Complexity: O( |S| ), where S is the given string
Auxiliary Space: O( |S| )
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