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 = “geeksforgeeks”, K = 3_
Output:_ “geeksforgeeksforgeeksforgeeks”_
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.
- Below is the implementation of the above approach.
C++
-
// C++ Program to implement -
// the above approach -
#include <bits/stdc++.h> -
**using****namespace**std; -
// KMP algorithm -
**int**``* 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 string -
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.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 Code -
**int**main() -
{ -
**int**k = 3; -
string s ="geeksforgeeks"``; -
cout << findString(s, k) << endl; -
} -
Output:
geeksforgeeksforgeeksforgeeks
- Time Complexity:_ O(N*K )_
- Auxiliary Space:_ O(K* N)_
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#dynamic programming #mathematical #pattern searching #searching #strings #prefix #substring #suffix
