Convert to a string that is repetition of a substring of k length
Given a string, find if it is possible to convert it to a string that is repetition of substring with k characters. To convert, we can replace one substring of length k with k characters.
Examples:
Input: str = "bdac", k = 2 Output: True We can either replace "bd" with "ac" or "ac" with "bd". Input: str = "abcbedabcabc", k = 3 Output: True Replace "bed" with "abc" so that the whole string becomes repetition of "abc". Input: str = "bcacc", k = 3 Output: False k doesn't divide string length i.e. 5%3 != 0 Input: str = "bcacbcac", k = 2 Output: False Input: str = "bcdbcdabcedcbcd", k = 3 Output: False
This can be used in compression. If we have a string where complete string is repetition except one substring, then we can use this algorithm to compress the string.
One observation is, length of string must be a multiple of k as we can replace only one substring.
The idea is declare a map mp which maps strings of length k to an integer denoting its count. So, if there are only two different sub-strings of length k in the map container and count of one of the sub-string is 1 then answer is true. Otherwise answer is false.
C++
// C++ program to check if a string can be converted to // a string that has repeated substrings of length k. #include<bits/stdc++.h> using namespace std; // Returns true if str can be coverted to a string // with k repeated substrings after replacing k // characters. bool checkString(string str, long k) { // Length of string must be a multiple of k int n = str.length(); if (n%k != 0) return false; // Map to store strings of length k and their counts unordered_map<string, int> mp; for (int i=0; i<n; i+=k) mp[str.substr(i, k)]++; // If string is already a repition of k substrings, // return true. if (mp.size() == 1) return true; // If number of distinct substrings is not 2, then // not possible to replace a string. if (mp.size() != 2) return false; // One of the two distinct must appear exactly once. // Either the first entry appears once, or it appears // n/k-1 times to make other substring appear once. if ((mp.begin()->second == (n/k - 1)) || mp.begin()->second == 1) return true; return false; } // Driver code int main() { checkString("abababcd", 2)? cout << "Yes" : cout << "No"; return 0; } |
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Java
// Java program to check if a string // can be converted to a string that has // repeated substrings of length k. import java.util.HashMap; import java.util.Iterator; class GFG { // Returns true if str can be coverted // to a string with k repeated substrings // after replacing k characters. static boolean checkString(String str, int k) { // Length of string must be // a multiple of k int n = str.length(); if (n % k != 0) return false; // Map to store strings of // length k and their counts HashMap<String, Integer> mp = new HashMap<>(); try { for (int i = 0; i < n; i += k) mp.put(str.substring(i, k), mp.get(str.substring(i, k)) == null ? 1 : mp.get(str.substring(i, k)) + 1); } catch (Exception e) { } // If string is already a repition // of k substrings, return true. if (mp.size() == 1) return true; // If number of distinct substrings is not 2, // then not possible to replace a string. if (mp.size() != 2) return false; HashMap.Entry<String, Integer> entry = mp.entrySet().iterator().next(); // One of the two distinct must appear // exactly once. Either the first entry // appears once, or it appears n/k-1 times // to make other substring appear once. if (entry.getValue() == (n / k - 1) || entry.getValue() == 1) return true; return false; } // Driver code public static void main(String[] args) { if (checkString("abababcd", 2)) System.out.println("Yes"); else System.out.println("No"); } } // This code is contributed by // sanjeev2552 |
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Python3
# Python3 program to check if a can be converted to # a that has repeated subs of length k. # Returns True if S can be coverted to a # with k repeated subs after replacing k # characters. def check( S, k): # Length of must be a multiple of k n = len(S) if (n % k != 0): return False # Map to store s of length k and their counts mp = {} for i in range(0, n, k): mp[S[i:k]] = mp.get(S[i:k], 0) + 1 # If is already a repition of k subs, # return True. if (len(mp) == 1): return True # If number of distinct subs is not 2, then # not possible to replace a . if (len(mp) != 2): return False # One of the two distinct must appear exactly once. # Either the first entry appears once, or it appears # n/k-1 times to make other sub appear once. for i in mp: if i == (n//k - 1) or mp[i] == 1: return True return False # Driver code if check("abababcd", 2): print("Yes") else: print("No") # This code is contributed by mohit kumar 29 |
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C#
// C# program to check if a string // can be converted to a string that has // repeated substrings of length k. using System; using System.Collections.Generic; class GFG { // Returns true if str can be coverted // to a string with k repeated substrings // after replacing k characters. static bool checkString(String str, int k) { // Length of string must be // a multiple of k int n = str.Length; if (n % k != 0) return false; // Map to store strings of // length k and their counts Dictionary<String, int> mp = new Dictionary<String, int>(); for (int i = 0; i < n; i += k) { if(!mp.ContainsKey(str.Substring(i, k))) mp.Add(str.Substring(i, k), 1); else mp[str.Substring(i, k)] = mp[str.Substring(i, k)] + 1; } // If string is already a repition // of k substrings, return true. if (mp.Count == 1) return true; // If number of distinct substrings is not 2, // then not possible to replace a string. if (mp.Count != 2) return false; foreach(KeyValuePair<String, int> entry in mp) { // One of the two distinct must appear // exactly once. Either the first entry // appears once, or it appears n/k-1 times // to make other substring appear once. if (entry.Value == (n / k - 1) || entry.Value == 1) return true; } return false; } // Driver code public static void Main(String[] args) { if (checkString("abababcd", 2)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } } // This code is contributed by Princi Singh |
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Output:
Yes
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