# 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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**Output:**

Yes

This article is contributed by **Himanshu Gupta(Bagri)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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