# Partiton the string in two parts such that both parts have at least k different characters

Given a string of lowercase English alphabets and an integer 0 < K <= 26. The task is to divide the string into two parts (also print them) such that both parts have at least k different characters. If there are more than one answers possible, print one having the smallest left part. If there is no such answers, print “Not Possible”.

Examples:

Input : str = “geeksforgeeks”, k = 4
Output : geeks , forgeeks
The string can be divided into two parts as “geeks” and “forgeeks”. Since “geeks” has four different characters ‘g’, ‘e’, ‘k’ and ‘s’ and this is the smallest left part, “forgeeks” has also at least four different characters.

Input : str = “aaaabbbb”, k = 2
Output :Not Possible

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach :

• Idea is to count the number of distinct characters using a Hashmap.
• If the count of the distinct variable becomes equal to k, then the left part of the string is found so store this index, break the loop and unmark all the characters.
• Now run a loop from where the left string ends to end of the given string and repeat the same process as it was done to find the left string.
• If count is greater than or equal to k, then right string could be found otherwise print “Not Possible”.
• If it is possible , then print the left string and right string.

Below is the implementation of the above approach

## C++

 `// C++ implemenattion of the above approach ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the partition of the ` `// string such that both parts have at ` `// least k different characters ` `void` `division_of_string(string str, ``int` `k) ` `{ ` `    ``// Length of the string ` `    ``int` `n = str.size(); ` ` `  `    ``// To check if the current  ` `    ``// character is already found ` `    ``map<``char``, ``bool``> has; ` ` `  `    ``int` `ans, cnt = 0, i = 0; ` ` `  `    ``// Count number of different ` `    ``// characters in the left part ` `    ``while` `(i < n) { ` ` `  `        ``// If current character is not  ` `        ``// already found, increase cnt by 1 ` `        ``if` `(!has[str[i]]) { ` `            ``cnt++; ` `            ``has[str[i]] = ``true``; ` `        ``} ` ` `  `        ``// If count becomes equal to k, we've  ` `        ``// got the first part, therefore, ` `        ``// store current index and break the loop ` `        ``if` `(cnt == k) { ` `            ``ans = i; ` `            ``break``; ` `        ``} ` ` `  `        ``i++; ` `    ``} ` ` `  `    ``// Clear the map ` `    ``has.clear(); ` ` `  `    ``// Assign cnt as 0 ` `    ``cnt = 0; ` ` `  `    ``while` `(i < n) { ` ` `  `        ``// If the current character is not  ` `        ``// already found, increase cnt by 1 ` `        ``if` `(!has[str[i]]) { ` `            ``cnt++; ` `            ``has[str[i]] = ``true``; ` `        ``} ` ` `  `        ``// If cnt becomes equal to k, the ` `        ``// second part also have k different ` `        ``// characters so break it ` `        ``if` `(cnt == k) { ` `            ``break``; ` `        ``} ` ` `  `        ``i++; ` `    ``} ` ` `  `    ``// If the second part has less than ` `    ``// k different characters, then  ` `    ``// print "Not Possible" ` `    ``if` `(cnt < k) { ` `        ``cout << ``"Not possible"` `<< endl; ` `    ``} ` ` `  `    ``// Otherwise print both parts ` `    ``else` `{ ` `        ``i = 0; ` `        ``while` `(i <= ans) { ` `            ``cout << str[i]; ` `            ``i++; ` `        ``} ` `        ``cout << endl; ` ` `  `        ``while` `(i < n) { ` `            ``cout << str[i]; ` `            ``i++; ` `        ``} ` `        ``cout << endl; ` `    ``} ` ` `  `    ``cout << endl; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"geeksforgeeks"``; ` `    ``int` `k = 4; ` ` `  `    ``// Function call ` `    ``division_of_string(str, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to find the partition of the ` `// string such that both parts have at ` `// least k different characters ` `static` `void` `division_of_string(``char``[] str, ``int` `k) ` `{ ` `    ``// Length of the string ` `    ``int` `n = str.length; ` ` `  `    ``// To check if the current  ` `    ``// character is already found ` `    ``Map has = ``new` `HashMap<>(); ` ` `  `    ``int` `ans = ``0``, cnt = ``0``, i = ``0``; ` ` `  `    ``// Count number of different ` `    ``// characters in the left part ` `    ``while` `(i < n) ` `    ``{ ` ` `  `        ``// If current character is not  ` `        ``// already found, increase cnt by 1 ` `        ``if` `(!has.containsKey(str[i]))  ` `        ``{ ` `            ``cnt++; ` `            ``has.put(str[i], ``true``); ` `        ``} ` ` `  `        ``// If count becomes equal to k, we've  ` `        ``// got the first part, therefore, ` `        ``// store current index and break the loop ` `        ``if` `(cnt == k)  ` `        ``{ ` `            ``ans = i; ` `            ``break``; ` `        ``} ` ` `  `        ``i++; ` `    ``} ` ` `  `    ``// Clear the map ` `    ``has.clear(); ` ` `  `    ``// Assign cnt as 0 ` `    ``cnt = ``0``; ` ` `  `    ``while` `(i < n)  ` `    ``{ ` ` `  `        ``// If the current character is not  ` `        ``// already found, increase cnt by 1 ` `        ``if` `(!has.containsKey(str[i])) ` `        ``{ ` `            ``cnt++; ` `            ``has.put(str[i], ``true``); ` `        ``} ` ` `  `        ``// If cnt becomes equal to k, the ` `        ``// second part also have k different ` `        ``// characters so break it ` `        ``if` `(cnt == k) ` `        ``{ ` `            ``break``; ` `        ``} ` ` `  `        ``i++; ` `    ``} ` ` `  `    ``// If the second part has less than ` `    ``// k different characters, then  ` `    ``// print "Not Possible" ` `    ``if` `(cnt < k) ` `    ``{ ` `        ``System.out.println(``"Not possible"``); ` `    ``} ` ` `  `    ``// Otherwise print both parts ` `    ``else`  `    ``{ ` `        ``i = ``0``; ` `        ``while` `(i <= ans) ` `        ``{ ` `            ``System.out.print(str[i]); ` `            ``i++; ` `        ``} ` `        ``System.out.println(``""``); ` ` `  `        ``while` `(i < n)  ` `        ``{ ` `            ``System.out.print(str[i]); ` `            ``i++; ` `        ``} ` `        ``System.out.println(``""``); ` `    ``} ` ` `  `    ``System.out.println(``""``); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``String str = ``"geeksforgeeks"``; ` `    ``int` `k = ``4``; ` ` `  `    ``// Function call ` `    ``division_of_string(str.toCharArray(), k); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implemenattion of the above approach ` ` `  `# Function to find the partition of the ` `# string such that both parts have at ` `# least k different characters ` `def` `division_of_string(string, k) : ` `     `  `    ``# Length of the string ` `    ``n ``=` `len``(string); ` ` `  `    ``# To check if the current  ` `    ``# character is already found ` `    ``has ``=` `{}; ` ` `  `    ``cnt ``=` `0``; i ``=` `0``; ` ` `  `    ``# Count number of different ` `    ``# characters in the left part ` `    ``while` `(i < n) : ` ` `  `        ``# If current character is not  ` `        ``# already found, increase cnt by 1 ` `        ``if` `string[i] ``not` `in` `has : ` `            ``cnt ``+``=` `1``; ` `            ``has[string[i]] ``=` `True``; ` ` `  `        ``# If count becomes equal to k, we've  ` `        ``# got the first part, therefore, ` `        ``# store current index and break the loop ` `        ``if` `(cnt ``=``=` `k) : ` `            ``ans ``=` `i; ` `            ``break``; ` ` `  `        ``i ``+``=` `1``; ` ` `  `    ``# Clear the map ` `    ``has.clear(); ` ` `  `    ``# Assign cnt as 0 ` `    ``cnt ``=` `0``; ` ` `  `    ``while` `(i < n) : ` ` `  `        ``# If the current character is not  ` `        ``# already found, increase cnt by 1 ` `        ``if` `(string[i] ``not` `in` `has) : ` `            ``cnt ``+``=` `1``; ` `            ``has[string[i]] ``=` `True``; ` ` `  `        ``# If cnt becomes equal to k, the ` `        ``# second part also have k different ` `        ``# characters so break it ` `        ``if` `(cnt ``=``=` `k) : ` `            ``break``; ` ` `  `        ``i ``+``=` `1``; ` ` `  `    ``# If the second part has less than ` `    ``# k different characters, then  ` `    ``# print "Not Possible" ` `    ``if` `(cnt < k) : ` `        ``print``(``"Not possible"``,end ``=` `""); ` ` `  `    ``# Otherwise print both parts ` `    ``else` `: ` `        ``i ``=` `0``; ` `        ``while` `(i <``=` `ans) : ` `            ``print``(string[i],end``=` `""); ` `            ``i ``+``=` `1``; ` `     `  `        ``print``(); ` ` `  `        ``while` `(i < n) : ` `            ``print``(string[i],end``=``""); ` `            ``i ``+``=` `1``; ` `             `  `        ``print``() ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``string ``=` `"geeksforgeeks"``; ` `    ``k ``=` `4``; ` ` `  `    ``# Function call ` `    ``division_of_string(string, k); ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to find the partition of the ` `// string such that both parts have at ` `// least k different characters ` `static` `void` `division_of_string(``char``[] str, ``int` `k) ` `{ ` `    ``// Length of the string ` `    ``int` `n = str.Length; ` ` `  `    ``// To check if the current  ` `    ``// character is already found ` `    ``Dictionary<``char``,``bool``> has = ``new` `Dictionary<``char``,``bool``> (); ` ` `  `    ``int` `ans = 0, cnt = 0, i = 0; ` ` `  `    ``// Count number of different ` `    ``// characters in the left part ` `    ``while` `(i < n) ` `    ``{ ` ` `  `        ``// If current character is not  ` `        ``// already found, increase cnt by 1 ` `        ``if` `(!has.ContainsKey(str[i]))  ` `        ``{ ` `            ``cnt++; ` `            ``has.Add(str[i], ``true``); ` `        ``} ` ` `  `        ``// If count becomes equal to k, we've  ` `        ``// got the first part, therefore, ` `        ``// store current index and break the loop ` `        ``if` `(cnt == k)  ` `        ``{ ` `            ``ans = i; ` `            ``break``; ` `        ``} ` ` `  `        ``i++; ` `    ``} ` ` `  `    ``// Clear the map ` `    ``has.Clear(); ` ` `  `    ``// Assign cnt as 0 ` `    ``cnt = 0; ` ` `  `    ``while` `(i < n)  ` `    ``{ ` ` `  `        ``// If the current character is not  ` `        ``// already found, increase cnt by 1 ` `        ``if` `(!has.ContainsKey(str[i])) ` `        ``{ ` `            ``cnt++; ` `            ``has.Add(str[i], ``true``); ` `        ``} ` ` `  `        ``// If cnt becomes equal to k, the ` `        ``// second part also have k different ` `        ``// characters so break it ` `        ``if` `(cnt == k) ` `        ``{ ` `            ``break``; ` `        ``} ` ` `  `        ``i++; ` `    ``} ` ` `  `    ``// If the second part has less than ` `    ``// k different characters, then  ` `    ``// print "Not Possible" ` `    ``if` `(cnt < k) ` `    ``{ ` `        ``Console.WriteLine(``"Not possible"``); ` `    ``} ` ` `  `    ``// Otherwise print both parts ` `    ``else` `    ``{ ` `        ``i = 0; ` `        ``while` `(i <= ans) ` `        ``{ ` `            ``Console.Write(str[i]); ` `            ``i++; ` `        ``} ` `        ``Console.WriteLine(``""``); ` ` `  `        ``while` `(i < n)  ` `        ``{ ` `            ``Console.Write(str[i]); ` `            ``i++; ` `        ``} ` `        ``Console.WriteLine(``""``); ` `    ``} ` ` `  `    ``Console.WriteLine(``""``); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``String str = ``"geeksforgeeks"``; ` `    ``int` `k = 4; ` ` `  `    ``// Function call ` `    ``division_of_string(str.ToCharArray(), k); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```geeks
forgeeks
```

Time Complexity: O(N) where N is the length of given string.

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