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 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 has = new Dictionary ();        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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