Check if a string can be split into two strings with same number of K-frequent characters

Given a string S and an integer K, the task is to check if it is possible to distribute these characters into two strings such that count of characters having a frequency K in both strings is equal.
If it is possible, then print a sequence consisting of 1 and 2, which denotes which character should be placed in which string. Otherwise, print NO.

Note: One of these new strings can be empty.

Examples:

Input: S = “abbbccc”, K = 1
Output: 1111211
Explanation:
The two strings are “abbbcc” and “c”.
Hence, both the strings have exactly 1 character having frequency K( = 1).

Input: S = “aaaa”, K = 3
Output: 1111
Explanation:
String can be split into “aaaa” and “”.
Hence, no character has frequency 3 in both the strings.



Approach:
Follow the steps below to solve the problem:

  • Check for the following three conditions to determine if a split is possible or not:

    1. If the total number of characters having a frequency K in the initial string is even, then these characters can be placed equally into two strings and the rest of the characters(having a frequency not equal to K) can be placed in any of the two groups.
    2. If the total number of characters having a frequency K in the initial string is odd, then if there is a character in the initial string having a frequency greater than K but not equal to 2K, then such a distribution is possible.

      Illustration:
      S =”abceeee”, K = 1

      Split into “abeee” and “ce”. Hence, both the strings have 2 characters with frequency 1.

    3. If the total number of characters having a frequency K in the initial string is odd, then if there is a character in the initial string having a frequency equal to 2K, then such distribution is possible.

      Illustration:
      S =”aaaabbccdde”, K = 2

      Split into “aabbc” and “aaddce” so that both the strings have two characters with frequency 2.

  • If all the three condition mentioned above fails then answer is “NO”.

Below is the implementation of the above approach:

C++

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// C++ implementation of the
// above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to print the
// arrangement of characters
void DivideString(string s, int n,
                  int k)
{
    int i, c = 0, no = 1;
    int c1 = 0, c2 = 0;
  
    // Stores frequency of
    // characters
    int fr[26] = { 0 };
  
    string ans = "";
  
    for (i = 0; i < n; i++) {
        fr[s[i] - 'a']++;
    }
  
    char ch, ch1;
    for (i = 0; i < 26; i++) {
  
        // Count the character
        // having frequency K
        if (fr[i] == k) {
            c++;
        }
  
        // Count the character
        // having frequency
        // greater than K and
        // not equal to 2K
        if (fr[i] > k
            && fr[i] != 2 * k) {
            c1++;
            ch = i + 'a';
        }
  
        if (fr[i] == 2 * k) {
            c2++;
            ch1 = i + 'a';
        }
    }
  
    for (i = 0; i < n; i++)
        ans = ans + "1";
  
    map<char, int> mp;
    if (c % 2 == 0 || c1 > 0 || c2 > 0) {
        for (i = 0; i < n; i++) {
  
            // Case 1
            if (fr[s[i] - 'a'] == k) {
                if (mp.find(s[i])
                    != mp.end()) {
                    ans[i] = '2';
                }
                else {
                    if (no <= (c / 2)) {
                        ans[i] = '2';
                        no++;
                        mp[s[i]] = 1;
                    }
                }
            }
        }
  
        // Case 2
        if (c % 2 == 1 && c1 > 0) {
            no = 1;
            for (i = 0; i < n; i++) {
                if (s[i] == ch && no <= k) {
  
                    ans[i] = '2';
                    no++;
                }
            }
        }
  
        // Case 3
        if (c % 2 == 1 && c1 == 0) {
            no = 1;
            int flag = 0;
            for (int i = 0; i < n; i++) {
                if (s[i] == ch1 && no <= k) {
                    ans[i] = '2';
                    no++;
                }
                if (fr[s[i] - 'a'] == k
                    && flag == 0
                    && ans[i] == '1') {
                    ans[i] = '2';
                    flag = 1;
                }
            }
        }
  
        cout << ans << endl;
    }
    else {
        // If all cases fail
        cout << "NO" << endl;
    }
}
  
// Driver Code
int main()
{
  
    string S = "abbbccc";
    int N = S.size();
    int K = 1;
  
    DivideString(S, N, K);
  
    return 0;
}

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Java

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// Java program for the above problem 
import java.util.*; 
  
class GFG{
      
// Function to print the 
// arrangement of characters 
public static void DivideString(String s, int n, 
                                          int k) 
    int i, c = 0, no = 1
    int c1 = 0, c2 = 0
  
    // Stores frequency of 
    // characters 
    int[] fr = new int[26]; 
  
    char[] ans = new char[n]; 
  
    for(i = 0; i < n; i++)
    
        fr[s.charAt(i) - 'a']++; 
    
  
    char ch = 'a', ch1 = 'a'
    for(i = 0; i < 26; i++)
    
          
        // Count the character 
        // having frequency K 
        if (fr[i] == k) 
        
            c++; 
        
  
        // Count the character 
        // having frequency 
        // greater than K and 
        // not equal to 2K 
        if (fr[i] > k && fr[i] != 2 * k)
        
            c1++; 
            ch = (char)(i + 'a'); 
        
  
        if (fr[i] == 2 * k) 
        
            c2++; 
            ch1 = (char)(i + 'a'); 
        
    
  
    for(i = 0; i < n; i++) 
        ans[i] = '1'
      
    HashMap<Character, Integer> mp = new HashMap<>(); 
  
    if (c % 2 == 0 || c1 > 0 || c2 > 0)
    
        for(i = 0; i < n; i++) 
        
  
            // Case 1 
            if (fr[s.charAt(i) - 'a'] == k) 
            
                if (mp.containsKey(s.charAt(i)))
                
                    ans[i] = '2'
                
                else 
                
                    if (no <= (c / 2))
                    
                        ans[i] = '2'
                        no++; 
                        mp.replace(s.charAt(i), 1);
                    
                
            
        
  
        // Case 2 
        if ( (c % 2 == 1) && (c1 > 0) )
        
            no = 1
            for(i = 0; i < n; i++)
            
                if (s.charAt(i) == ch && no <= k) 
                
                    ans[i] = '2'
                    no++; 
                
            
        
  
        // Case 3 
        if (c % 2 == 1 && c1 == 0
        
            no = 1
            int flag = 0
              
            for(i = 0; i < n; i++)
            
                if (s.charAt(i) == ch1 && no <= k) 
                
                    ans[i] = '2'
                    no++; 
                
                if (fr[s.charAt(i) - 'a'] == k && 
                      flag == 0 && ans[i] == '1')
                
                    ans[i] = '2'
                    flag = 1
                
            
        
        System.out.println(ans);
    
    else
    {
          
        // If all cases fail
        System.out.println("NO");
    
  
// Driver code
public static void main(String[] args)
{
    String S = "abbbccc"
    int N = S.length(); 
    int K = 1
  
    DivideString(S, N, K); 
}
}
  
// This code is contributed by divyeshrabadiya07

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Python3

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# Python3 implementation of the
# above approach
  
# Function to print the
# arrangement of characters
def DivideString(s, n, k):
      
    c = 0
    no = 1
    c1 = 0
    c2 = 0
  
    # Stores frequency of
    # characters
    fr = [0] * 26
  
    ans = []
    for i in range(n):
        fr[ord(s[i]) - ord('a')] += 1
  
    for i in range(26):
  
        # Count the character
        # having frequency K
        if (fr[i] == k):
            c += 1
  
        # Count the character having
        # frequency greater than K and
        # not equal to 2K
        if (fr[i] > k and fr[i] != 2 * k):
            c1 += 1
            ch = chr(ord('a') + i)
  
        if (fr[i] == 2 * k):
            c2 += 1
            ch1 = chr(ord('a') + i)
  
    for i in range(n):
        ans.append("1")
  
    mp = {}
    if (c % 2 == 0 or c1 > 0 or c2 > 0):
        for i in range(n):
              
            # Case 1
            if (fr[ord(s[i]) - ord('a')] == k):
                if (s[i] in mp):
                    ans[i] = '2'
  
                else:
                    if (no <= (c // 2)):
                        ans[i] = '2'
                        no += 1
                        mp[s[i]] = 1
                          
        # Case 2
        if (c % 2 == 1 and c1 > 0):
            no = 1
            for i in range(n):
                if (s[i] == ch and no <= k):
                    ans[i] = '2'
                    no += 1
                      
        # Case 3
        if (c % 2 == 1 and c1 == 0):
            no = 1
            flag = 0
              
            for i in range(n):
                if (s[i] == ch1 and no <= k):
                    ans[i] = '2'
                    no += 1
                      
                if (fr[s[i] - 'a'] == k and 
                              flag == 0 and 
                            ans[i] == '1'):
                    ans[i] = '2'
                    flag = 1
  
        print("".join(ans))
    else:
          
        # If all cases fail
        print("NO")
  
# Driver Code
if __name__ == '__main__':
  
    S = "abbbccc"
    N = len(S)
    K = 1
  
    DivideString(S, N, K)
  
# This code is contributed by mohit kumar 29

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Output:

1111211

Time Complexity: O(N)
Auxiliary Space: O(N)

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