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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


Output:

1111211

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Recommended Posts:


Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : mohit kumar 29