Data Modelling & AI Data Structure & Algorithm

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

24 June 2025

2

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 the 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:
Strings 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 a 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 conditions mentioned above fail, then the answer is “NO”.

Below is the implementation of the above approach:

C++

// 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;
}

Java

// 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

Python3

# 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

C#

// C# program for the above problem 
using System; 
using System.Collections.Generic; 
 
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[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'; 
     
    Dictionary<char,
               int> mp = new Dictionary<char,
                                        int>(); 
 
    if (c % 2 == 0 || c1 > 0 || c2 > 0) 
    { 
        for(i = 0; i < n; i++) 
        { 
 
            // Case 1 
            if (fr[s[i] - 'a'] == k) 
            { 
                if (mp.ContainsKey(s[i])) 
                { 
                    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(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; 
                } 
            } 
        } 
        Console.Write(ans); 
    } 
    else
    { 
         
        // If all cases fail 
        Console.Write("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 rutvik_56

Javascript

<script>
      // JavaScript program for the above problem
      // Function to print the
      // arrangement of characters
      function DivideString(s, n, k) {
        var i,
          c = 0,
          no = 1;
        var c1 = 0,
          c2 = 0;
 
        // Stores frequency of
        // characters
        var fr = new Array(26).fill(0);
 
        var ans = [];
 
        for (i = 0; i < n; i++) {
          fr[s[i].charCodeAt(0) - "a".charCodeAt(0)]++;
        }
 
        var 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 = String.fromCharCode(i + "a".charCodeAt(0));
          }
 
          if (fr[i] === 2 * k) {
            c2++;
            ch1 = String.fromCharCode(i + "a".charCodeAt(0));
          }
        }
 
        for (i = 0; i < n; i++) ans.push("1");
 
        var mp = {};
 
        if (c % 2 === 0 || c1 > 0 || c2 > 0) {
          for (i = 0; i < n; i++) {
            // Case 1
            if (fr[s[i].charCodeAt(0) - "a".charCodeAt(0)] === k) {
              if (mp.hasOwnProperty(s[i])) {
                ans[i] = "2";
              } 
              else {
                if (no <= parseInt(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;
            var flag = 0;
 
            for (i = 0; i < n; i++) {
              if (s[i] === ch1 && no <= k) {
                ans[i] = "2";
                no++;
              }
              if (
                fr[s[i].charCodeAt(0) - "a".charCodeAt(0)] === k &&
                flag === 0 &&
                ans[i] === "1"
              ) {
                ans[i] = "2";
                flag = 1;
              }
            }
          }
          document.write(ans.join(""));
        } 
        else {
          // If all cases fail
          document.write("NO");
        }
      }
 
      // Driver code
      var S = "abbbccc";
      var N = S.length;
      var K = 1;
 
      DivideString(S, N, K);
</script>

Output:

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

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

C++

Java

Python3

C#

Javascript

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