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:
 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.

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

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 = 2Split 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++
// 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; } 
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 
1111211
Time Complexity: O(N)
Auxiliary Space: O(N)
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Improved By : mohit kumar 29