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 charactersvoid 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 Codeint 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 codepublic 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 charactersdef 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 Codeif __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:
1111211
Time Complexity: O(N)
Auxiliary Space: O(N)
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