Check if a string can be split into two strings with same number of K-frequent characters
Last Updated :
03 Jun, 2021
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 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 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# 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
<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>
Time Complexity: O(N)
Auxiliary Space: O(N)
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