Find Binary String of size at most 3N containing at least 2 given strings of size 2N as subsequences

Last Updated : 21 Feb, 2022

Given three binary strings a, b, and c each having 2*N characters each, the task is to find a string having almost 3*N characters such that at least two of the given three strings occur as its one of the subsequence.

Examples:

Input: a = "00", b = "11", c = "01"
Output: "010"
Explanation: The strings "00" and "01" are subsequences of string "010" and is has not more than 3*N characters. Also, "001", "011" can be the possible answers.

Input: a = "011001", b = "111010", c = "010001"
Output: "011001010"
Explanation: Here, all the three given strings occur as the subsequences of the output string.

Approach: The given problem can be solved using the following observations:

It can be observed that according to the Pigeonhole Principle, there must exist a set of two strings such that the most frequent character in both the strings is the same and its frequency is >=N.
Hence, for two such strings, a string of N most frequent characters can be created. The other remaining N characters of both the strings can be appended into the string respectively according to the order they occur. Therefore, the maximum size of the resultant string will be at most 3*N.

Therefore, after finding the set of strings with the same most frequent element, the rest can be solved using the two-pointer approach. Maintain two pointers, one for each string, and follow the below steps:

Initially, i =0 and j = 0, where i represent the 1st string s1 and j represent the second string s2.
If s1[i] is not equal to the most frequent character, print s1[i] and increment i.
If s2[j] is not equal to the most frequent character, print s2[i] and increment j.
If both s1[i] and s2[j] is representing the most frequent character, print s1[i] and increment both i and j.

Below is the implementation of the above approach:

C++

// C++ program for the above approach #include <bits/stdc++.h> using namespace std;  // Function to find most frequent // character in the given string char MostFrequent(string s) {     // Stores the frequency of     // 0 and 1 respectively     int arr[] = { 0, 0 };      for (char ch : s) {         arr[ch - '0']++;     }      // Stores most frequent character     char result = arr[0] > arr[1] ? '0' : '1';      // Return Answer     return result; }  // Function to find a Binary String of // at most 3N characters having at least // two of the given three strings of 2N // characters as one of its subsequence void findStr(string& a, string& b, string& c) {     // Stores most frequent char     char freq;      // Stores the respective string     // with most frequent values     string s1, s2;      // Code to find the set of     // two strings with same     // most frequent characters     if (MostFrequent(a) == MostFrequent(b)) {         s1 = a;         s2 = b;         freq = MostFrequent(a);     }     else if (MostFrequent(a) == MostFrequent(c)) {         s1 = a;         s2 = c;         freq = MostFrequent(a);     }     else {         s1 = b;         s2 = c;         freq = MostFrequent(b);     }      // Pointer to iterate strings     int i = 0, j = 0;      // Traversal using the     // two-pointer approach     while (i < s1.size() && j < s2.size()) {          // if current character         // is not most frequent         while (i < s1.size() && s1[i] != freq) {             cout << s1[i++];         }         // if current character         // is not most frequent         while (j < s2.size() && s2[j] != freq) {             cout << s2[j++];         }          // If end of string is reached         if (i == s1.size() || j == s2.size())             break;          // If both string character         // are same as most frequent         if (s1[i] == s2[j]) {             cout << s1[i];             i++;             j++;         }         else {             cout << s1[i];             i++;         }     }      // Print leftover characters     // of the string s1     while (i < s1.size()) {         cout << s1[i++];     }      // Print leftover characters     // of the string s2     while (j < s2.size()) {         cout << s2[j++];     } }  // Driver Code int main() {     string a, b, c;     a = "00";     b = "11";     c = "01";      findStr(a, b, c);     return 0; }

Java

// Java program for the above approach class GFG {   // Function to find most frequent   // character in the given String   static char MostFrequent(String s)   {     // Stores the frequency of     // 0 and 1 respectively     int []arr = { 0, 0 };      for(char ch : s.toCharArray()) {       arr[ch - '0']++;     }      // Stores most frequent character     char result = arr[0] > arr[1] ? '0' : '1';      // Return Answer     return result;   }    // Function to find a Binary String of   // at most 3N characters having at least   // two of the given three Strings of 2N   // characters as one of its subsequence   static void findStr(String a, String b, String c)   {     // Stores most frequent char     char freq;      // Stores the respective String     // with most frequent values     String s1 = "", s2 = "";      // Code to find the set of     // two Strings with same     // most frequent characters     if (MostFrequent(a) == MostFrequent(b)) {       s1 = a;       s2 = b;       freq = MostFrequent(a);     }     else if (MostFrequent(a) == MostFrequent(c)) {       s1 = a;       s2 = c;       freq = MostFrequent(a);     }     else {       s1 = b;       s2 = c;       freq = MostFrequent(b);     }      // Pointer to iterate Strings     int i = 0, j = 0;      // Traversal using the     // two-pointer approach     while (i < s1.length()&& j < s2.length()) {        // if current character       // is not most frequent       while (i < s1.length() && s1.charAt(i) != freq) {         System.out.print(s1.charAt(i++));       }       // if current character       // is not most frequent       while (j < s2.length() && s2.charAt(j) != freq) {         System.out.print(s2.charAt(j++));       }        // If end of String is reached       if (i == s1.length()|| j == s2.length())         break;        // If both String character       // are same as most frequent       if (s1.charAt(i) == s2.charAt(j)) {         System.out.print(s1.charAt(i));         i++;         j++;       }       else {         System.out.print(s1.charAt(i));         i++;       }     }      // Print leftover characters     // of the String s1     while (i < s1.length()) {       System.out.print(s1.charAt(i++));     }      // Print leftover characters     // of the String s2     while (j < s2.length()) {       System.out.print(s2.charAt(j++));     }   }    // Driver Code   public static void main(String args[])   {     String a = "00";     String b = "11";     String c = "01";      findStr(a, b, c);   } }  // This code is contributed Saurabh Jaiswal

// C# program for the above approach using System; class GFG {   // Function to find most frequent   // character in the given string   static char MostFrequent(string s)   {     // Stores the frequency of     // 0 and 1 respectively     int []arr = { 0, 0 };      foreach (char ch in s) {       arr[ch - '0']++;     }      // Stores most frequent character     char result = arr[0] > arr[1] ? '0' : '1';      // Return Answer     return result;   }    // Function to find a Binary String of   // at most 3N characters having at least   // two of the given three strings of 2N   // characters as one of its subsequence   static void findStr(string a, string b, string c)   {     // Stores most frequent char     char freq;      // Stores the respective string     // with most frequent values     string s1 = "", s2 = "";      // Code to find the set of     // two strings with same     // most frequent characters     if (MostFrequent(a) == MostFrequent(b)) {       s1 = a;       s2 = b;       freq = MostFrequent(a);     }     else if (MostFrequent(a) == MostFrequent(c)) {       s1 = a;       s2 = c;       freq = MostFrequent(a);     }     else {       s1 = b;       s2 = c;       freq = MostFrequent(b);     }      // Pointer to iterate strings     int i = 0, j = 0;      // Traversal using the     // two-pointer approach     while (i < s1.Length && j < s2.Length) {        // if current character       // is not most frequent       while (i < s1.Length && s1[i] != freq) {         Console.Write(s1[i++]);       }       // if current character       // is not most frequent       while (j < s2.Length && s2[j] != freq) {         Console.Write(s2[j++]);       }        // If end of string is reached       if (i == s1.Length || j == s2.Length)         break;        // If both string character       // are same as most frequent       if (s1[i] == s2[j]) {         Console.Write(s1[i]);         i++;         j++;       }       else {         Console.Write(s1[i]);         i++;       }     }      // Print leftover characters     // of the string s1     while (i < s1.Length) {       Console.Write(s1[i++]);     }      // Print leftover characters     // of the string s2     while (j < s2.Length) {       Console.Write(s2[j++]);     }   }    // Driver Code   public static void Main()   {     string a = "00";     string b = "11";     string c = "01";      findStr(a, b, c);   } }  // This code is contributed Samim Hossain Mondal.

Python3

# python3 program for the above approach  # Function to find most frequent # character in the given string def MostFrequent(s):      # Stores the frequency of     # 0 and 1 respectively     arr = [0, 0]      for ch in s:         arr[ord(ch) - ord('0')] += 1      # Stores most frequent character     result = '0' if arr[0] > arr[1] else '1'      # Return Answer     return result  # Function to find a Binary String of # at most 3N characters having at least # two of the given three strings of 2N # characters as one of its subsequence def findStr(a, b, c):      # Stores most frequent char     freq = ''      # Stores the respective string     # with most frequent values     s1, s2 = "", ""      # Code to find the set of     # two strings with same     # most frequent characters     if (MostFrequent(a) == MostFrequent(b)):         s1 = a         s2 = b         freq = MostFrequent(a)      elif (MostFrequent(a) == MostFrequent(c)):         s1 = a         s2 = c         freq = MostFrequent(a)      else:         s1 = b         s2 = c         freq = MostFrequent(b)      # Pointer to iterate strings     i, j = 0, 0      # Traversal using the     # two-pointer approach     while (i < len(s1) and j < len(s2)):          # if current character         # is not most frequent         while (i < len(s1) and s1[i] != freq):             print(s1[i], end="")             i += 1          # if current character         # is not most frequent         while (j < len(s2) and s2[j] != freq):             print(s2[j], end="")             j += 1          # If end of string is reached         if (i == len(s1) or j == len(s2)):             break          # If both string character         # are same as most frequent         if (s1[i] == s2[j]):             print(s1[i], end="")             i += 1             j += 1          else:             print(s1[i], end="")             i += 1      # Print leftover characters     # of the string s1     while (i < len(s1)):         print(s1[i], end="")         i += 1      # Print leftover characters     # of the string s2     while (j < len(s2)):         print(s2[j], end="")         j += 1  # Driver Code if __name__ == "__main__":      a = "00"     b = "11"     c = "01"      findStr(a, b, c)      # This code is contributed by rakeshsahni

JavaScript

    <script>         // JavaScript code for the above approach           // Function to find most frequent         // character in the given string         function MostFrequent(s)          {                      // Stores the frequency of             // 0 and 1 respectively             let arr = new Array(2).fill(0)              for (let i = 0; i < s.length; i++) {                 let ch = s[i];                 arr[ch.charCodeAt(0) - '0'.charCodeAt(0)]++;             }              // Stores most frequent character             let result = arr[0] > arr[1] ? '0' : '1';              // Return Answer             return result;         }          // Function to find a Binary String of         // at most 3N characters having at least         // two of the given three strings of 2N         // characters as one of its subsequence         function findStr(a, b, c)          {                      // Stores most frequent char             let freq;              // Stores the respective string             // with most frequent values             let s1, s2;              // Code to find the set of             // two strings with same             // most frequent characters             if (MostFrequent(a) == MostFrequent(b)) {                 s1 = a;                 s2 = b;                 freq = MostFrequent(a);             }             else if (MostFrequent(a) == MostFrequent(c)) {                 s1 = a;                 s2 = c;                 freq = MostFrequent(a);             }             else {                 s1 = b;                 s2 = c;                 freq = MostFrequent(b);             }              // Pointer to iterate strings             let i = 0, j = 0;              // Traversal using the             // two-pointer approach             while (i < s1.length && j < s2.length) {                  // if current character                 // is not most frequent                 while (i < s1.length && s1[i] != freq) {                     document.write(s1[i++]);                 }                 // if current character                 // is not most frequent                 while (j < s2.length && s2[j] != freq) {                     document.write(s2[j++]);                 }                  // If end of string is reached                 if (i == s1.length || j == s2.length)                     break;                  // If both string character                 // are same as most frequent                 if (s1[i] == s2[j]) {                     document.write(s1[i]);                     i++;                     j++;                 }                 else {                     document.write(s1[i]);                     i++;                 }             }              // Print leftover characters             // of the string s1             while (i < s1.length) {                 document.write(s1[i++]);             }              // Print leftover characters             // of the string s2             while (j < s2.length) {                 document.write(s2[j++]);             }         }          // Driver Code         let a, b, c;         a = "00";         b = "11";         c = "01";          findStr(a, b, c);         // This code is contributed by Potta Lokesh     </script>

Output

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

Minimum number of socks required to picked to have at least K pairs of the same color

singhh3010

Improve

Article Tags :

Practice Tags :

Find Binary String of size at most 3N containing at least 2 given strings of size 2N as subsequences

Similar Reads