# Most frequent character in a string after replacing all occurrences of X in a Binary String

Last Updated : 28 Sep, 2022

Given a string S of length N consisting of 1, 0, and X, the task is to print the character (‘1’ or ‘0’) with the maximum frequency after replacing every occurrence of X as per the following conditions:

• If the character present adjacently to the left of X is 1, replace X with 1.
• If the character present adjacently to the right of X is 0, replace X with 0.
• If both the above conditions are satisfied, X remains unchanged.

Note: If the frequency of 1 and 0 is the same after replacements, then print X.

Examples:

Input: S = “XX10XX10XXX1XX”
Output: 1
Explanation:
Operation 1: S = “X11001100X1XX”
Operation 2: S = “111001100X1XX”
No further replacements are possible.
Hence, the frequencies of ‘1’ and ‘0’ are 6 and 4 respectively.

Input: S = “0XXX1”
Output: X
Explanation:
Operation 1: S = “00X11”
No further replacements are possible.
Hence, the frequencies of both ‘1’ and ‘0’ are 2.

Approach: The given problem can be solved based on the following observations:

• All the ‘X’s lying between ‘1’ and ‘0’ (e.g. 1XXX0) is of no significance because neither of ‘1’ and ‘0’ can convert it.
• All the ‘X’s lying between ‘0’ and ‘1’ (e.g. 0XXX1) is also of no significance because it will contribute equally to both 1 and 0. Consider any substring of the form “0X….X1”, then after changing the first occurrence of X from the start and the end of the string, the actual frequency of 0 and 1 in the substring remains unchanged.

From the above observations it can be concluded that the result depends upon the following conditions:

• The count of ‘1’ and ‘0’ in the original string.
• The frequency of X that are present between two consecutive 0s or two consecutive 1s, i.e. “0XXX0” and “1XXXX1” respectively.
• The number of continuous ‘X’ which are present at the starting of string and has a right end ‘1’, i.e. “XXXX1…..”.
• The number of continuous ‘X’s which are present at end of the string and has a left end ‘0’ i.e., …..0XXX.

Hence, count the number of 1s and 0s as per the above conditions and print the resultant count.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std;   // Function to find the most frequent // character after replacing X with // either '0' or '1' according as per // the given conditions void maxOccurringCharacter(string s) {       // Store the count of 0s and     // 1s in the string S     int count0 = 0, count1 = 0;       // Count the frequency of     // 0 and 1     for (int i = 0; i < s.length(); i++) {           // If the character is 1         if (s[i] == '1') {             count1++;         }           // If the character is 0         else if (s[i] == '0') {             count0++;         }     }       // Stores first occurrence of 1     int prev = -1;     for (int i = 0; i < s.length(); i++) {           if (s[i] == '1') {             prev = i;             break;         }     }       // Traverse the string to count     // the number of X between two     // consecutive 1s     for (int i = prev + 1; i < s.length(); i++) {           // If the current character         // is not X         if (s[i] != 'X') {               // If the current character             // is 1, add the number of             // Xs to count1 and set             // prev to i             if (s[i] == '1') {                 count1 += i - prev - 1;                 prev = i;             }               // Otherwise             else {                   // Find next occurrence                 // of 1 in the string                 bool flag = true;                   for (int j = i + 1; j < s.length(); j++) {                     if (s[j] == '1') {                         flag = false;                         prev = j;                         break;                     }                 }                   // If it is found,                 // set i to prev                 if (!flag) {                     i = prev;                 }                   // Otherwise, break                 // out of the loop                 else {                     i = s.length();                 }             }         }     }       // Store the first occurrence of 0     prev = -1;     for (int i = 0; i < s.length(); i++) {           if (s[i] == '0') {             prev = i;             break;         }     }       // Repeat the same procedure to     // count the number of X between     // two consecutive 0s     for (int i = prev + 1; i < s.length(); i++) {           // If the current character is not X         if (s[i] != 'X') {               // If the current character is 0             if (s[i] == '0') {                   // Add the count of Xs to count0                 count0 += i - prev - 1;                   // Set prev to i                 prev = i;             }               // Otherwise             else {                   // Find the next occurrence                 // of 0 in the string                 bool flag = true;                   for (int j = i + 1; j < s.length(); j++) {                       if (s[j] == '0') {                         prev = j;                         flag = false;                         break;                     }                 }                   // If it is found,                 // set i to prev                 if (!flag) {                     i = prev;                 }                   // Otherwise, break out                 // of the loop                 else {                     i = s.length();                 }             }         }     }       // Count number of X present in     // the starting of the string     // as XXXX1...     if (s[0] == 'X') {           // Store the count of X         int count = 0;         int i = 0;         while (s[i] == 'X') {             count++;             i++;         }           // Increment count1 by         // count if the condition         // is satisfied         if (s[i] == '1') {             count1 += count;         }     }       // Count the number of X     // present in the ending of     // the string as ...XXXX0     if (s[(s.length() - 1)] == 'X') {           // Store the count of X         int count = 0;         int i = s.length() - 1;         while (s[i] == 'X') {             count++;             i--;         }           // Increment count0 by         // count if the condition         // is satisfied         if (s[i] == '0') {             count0 += count;         }     }       // If count of 1 is equal to     // count of 0, print X     if (count0 == count1) {         cout << "X" << endl;     }       // Otherwise, if count of 1     // is greater than count of 0     else if (count0 > count1) {         cout << 0 << endl;     }       // Otherwise, print 0     else         cout << 1 << endl; }   // Driver Code int main() {     string S = "XX10XX10XXX1XX";     maxOccurringCharacter(S); }   // This code is contributed by SURENDAR_GANGWAR.

## Java

 // Java program for the above approach import java.io.*;   class GFG {       // Function to find the most frequent     // character after replacing X with     // either '0' or '1' according as per     // the given conditions     public static void     maxOccurringCharacter(String s)     {         // Store the count of 0s and         // 1s in the string S         int count0 = 0, count1 = 0;           // Count the frequency of         // 0 and 1         for (int i = 0;              i < s.length(); i++) {               // If the character is 1             if (s.charAt(i) == '1') {                 count1++;             }               // If the character is 0             else if (s.charAt(i) == '0') {                 count0++;             }         }           // Stores first occurrence of 1         int prev = -1;           for (int i = 0;              i < s.length(); i++) {               if (s.charAt(i) == '1') {                 prev = i;                 break;             }         }           // Traverse the string to count         // the number of X between two         // consecutive 1s         for (int i = prev + 1;              i < s.length(); i++) {               // If the current character             // is not X             if (s.charAt(i) != 'X') {                   // If the current character                 // is 1, add the number of                 // Xs to count1 and set                 // prev to i                 if (s.charAt(i) == '1') {                     count1 += i - prev - 1;                     prev = i;                 }                   // Otherwise                 else {                       // Find next occurrence                     // of 1 in the string                     boolean flag = true;                       for (int j = i + 1;                          j < s.length();                          j++) {                         if (s.charAt(j) == '1') {                             flag = false;                             prev = j;                             break;                         }                     }                       // If it is found,                     // set i to prev                     if (!flag) {                         i = prev;                     }                       // Otherwise, break                     // out of the loop                     else {                         i = s.length();                     }                 }             }         }           // Store the first occurrence of 0         prev = -1;         for (int i = 0; i < s.length(); i++) {               if (s.charAt(i) == '0') {                 prev = i;                 break;             }         }           // Repeat the same procedure to         // count the number of X between         // two consecutive 0s         for (int i = prev + 1;              i < s.length(); i++) {               // If the current character is not X             if (s.charAt(i) != 'X') {                   // If the current character is 0                 if (s.charAt(i) == '0') {                       // Add the count of Xs to count0                     count0 += i - prev - 1;                       // Set prev to i                     prev = i;                 }                   // Otherwise                 else {                       // Find the next occurrence                     // of 0 in the string                     boolean flag = true;                       for (int j = i + 1;                          j < s.length(); j++) {                           if (s.charAt(j) == '0') {                             prev = j;                             flag = false;                             break;                         }                     }                       // If it is found,                     // set i to prev                     if (!flag) {                         i = prev;                     }                       // Otherwise, break out                     // of the loop                     else {                         i = s.length();                     }                 }             }         }           // Count number of X present in         // the starting of the string         // as XXXX1...         if (s.charAt(0) == 'X') {               // Store the count of X             int count = 0;             int i = 0;             while (s.charAt(i) == 'X') {                 count++;                 i++;             }               // Increment count1 by             // count if the condition             // is satisfied             if (s.charAt(i) == '1') {                 count1 += count;             }         }           // Count the number of X         // present in the ending of         // the string as ...XXXX0         if (s.charAt(s.length() - 1)             == 'X') {               // Store the count of X             int count = 0;             int i = s.length() - 1;             while (s.charAt(i) == 'X') {                 count++;                 i--;             }               // Increment count0 by             // count if the condition             // is satisfied             if (s.charAt(i) == '0') {                 count0 += count;             }         }           // If count of 1 is equal to         // count of 0, print X         if (count0 == count1) {             System.out.println("X");         }           // Otherwise, if count of 1         // is greater than count of 0         else if (count0 > count1) {             System.out.println(0);         }           // Otherwise, print 0         else             System.out.println(1);     }       // Driver Code     public static void main(String[] args)     {         String S = "XX10XX10XXX1XX";         maxOccurringCharacter(S);     } }

## Python3

 # Python program for the above approach   # Function to find the most frequent # character after replacing X with # either '0' or '1' according as per # the given conditions def maxOccurringCharacter(s):       # Store the count of 0s and   # 1s in the S   count0 = 0   count1 = 0     # Count the frequency of   # 0 and 1   for i in range(len(s)):       # If the character is 1     if (s[i] == '1') :       count1 += 1           # If the character is 0     elif (s[i] == '0') :       count0 += 1         # Stores first occurrence of 1   prev = -1   for i in range(len(s)):     if (s[i] == '1') :       prev = i       break         # Traverse the to count   # the number of X between two   # consecutive 1s   for i in range(prev + 1, len(s)):       # If the current character     # is not X     if (s[i] != 'X') :         # If the current character       # is 1, add the number of       # Xs to count1 and set       # prev to i       if (s[i] == '1') :         count1 += i - prev - 1         prev = i               # Otherwise       else :           # Find next occurrence         # of 1 in the string         flag = True         for j in range(i+1, len(s)):           if (s[j] == '1') :             flag = False             prev = j             break                     # If it is found,         # set i to prev         if (flag == False) :           i = prev                   # Otherwise, break         # out of the loop         else :           i = len(s)             # Store the first occurrence of 0   prev = -1   for i in range(0, len(s)):       if (s[i] == '0') :       prev = i       break         # Repeat the same procedure to   # count the number of X between   # two consecutive 0s   for i in range(prev + 1, len(s)):       # If the current character is not X     if (s[i] != 'X') :         # If the current character is 0       if (s[i] == '0') :           # Add the count of Xs to count0         count0 += i - prev - 1           # Set prev to i         prev = i               # Otherwise       else :           # Find the next occurrence         # of 0 in the string         flag = True           for j in range(i + 1, len(s)):           if (s[j] == '0') :             prev = j             flag = False             break                    # If it is found,         # set i to prev         if (flag == False) :           i = prev                   # Otherwise, break out         # of the loop         else :           i = len(s)            # Count number of X present in   # the starting of the string   # as XXXX1...   if (s[0] == 'X') :       # Store the count of X     count = 0     i = 0     while (s[i] == 'X') :       count += 1       i += 1           # Increment count1 by     # count if the condition     # is satisfied     if (s[i] == '1') :       count1 += count        # Count the number of X   # present in the ending of   # the as ...XXXX0   if (s[(len(s) - 1)]       == 'X') :       # Store the count of X     count = 0     i = len(s) - 1     while (s[i] == 'X') :       count += 1       i -= 1           # Increment count0 by     # count if the condition     # is satisfied     if (s[i] == '0') :       count0 += count         # If count of 1 is equal to   # count of 0, print X   if (count0 == count1) :     print("X")       # Otherwise, if count of 1   # is greater than count of 0   elif (count0 > count1) :     print( 0 )       # Otherwise, print 0   else:     print(1)   # Driver Code   S = "XX10XX10XXX1XX" maxOccurringCharacter(S)   # This code is contributed by sanjoy_62.

## C#

 // C# program for the above approach using System; public class GFG {     // Function to find the most frequent   // character after replacing X with   // either '0' or '1' according as per   // the given conditions   public static void maxOccurringCharacter(string s)   {       // Store the count of 0s and     // 1s in the string S     int count0 = 0, count1 = 0;       // Count the frequency of     // 0 and 1     for (int i = 0;          i < s.Length; i++) {         // If the character is 1       if (s[i] == '1') {         count1++;       }         // If the character is 0       else if (s[i] == '0') {         count0++;       }     }       // Stores first occurrence of 1     int prev = -1;       for (int i = 0;          i < s.Length; i++) {         if (s[i] == '1') {         prev = i;         break;       }     }       // Traverse the string to count     // the number of X between two     // consecutive 1s     for (int i = prev + 1;          i < s.Length; i++) {         // If the current character       // is not X       if (s[i] != 'X') {           // If the current character         // is 1, add the number of         // Xs to count1 and set         // prev to i         if (s[i] == '1') {           count1 += i - prev - 1;           prev = i;         }           // Otherwise         else {             // Find next occurrence           // of 1 in the string           bool flag = true;             for (int j = i + 1;                j < s.Length;                j++) {             if (s[j] == '1') {               flag = false;               prev = j;               break;             }           }             // If it is found,           // set i to prev           if (!flag) {             i = prev;           }             // Otherwise, break           // out of the loop           else {             i = s.Length;           }         }       }     }       // Store the first occurrence of 0     prev = -1;     for (int i = 0; i < s.Length; i++) {         if (s[i] == '0') {         prev = i;         break;       }     }       // Repeat the same procedure to     // count the number of X between     // two consecutive 0s     for (int i = prev + 1;          i < s.Length; i++) {         // If the current character is not X       if (s[i] != 'X') {           // If the current character is 0         if (s[i] == '0') {             // Add the count of Xs to count0           count0 += i - prev - 1;             // Set prev to i           prev = i;         }           // Otherwise         else {             // Find the next occurrence           // of 0 in the string           bool flag = true;             for (int j = i + 1;                j < s.Length; j++) {               if (s[j] == '0') {               prev = j;               flag = false;               break;             }           }             // If it is found,           // set i to prev           if (!flag) {             i = prev;           }             // Otherwise, break out           // of the loop           else {             i = s.Length;           }         }       }     }       // Count number of X present in     // the starting of the string     // as XXXX1...     if (s[0] == 'X') {         // Store the count of X       int count = 0;       int i = 0;       while (s[i] == 'X') {         count++;         i++;       }         // Increment count1 by       // count if the condition       // is satisfied       if (s[i] == '1') {         count1 += count;       }     }       // Count the number of X     // present in the ending of     // the string as ...XXXX0     if (s[s.Length - 1]         == 'X') {         // Store the count of X       int count = 0;       int i = s.Length - 1;       while (s[i] == 'X') {         count++;         i--;       }         // Increment count0 by       // count if the condition       // is satisfied       if (s[i] == '0') {         count0 += count;       }     }       // If count of 1 is equal to     // count of 0, print X     if (count0 == count1) {       Console.WriteLine("X");     }       // Otherwise, if count of 1     // is greater than count of 0     else if (count0 > count1) {       Console.WriteLine(0);     }       // Otherwise, print 0     else       Console.WriteLine(1);   }     // Driver Code   public static void Main(string[] args)   {     string S = "XX10XX10XXX1XX";     maxOccurringCharacter(S);   } }   // This code is contributed by AnkThon

## Javascript



Output:

1

Time Complexity: O(N)
Auxiliary Space: O(1)