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 <bits/stdc++.h>
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

<script>
 
// javascript 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
 
function maxOccurringCharacter(s)
{
    // Store the count of 0s and
    // 1s in the string S
    var count0 = 0, count1 = 0;
 
    // Count the frequency of
    // 0 and 1
    for (var 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
    var prev = -1;
 
    for (var 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 (var 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
                flag = true;
 
                for (var 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 (var 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 (var 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
                flag = true;
 
                for (var 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
        var count = 0;
        var 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
        var count = 0;
        var 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) {
        document.write("X");
    }
 
    // Otherwise, if count of 1
    // is greater than count of 0
    else if (count0 > count1) {
        document.write(0);
    }
 
    // Otherwise, print 0
    else
        document.write(1);
}
 
// Driver Code
var S = "XX10XX10XXX1XX";
maxOccurringCharacter(S);
// This code is contributed by 29AjayKumar 
</script>

Output:

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

Find all Ramanujan Numbers that can be formed by numbers upto L

aditya7409

Improve

Article Tags :

Practice Tags :

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

C++

Java

Python3

C#

Javascript

Similar Reads

Thank You!

What kind of Experience do you want to share?