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CoCubes Coding Questions and Answers 

Q) Given a string consisting of only 0, 1, A, B, C where
A = AND
B = OR
C = XOR
Calculate the value of the string assuming no order of precedence and evaluation is done from left to right.

Constraints – The length of string will be odd. It will always be a valid string.
Example, 1AA0 will not be given as an input.

Examples:

Input: 1A0B1

Output : 1 1 AND 0 OR 1 = 1 Input : 1C1B1B0A0 Output : 0
C Program 

#include <stdio.h>
#include <string.h>
int evaluateBoolExpr(char* s)
{
    int n = strlen(s);
    for (int i = 0; i < n; i += 2)
    {
        // If operator next to current operand
        // is AND.
        if (i + 1 < n && i + 2 < n)
        {
            if (s[i+1] == 'A')
            {
                if (s[i+2] == '0'||s[i] == '0')
                    s[i+2] = '0';
                else
                    s[i+2] = '1';
            }
            // If operator next to current operand
            // is OR.
            else if ((i + 1) < n && s[i+1] == 'B')
            {
                if (s[i+2] == '1'||s[i] == '1')
                    s[i+2] = '1';
                else
                    s[i+2] = '0';
            }
            // If operator next to current operand
            // is XOR (Assuming a valid input)
            else
            {
                if (s[i+2] == s[i])
                    s[i+2] = '0';
                    
                else
                    s[i+2] = '1';
            }
        }
    }
    return s[n-1] - '0';
} 
int main()
{
    char str[100];
    scanf("%[^\n]s",str);
    int result = evaluateBoolExpr(str);
    printf("%d",result);
}

Java Program

// Java program to evaluate value of an expression.
class Main
{

// Evaluates boolean expression
// and returns the result
  static int evaluateBoolExpr (StringBuffer s)
  {
    int n = s.length ();

// Traverse all operands by jumping
// a character after every iteration.
    for (int i = 0; i < n; i += 2)
      {

// If operator next to current operand
// is AND.
	if (i + 1 < n && i + 2 < n)
	  {
	    if (s.charAt (i + 1) == 'A')
	      {
		if (s.charAt (i + 2) == '0'||s.charAt (i) == 0)
		  s.setCharAt (i + 2, '0');
		else
		  s.setCharAt (i + 2, '1');
	      }

// If operator next to current operand
// is OR.
	    else if ((i + 1) < n && s.charAt (i + 1) == 'B')
	      {
		if (s.charAt (i + 2) == '1'||s.charAt (i) == '1')
		  s.setCharAt (i + 2, '1');
		else
		  s.setCharAt (i + 2, '0');
	      }

// If operator next to current operand
// is XOR (Assuming a valid input)
	    else
	      {
		if (s.charAt (i + 2) == s.charAt (i))
		  s.setCharAt (i + 2, '0');
		else
		  s.setCharAt (i + 2, '1');
	      }
	  }
      }
    return s.charAt (n - 1) - '0';
  }

// Driver code
  public static void main (String[]args)
  {
    String s = "1C1B1B0A0";
    StringBuffer sb = new StringBuffer (s);
    System.out.println (evaluateBoolExpr (sb));
  }
}

Python Program

def solve(s):
    s=s.replace("A","&").replace("B","|").replace("C","^")
    return eval(s)

s=input()
print(solve(s))

Q) Make a function which accepts a string as an argument that may contain repetitive characters. Implement the function to modify and return the input string, such that each character once, along with the count of consecutive occurrence. Do not append count if the character occurs only once.

Note – 

• The string will only contain lowercase English Alphabets
• If you have to manipulate the input string in place you cant use another string

Assumption –
No character will occur consecutively more than 9 times.

Example – 

Input
aaaaabbbccccccccdaa

Output
a4b3c8da2

Java Program 

import java.util.Scanner;
public class Main
{
  public static void main (String[]args)
  {
    try (Scanner sc = new Scanner (System.in);
      )
    {
      StringBuilder sb = new StringBuilder (sc.nextLine ());
      int count = 1;
      char current = sb.charAt (0), next;
      for (int i = 1; i < sb.length (); i++)
	{
	  next = sb.charAt (i);
	  if (next == current)
	    {
	      sb.deleteCharAt (i);
	      i = i - 1;
	      count++;
	    }
	  else
	    {
	      sb.insert (i, count);
	      count = 1;
	      current = next;
	      i = i + 1;
	    }
	}
      sb.append (count);
      System.out.println (sb);
    }
  }
}

def solve(s):
    ans=""
    c=1
    for i in range(len(s)-1):
        if(s[i]==s[i+1]):
            c+=1
        else:
            if(c==1):
                ans+=s[i]
            else:
                ans+=s[i]+str(c)
            c=1 
    if(c==1):
        ans+=s[i+1]
    else:
        ans+=s[i+1]+str(c)
    return ans


s=input()
print(solve(s))

import java.util.Scanner;
public class Main
{

  public static void main (String[]args)
  {
    try (Scanner sc = new Scanner (System.in);
      )
    {
      StringBuilder sb = new StringBuilder (sc.nextLine ());
      StringBuilder s = new StringBuilder ();
      Boolean finish = true;
      while (finish)
	{
	  if (isPalindrome (sb))
	  {
	      System.out.println("NULL");
	  finish = false;
	  }

	  else
	    {
	      s.append (sb.charAt (0));
	      sb = new StringBuilder (sb.substring (1));
	    }
	}
      System.out.println (s.reverse ());
    }
  }
  public static Boolean isPalindrome (StringBuilder sb)
  {
    int l = sb.length () / 2;
    int m = sb.length () - 1;
    for (int i = 0; i <= l; i++)
      {
	if (!(sb.charAt (i) == sb.charAt (m - i)))
	  return false;
      }
    return true;
  }

}

def ispalindrome(s):
    return s==s[::-1]

def solve(s):
    if(ispalindrome(s)):
        return None
    for i in range(len(s)):
        x=s[:i][::-1]
        if(ispalindrome(s+x)):
            return x 
        
s=input()
print(solve(s))

#include <stdio.h>
#include <string.h>
int solve(int n,int m)
{
    if(n>m)
        return (n*m)-(n-m);
    else
        return (m%n)-(m+n);
} 
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    int result = solve(n,m);
    printf("%d",result);
}

C++ Program 

#include <iostream>

using namespace std;
int solve(int n,int m)
{
    if(n>m)
        return (n*m)-(n-m);
    else
        return (m%n)-(m+n);
} 
int main()
{
    int n,m;
    cin>>n>>m;
    int result = solve(n,m);
    cout<< result;
    return 0;
}

Java Program 

import java.util.Scanner;
import java.io.*;

class Main {
    public static void main (String[] args) {
        Scanner prep = new Scanner( System.in );
          int n,m;
          n= prep.nextInt();
          m= prep.nextInt();
        System.out.println(solve(n,m));
    }
    public static int solve(int n, int m)
    {
        if(n>m)
        return (n*m)-(n-m);
        else
        return (m%n)-(m+n);
    }
}

def solve(n,m):
    if(n>m):
        return (n*m)-(n-m)
    else:
        return (m%n)-(m+n)
n=int(input())
m=int(input())
print(solve(n,m))

def solve(arr,n):
    count=0 
    for i in set(arr):
        if(arr.count(i)%2==1):
            count+=i
    return count

n=int(input())
arr=list(map(int,input().split()))
print(solve(arr,n))

C Program 

#include <stdio.h>
#include<string.h>
int solve(char str[])
{

    int count= 0;
    for (int i = 0; str[i]!='\0'; i++) 
    {
        if ((str[i] >= 'A' && str[i] <= 'Z')|| (str[i] >= 'a' && str[i] <= 'z') )
            count++;
        if(str[i] >= '0' && str[i] <= '9')
            count++;
    }

    return count;
}

int main()
{
    char str[]= "Hello World!123";
    printf("%d",solve(str));
    return 0;
}

#include <iostream>
using namespace std;

int solve(string str)
{

    int count= 0;
    for (int i = 0; i < str.length(); i++) 
    {
        if ((str[i] >= 'A' && str[i] <= 'Z')|| (str[i] >= 'a' && str[i] <= 'z') )
            count++;
        if(str[i] >= '0' && str[i] <= '9')
            count++;
    }

    return count;
}

int main()
{
    string str = "Hello World!123";
    cout<< solve(str);
    return 0;
}

import java.util.Scanner;
import java.io.*;

class Main {
    public static void main (String[] args) {
        System.out.println(solve("Hello World!123"));
    }
    public static int solve(String str)
    {
        int count= 0;
        char[] str1 = str.toCharArray();
        for (int i = 0;i< str1.length; i++) 
        {
            if ((str1[i] >= 'A' && str1[i] <= 'Z')|| (str1[i] >= 'a' && str1[i] <= 'z') )
                count++;
            if(str1[i] >= '0' && str1[i] <= '9')
                count++;
        }
    
        return count;
    }
}

Python Program 

def solve(str):
    count=0 
    for i in str:
        if(i.isalpha() or i.isnumeric()):
            count+=1 
    return count

n=input()
print(solve(n))

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