# Bitwise Operations on Digits of a Number

• Last Updated : 14 Aug, 2021

Given a number N, the task is to perform the bitwise operations on digits of the given number N. The bitwise operations include:

• Finding the XOR of all digits of the given number N
• Finding the OR of all digits of the given number N
• Finding the AND of all digits of the given number N

Examples:

Input: N = 486
Output:
XOR = 10
OR = 14
AND = 0

Input: N = 123456
Output:
XOR = 10
OR = 14
AND = 0

Approach:

1. Get the number

2. Find the digits of the number and store it in an array for computation purpose.

3. Now perform the various bitwise operations (XOR, OR, and AND) on this array one by one.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std; int digit[100000]; // Function to find the digitsint findDigits(int n){    int count = 0;     while (n != 0) {         digit[count] = n % 10;        n = n / 10;        ++count;    }     return count;} // Function to Find OR// of all digits of a numberint OR_of_Digits(int n, int count){     int ans = 0;     for (int i = 0; i < count; i++) {        // Find OR of all digits        ans = ans | digit[i];    }     // return OR of digits    return ans;} // Function to Find AND// of all digits of a numberint AND_of_Digits(int n, int count){     int ans = 0;     for (int i = 0; i < count; i++) {        // Find AND of all digits        ans = ans & digit[i];    }     // return AND of digits    return ans;} // Function to Find XOR// of all digits of a numberint XOR_of_Digits(int n, int count){     int ans = 0;     for (int i = 0; i < count; i++) {        // Find XOR of all digits        ans = ans ^ digit[i];    }     // return XOR of digits    return ans;} // Driver codevoid bitwise_operation(int N){     // Find and store all digits    int countOfDigit = findDigits(N);     // Find XOR of digits    cout << "XOR = "         << XOR_of_Digits(N, countOfDigit)         << endl;     // Find OR of digits    cout << "OR = "         << OR_of_Digits(N, countOfDigit)         << endl;     // Find AND of digits    cout << "AND = "         << AND_of_Digits(N, countOfDigit)         << endl;} // Driver codeint main(){     int N = 123456;     bitwise_operation(N);     return 0;}

## Java

 // Java implementation of the approachimport java.util.*; class GFG{ static int []digit = new int[100000]; // Function to find the digitsstatic int findDigits(int n){    int count = 0;     while (n != 0) {         digit[count] = n % 10;        n = n / 10;        ++count;    }     return count;} // Function to Find OR// of all digits of a numberstatic int OR_of_Digits(int n, int count){     int ans = 0;     for (int i = 0; i < count; i++) {                 // Find OR of all digits        ans = ans | digit[i];    }     // return OR of digits    return ans;} // Function to Find AND// of all digits of a numberstatic int AND_of_Digits(int n, int count){     int ans = 0;     for (int i = 0; i < count; i++) {                 // Find AND of all digits        ans = ans & digit[i];    }     // return AND of digits    return ans;} // Function to Find XOR// of all digits of a numberstatic int XOR_of_Digits(int n, int count){     int ans = 0;     for (int i = 0; i < count; i++) {                 // Find XOR of all digits        ans = ans ^ digit[i];    }     // return XOR of digits    return ans;} // Driver codestatic void bitwise_operation(int N){     // Find and store all digits    int countOfDigit = findDigits(N);     // Find XOR of digits    System.out.print("XOR = "        + XOR_of_Digits(N, countOfDigit)        +"\n");     // Find OR of digits    System.out.print("OR = "        + OR_of_Digits(N, countOfDigit)        +"\n");     // Find AND of digits    System.out.print("AND = "        + AND_of_Digits(N, countOfDigit)        +"\n");} // Driver codepublic static void main(String[] args){     int N = 123456;     bitwise_operation(N);}} // This code is contributed by sapnasingh4991

## Python 3

 # Python 3 implementation of the approachdigit = [0]*(100000) # Function to find the digitsdef findDigits(n):    count = 0     while (n != 0):         digit[count] = n % 10;        n = n // 10;        count += 1     return count # Function to Find OR# of all digits of a numberdef OR_of_Digits( n,count):    ans = 0     for i in range(count):                 # Find OR of all digits        ans = ans | digit[i]     # return OR of digits    return ans # Function to Find AND# of all digits of a numberdef AND_of_Digits(n, count):     ans = 0     for i in range(count):                 # Find AND of all digits        ans = ans & digit[i]     # return AND of digits    return ans # Function to Find XOR# of all digits of a numberdef XOR_of_Digits(n, count):     ans = 0     for i in range(count):                 # Find XOR of all digits        ans = ans ^ digit[i]     # return XOR of digits    return ans # Driver codedef bitwise_operation( N):     # Find and store all digits    countOfDigit = findDigits(N)     # Find XOR of digits    print("XOR = ",XOR_of_Digits(N, countOfDigit))     # Find OR of digits    print("OR = ",OR_of_Digits(N, countOfDigit))     # Find AND of digits    print("AND = ",AND_of_Digits(N, countOfDigit)) # Driver codeN = 123456;bitwise_operation(N) # This code is contributed by apurva raj

## C#

 // C# implementation of the approachusing System; class GFG{  static int []digit = new int[100000];  // Function to find the digitsstatic int findDigits(int n){    int count = 0;      while (n != 0) {          digit[count] = n % 10;        n = n / 10;        ++count;    }      return count;}  // Function to Find OR// of all digits of a numberstatic int OR_of_Digits(int n, int count){      int ans = 0;      for (int i = 0; i < count; i++) {                  // Find OR of all digits        ans = ans | digit[i];    }      // return OR of digits    return ans;}  // Function to Find AND// of all digits of a numberstatic int AND_of_Digits(int n, int count){      int ans = 0;      for (int i = 0; i < count; i++) {                  // Find AND of all digits        ans = ans & digit[i];    }      // return AND of digits    return ans;}  // Function to Find XOR// of all digits of a numberstatic int XOR_of_Digits(int n, int count){      int ans = 0;      for (int i = 0; i < count; i++) {                  // Find XOR of all digits        ans = ans ^ digit[i];    }      // return XOR of digits    return ans;}  // Driver codestatic void bitwise_operation(int N){      // Find and store all digits    int countOfDigit = findDigits(N);      // Find XOR of digits    Console.Write("XOR = "        + XOR_of_Digits(N, countOfDigit)        +"\n");      // Find OR of digits    Console.Write("OR = "        + OR_of_Digits(N, countOfDigit)        +"\n");      // Find AND of digits    Console.Write("AND = "        + AND_of_Digits(N, countOfDigit)        +"\n");}  // Driver codepublic static void Main(String[] args){      int N = 123456;      bitwise_operation(N);}} // This code is contributed by 29AjayKumar

## Javascript


Output:
XOR = 7
OR = 7
AND = 0

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

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