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Binary representation of a given number

Last Updated : 17 Mar, 2025
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Given an integer n, the task is to print the binary representation of the number. 

Note: The given number will be maximum of 32 bits, so append 0's to the left if the result string is smaller than 30 length.

Examples:

Input: n = 2
Output: 00000000000000000000000000000010

Input: n = 0
Output: 00000000000000000000000000000000

Approach - Using Bit Manipulation - O(1) time and O(1) space

The idea is to check each bit position of the number from left to right (most significant bit to least significant bit). For each bit position, we use the bitwise AND operation with a mask created by left-shifting 1 to that position (1<<i). If the result is non-zero, it means that bit is set to 1 in the original number, so we append '1' to our answer string; otherwise, we append '0'.

C++
// C++ program to find binary  // representation of a given number #include <iostream> using namespace std;  string getBinaryRep(int n) {      string ans = "";          // Check for each bit.     for (int i=31; i>=0; i--) {                  // If i'th bit is set          if (n&(1<<i)) ans += '1';         else ans += '0';     }          return ans; }  int main() {     int n = 2;     cout << getBinaryRep(n);      return 0; } 
Java
// Java program to find binary  // representation of a given number  class GfG {          static String getBinaryRep(int n) {          String ans = "";                  // Check for each bit.         for (int i = 31; i >= 0; i--) {                          // If i'th bit is set              if ((n & (1 << i)) != 0) ans += '1';             else ans += '0';         }                  return ans;     }          public static void main(String[] args) {         int n = 2;         System.out.println(getBinaryRep(n));     } } 
Python
# Python program to find binary  # representation of a given number  def getBinaryRep(n):     ans = ""          # Check for each bit.     for i in range(31, -1, -1):                  # If i'th bit is set          if (n & (1 << i)):             ans += '1'         else:             ans += '0'          return ans  if __name__ == "__main__":     n = 2     print(getBinaryRep(n)) 
C#
// C# program to find binary  // representation of a given number  using System;  class GfG {          static string getBinaryRep(int n) {          string ans = "";                  // Check for each bit.         for (int i = 31; i >= 0; i--) {                          // If i'th bit is set              if ((n & (1 << i)) != 0) ans += '1';             else ans += '0';         }                  return ans;     }          static void Main() {         int n = 2;         Console.WriteLine(getBinaryRep(n));     } } 
JavaScript
// JavaScript program to find binary  // representation of a given number  function getBinaryRep(n) {      let ans = "";          // Check for each bit.     for (let i = 31; i >= 0; i--) {                  // If i'th bit is set          if ((n & (1 << i)) !== 0) ans += '1';         else ans += '0';     }          return ans; }  let n = 2; console.log(getBinaryRep(n)); 

Output
00000000000000000000000000000010

Approach - Using Modulus and Division Operator - O(1) time and O(1) space

The idea is to build a 32-bit binary representation by examining each bit of the number from right to left. We start by creating a string of 32 zeros, then iterate through all 32 bit positions. For each position, we check if the current rightmost bit is set (using n%2), and if so, we place a '1' at the appropriate position in our string. After checking each bit, we right-shift the number (using integer division by 2) to examine the next bit in the subsequent iteration.

The expression n%2 (n modulo 2) reveals the rightmost bit because any binary number ends with either 0 or 1. If the number is odd, it ends with 1 and n%2 equals 1; if it's even, it ends with 0 and n%2 equals 0. Meanwhile, the integer division n/2 effectively performs a right shift operation in binary. When we divide by 2, we're removing the rightmost bit and shifting all other bits one position to the right. For example, 10 (binary 1010) divided by 2 gives 5 (binary 101), effectively shifting all bits right by one position.

C++
// C++ program to find binary  // representation of a given number #include <iostream> using namespace std;  string getBinaryRep(int n) {      // Create a 32 length string      // of 0's      string ans = "";     for (int i=0; i<32; i++) ans += '0';          for (int i=0; i<32; i++) {                  // Set 1 f rightmost bit is set          if (n%2 == 1) ans[31 - i] = '1';                  // Right shift bits using divison          // operator          n = n / 2;     }          return ans; }  int main() {     int n = 2;     cout << getBinaryRep(n);      return 0; } 
Java
// Java program to find binary  // representation of a given number class GfG {      static String getBinaryRep(int n) {          // Create a 32 length char array          // of '0's          char[] ans = new char[32];         for (int i = 0; i < 32; i++) ans[i] = '0';          for (int i = 0; i < 32; i++) {                          // Set 1 if rightmost bit is set              if (n % 2 == 1) ans[31 - i] = '1';                          // Right shift bits using division              // operator              n = n / 2;         }          return new String(ans);     }      public static void main(String[] args) {         int n = 2;         System.out.println(getBinaryRep(n));     } } 
Python
# Python program to find binary  # representation of a given number  def getBinaryRep(n):      # Create a 32 length string      # of 0's      ans = ['0'] * 32          for i in range(32):                  # Set 1 if rightmost bit is set          if n % 2 == 1:             ans[31 - i] = '1'                  # Right shift bits using division          # operator          n = n // 2          return "".join(ans)  if __name__ == "__main__":     n = 2     print(getBinaryRep(n)) 
C#
// C# program to find binary  // representation of a given number  using System;  class GfG {          static string getBinaryRep(int n) {          // Create a 32 length string          // of 0's          char[] ans = new char[32];         for (int i = 0; i < 32; i++) ans[i] = '0';                  for (int i = 0; i < 32; i++) {                          // Set 1 if rightmost bit is set              if (n % 2 == 1) ans[31 - i] = '1';                          // Right shift bits using division              // operator              n = n / 2;         }                  return new string(ans);     }          static void Main() {         int n = 2;         Console.WriteLine(getBinaryRep(n));     } } 
JavaScript
// JavaScript program to find binary  // representation of a given number  function getBinaryRep(n) {      // Create a 32 length string      // of 0's      let ans = Array(32).fill('0');          for (let i = 0; i < 32; i++) {                  // Set 1 if rightmost bit is set          if (n % 2 === 1) ans[31 - i] = '1';                  // Right shift bits using division          // operator          n = Math.floor(n / 2);     }          return ans.join(''); }  let n = 2; console.log(getBinaryRep(n)); 

Output
00000000000000000000000000000010

Next Article
Count set bits in an integer

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Article Tags :
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Practice Tags :
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