Decimal to Binary

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Given a decimal number N, compute its binary equivalent.

Examples:

Input: N = 7

Output: 111

Input: N = 33

Output: 100001

Input: N = 0

Output: 0

Constraints

0 ≤ N ≤ 10⁶

Editorial

Intuition:

The binary representation of a decimal number is obtained by repeatedly dividing the number by 2 and collecting the remainders in reverse order.

Approach:

If N is 0, output "0" directly.
Initialize an empty string to store binary digits.
While N > 0, compute remainder = N % 2, prepend it to the result string, and update N = N // 2.
The result string now contains the binary equivalent.

#include <iostream>
#include <string>

class Solution {
public:
    void decimalToBinary(int N) {
        if (N == 0) {
            std::cout << "0";
            return;
        }
        std::string binary = "";
        while (N > 0) {
            int rem = N % 2;
            binary = std::to_string(rem) + binary;
            N = N / 2;
        }
        std::cout << binary;
    }
};
class Solution {
    public void decimalToBinary(int N) {
        if (N == 0) {
            System.out.print("0");
            return;
        }
        StringBuilder binary = new StringBuilder();
        while (N > 0) {
            int rem = N % 2;
            binary.insert(0, rem);
            N = N / 2;
        }
        System.out.print(binary.toString());
    }
}
class Solution:
    def decimal_to_binary(self, N):
        if N == 0:
            print("0")
            return
        binary = ""
        while N > 0:
            rem = N % 2
            binary = str(rem) + binary
            N = N // 2
        print(binary)
class Solution {
    decimalToBinary(N) {
        if (N === 0) {
            console.log("0");
            return;
        }
        let binary = "";
        while (N > 0) {
            let rem = N % 2;
            binary = rem.toString() + binary;
            N = Math.floor(N / 2);
        }
        console.log(binary);
    }
}

Complexity Analysis:

Time Complexity: O(log N), as the loop runs for the number of bits in N.

Space Complexity: O(log N), for storing the binary string of length log₂(N).

Notes

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