LeetCode #67 — EASY

Add Binary

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

Problem Statement

Given two binary strings a and b, return their sum as a binary string.

Example 1:

Input: a = "11", b = "1"
Output: "100"

Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

Constraints:

1 <= a.length, b.length <= 10⁴
a and b consist only of '0' or '1' characters.
Each string does not contain leading zeros except for the zero itself.

Step 01

Brute Force Baseline

Problem summary: Given two binary strings a and b, return their sum as a binary string. Example 1: Input: a = "11", b = "1" Output: "100" Example 2: Input: a = "1010", b = "1011" Output: "10101"

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Bit Manipulation

Example 1

"11"
"1"

Example 2

"1010"
"1011"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

class Solution {
    public String addBinary(String a, String b) {
        StringBuilder sb = new StringBuilder();
        int i = a.length() - 1, j = b.length() - 1, carry = 0;

        while (i >= 0 || j >= 0 || carry != 0) {
            int sum = carry;
            if (i >= 0) sum += a.charAt(i--) - '0';
            if (j >= 0) sum += b.charAt(j--) - '0';
            sb.append(sum % 2);
            carry = sum / 2;
        }

        return sb.reverse().toString();
    }
}

func addBinary(a string, b string) string {
    i, j, carry := len(a)-1, len(b)-1, 0
    out := make([]byte, 0)

    for i >= 0 || j >= 0 || carry != 0 {
        sum := carry
        if i >= 0 {
            sum += int(a[i] - '0')
            i--
        }
        if j >= 0 {
            sum += int(b[j] - '0')
            j--
        }
        out = append(out, byte(sum%2)+'0')
        carry = sum / 2
    }

    for l, r := 0, len(out)-1; l < r; l, r = l+1, r-1 {
        out[l], out[r] = out[r], out[l]
    }
    return string(out)
}

class Solution:
    def addBinary(self, a: str, b: str) -> str:
        i, j = len(a) - 1, len(b) - 1
        carry = 0
        out = []

        while i >= 0 or j >= 0 or carry:
            total = carry
            if i >= 0:
                total += ord(a[i]) - ord('0')
                i -= 1
            if j >= 0:
                total += ord(b[j]) - ord('0')
                j -= 1

            out.append(str(total % 2))
            carry = total // 2

        return ''.join(reversed(out))

impl Solution {
    pub fn add_binary(a: String, b: String) -> String {
        let bytes_a = a.as_bytes();
        let bytes_b = b.as_bytes();
        let mut i = bytes_a.len() as i32 - 1;
        let mut j = bytes_b.len() as i32 - 1;
        let mut carry = 0;
        let mut out: Vec<u8> = Vec::new();

        while i >= 0 || j >= 0 || carry != 0 {
            let mut sum = carry;
            if i >= 0 {
                sum += (bytes_a[i as usize] - b'0') as i32;
                i -= 1;
            }
            if j >= 0 {
                sum += (bytes_b[j as usize] - b'0') as i32;
                j -= 1;
            }
            out.push((sum % 2) as u8 + b'0');
            carry = sum / 2;
        }

        out.reverse();
        String::from_utf8(out).unwrap()
    }
}

function addBinary(a: string, b: string): string {
  let i = a.length - 1;
  let j = b.length - 1;
  let carry = 0;
  const out: string[] = [];

  while (i >= 0 || j >= 0 || carry !== 0) {
    let sum = carry;
    if (i >= 0) sum += a.charCodeAt(i--) - 48;
    if (j >= 0) sum += b.charCodeAt(j--) - 48;
    out.push(String(sum % 2));
    carry = Math.floor(sum / 2);
  }

  return out.reverse().join('');
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.