Sum of Two Integers Without + Operator

▶

Steps

setup

Initialize Inputs

We start with the input values a = 1 and b = 2. Both numbers are represented in 32-bit binary form for bitwise operations.

💡 Representing numbers in 32 bits allows us to visualize each bit's contribution to the sum and carry.

Line:

def getSum(a: int, b: int) -> int:

💡 Understanding the binary representation of inputs is essential to follow the bitwise addition process.

compare

Check if b is zero

The algorithm checks if b is zero. Since b=2, which is not zero, the function proceeds to calculate sum and carry.

💡 This check determines if the recursion should stop; if b is zero, the sum is complete.

Line:

if b == 0:
    return a if a <= 0x7FFFFFFF else ~(a ^ MASK)

💡 The presence of a non-zero carry means addition is not finished yet.

fill_row

Calculate sum as XOR of a and b

Compute sum_ = (a ^ b) & MASK. XOR adds bits without carry, showing bits where exactly one of a or b has a 1.

💡 XOR is the core operation that adds bits ignoring carry, revealing the partial sum.

Line:

sum_ = (a ^ b) & MASK

💡 XOR highlights bits that differ between a and b, forming the sum bits before carry is added.

fill_row

Calculate carry as AND of a and b, shifted left by 1

Compute carry = ((a & b) << 1) & MASK. AND finds bits where both a and b have 1, shifted left to represent carry to next bit.

💡 Carry bits indicate where addition causes a bit to overflow to the next higher bit.

Line:

carry = ((a & b) << 1) & MASK

💡 Carry bits must be added in the next recursion to complete the sum.

compare

Check if carry is zero

Since carry is zero, the recursion will stop and the sum is returned as the final result.

💡 A zero carry means no further addition is needed; the sum is complete.

Line:

if b == 0:
    return a if a <= 0x7FFFFFFF else ~(a ^ MASK)

💡 The algorithm terminates when there is no carry left to add.

reconstruct

Return final sum

The function returns the final sum value 3, which is the sum of 1 and 2 without using the '+' operator.

💡 Returning the final sum completes the recursive addition process.

Line:

return a if a <= 0x7FFFFFFF else ~(a ^ MASK)

💡 The final sum is correctly computed using bitwise operations.

traverse

Recursive call with sum and carry (if carry not zero)

If carry was not zero, the function would recursively call getSum with sum_ and carry as new inputs to continue addition.

💡 Recursion allows the algorithm to add carry bits until no carry remains.

Line:

return getSum(sum_, carry)

💡 Recursion is the mechanism to handle carry propagation.

fill_cells

Visualize bitwise XOR operation in detail

Highlight bits 30 and 31 where XOR sets bits in sum because bits differ between a and b.

💡 Seeing which bits XOR affects clarifies how partial sums form.

Line:

sum_ = (a ^ b) & MASK

💡 XOR sets bits where exactly one input bit is 1.

fill_cells

Visualize bitwise AND and shift left for carry

Highlight that no bits are set in carry because a and b have no overlapping 1 bits, so carry is zero.

💡 Understanding carry bits is crucial to see when addition needs further recursion.

Line:

carry = ((a & b) << 1) & MASK

💡 No overlapping bits means no carry to add.

reconstruct

Final result confirmed

The final sum 3 is confirmed as the correct output of the addition without using '+'.

💡 This step confirms the algorithm's correctness and completion.

Line:

print(getSum(1, 2))  # Output: 3

💡 Bitwise operations can fully replace arithmetic addition.

def getSum(a: int, b: int) -> int:
    MASK = 0xFFFFFFFF
    # STEP 2: Check if b is zero
    if b == 0:
        # STEP 6: Return final sum
        return a if a <= 0x7FFFFFFF else ~(a ^ MASK)
    # STEP 3: Calculate sum as XOR
    sum_ = (a ^ b) & MASK
    # STEP 4: Calculate carry as AND shifted left
    carry = ((a & b) << 1) & MASK
    # STEP 7: Recursive call with sum and carry
    return getSum(sum_, carry)

if __name__ == "__main__":
    print(getSum(1, 2))  # Output: 3  # STEP 10: Final output

📊

Sum of Two Integers Without + Operator - Watch the Algorithm Execute, Step by Step

Watching this step-by-step visualization reveals how addition can be decomposed into bitwise operations, making the process transparent and intuitive without relying on arithmetic operators.

Start

A

0

1

=1

B

0

1

0

=2

Check if b == 0

A

0

1

=1

B

0

1

0

=2

💡 b is not zero, so continue

sum = a XOR b

A

0

1

=1

B

0

1

0

=2

──────

=

0

1

1

=3

💡 Bits where a and b differ are set in sum

carry = (a AND b) << 1

A

0

1

=1

B

0

1

0

=2

────────────────

=

0

0

0

0

0

0

0

0

=0

💡 No bits set in carry since a and b have no overlapping 1s

Check if carry == 0

A

0

1

1

=3

B

=0

💡 Carry is zero, so recursion ends

Return sum

A

0

1

1

=3

B

💡 Final sum is 3

No recursion needed (carry is zero)

A

0

1

1

=3

B

=0

💡 No recursion needed here

XOR operation detail

A

0

1

=1

B

0

1

0

=2

──────

=

0

1

1

=3

💡 Bits 30 and 31 differ between a and b, so XOR sets them

AND and shift left for carry

A

0

1

=1

B

0

1

0

=2

────────────────

=

0

0

0

0

0

0

0

0

=0

💡 No bits set in carry

Final sum

A

0

1

1

=3

B

💡 Sum of 1 and 2 is 3

Key Takeaways

✓ Addition can be performed using only bitwise XOR and AND operations combined with bit shifts.

This insight is difficult to grasp from code alone because the recursive interplay of sum and carry is abstract without visualization.

✓ XOR computes the partial sum ignoring carry, while AND followed by a left shift computes the carry bits that must be added next.

Seeing the bits set by XOR and AND operations side-by-side clarifies their distinct roles in addition.

✓ The recursion stops when carry becomes zero, meaning no further bit overflow exists.

Understanding the stopping condition is easier when you see carry bits vanish in the visualization.

Practice

1. Consider the following code snippet implementing the trie-based maximum XOR finder. Given the input array [2, 8], what is the return value of findMaximumXOR([2, 8])?

easy

A. 2

B. 8

C. 10

D. 0

Solution

1. Step 1: Insert numbers 2 and 8 into trie

   2 in binary: 000...0010, 8 in binary: 000...1000. Trie stores bits for both numbers.

2. Step 2: For each number, find max XOR partner

   For 2, try to find bits opposite to 2's bits in trie. At bit 3 (8's highest set bit), toggled bit exists, so max_xor |= 8. At bit 1, toggled bit exists, max_xor |= 2. Total max_xor = 10.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   2 XOR 8 = 10, which is maximum XOR [OK]

Hint: XOR of 2 and 8 is 10, max XOR found [OK]

Common Mistakes:

• Confusing max_xor with input number
• Off-by-one bit indexing
• Returning XOR with self

2. You are given a sequence of integers representing bytes. You need to verify if this sequence forms a valid encoding where each character can be 1 to 4 bytes long, and continuation bytes must start with bits '10'. Which approach best guarantees an efficient and correct validation?

easy

A. Use a greedy approach that always assumes the longest possible character at each step.

B. Use a brute force approach checking each byte with nested conditions for all possible byte patterns.

C. Use dynamic programming to store valid prefixes and build up to the full sequence.

D. Use a single pass with a counter tracking how many continuation bytes are expected next.

Solution

1. Step 1: Understand the problem constraints

   The problem requires validating UTF-8 encoding where each character can be 1 to 4 bytes, and continuation bytes must start with '10'.

2. Step 2: Identify the approach that efficiently tracks continuation bytes

   The single pass with a continuation byte counter efficiently tracks how many continuation bytes are expected, ensuring correctness and O(n) time.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Single pass with counter matches UTF-8 rules and is optimal [OK]

Hint: Track continuation bytes with a counter for linear validation [OK]

Common Mistakes:

• Assuming greedy longest match always works
• Using DP unnecessarily complicates the problem

3. What is the time complexity of the Brian Kernighan's bit counting trick approach when counting set bits for all numbers from 0 to n?

medium

A. O(n * k) where k is the average number of set bits per number

B. O(n) because each number is processed in constant time

C. O(n * log n) because each number requires checking all bits

D. O(n^2) due to nested loops over bits and numbers

Solution

1. Step 1: Identify loops and operations

   Outer loop runs n+1 times; inner loop runs as many times as set bits in each number.

2. Step 2: Analyze average inner loop iterations

   Each iteration clears one set bit, so total inner loop iterations over all numbers is proportional to total set bits, roughly n * k.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Counting bits per number depends on number of set bits, not all bits [OK]

Hint: Time depends on total set bits, not total bits [OK]

Common Mistakes:

• Assuming O(n log n) because of bit length
• Thinking each number is constant time
• Confusing nested loops as quadratic

4. Suppose the problem is extended to find the bitwise AND of all numbers in the range [left, right], but now left and right can be negative integers. Which modification to the optimal algorithm is necessary?

hard

A. Convert negative numbers to their 2's complement positive equivalents and then apply the algorithm.

B. Use a mask to limit the bit-length and shift both numbers until equal, considering sign bits.

C. No modification needed; the existing bit-clearing approach works for negative numbers.

D. Iterate through the entire range because bitwise AND with negatives is undefined.

Solution

1. Step 1: Understand negative number representation

   Negative integers use 2's complement, so sign bits affect bitwise AND results.

2. Step 2: Modify approach

   Masking to limit bit-length and shifting both numbers until equal handles sign bits correctly and finds common prefix.

3. Step 3: Reject naive iteration

   Iterating entire range is inefficient and unnecessary; direct bitwise manipulation with sign awareness is needed.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Handling sign bits requires masking and shifting, not naive bit-clearing [OK]

Hint: Handle sign bits with masking and shifting for negatives [OK]

Common Mistakes:

• Assuming algorithm works unchanged for negatives
• Trying to convert negatives to positives incorrectly
• Iterating entire range for negatives

5. Suppose you want to check if a number is a power of two but the input can be negative or zero, and you want to handle 64-bit integers efficiently. Which modification to the bit manipulation approach is correct?

hard

A. Use the condition n > 0 and (n & (n - 1)) == 0 to handle all cases correctly.

B. Use n & (n + 1) == 0 to handle negative numbers as well.

C. Use floating point division to check if log2(n) is an integer for negative inputs.

D. Ignore negative inputs and only check positive numbers with n & (n - 1) == 0.

Solution

1. Step 1: Understand input constraints

   Negative and zero inputs are not powers of two, so must be excluded explicitly.

2. Step 2: Apply correct bit manipulation check

   Check n > 0 first, then use n & (n - 1) == 0 to verify exactly one set bit.

3. Step 3: Confirm correctness for 64-bit integers

   This approach works efficiently for 64-bit integers as well, since bitwise ops are constant time.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Explicit positive check avoids errors on zero/negative inputs [OK]

Hint: Always check n > 0 before bitwise power-of-two test [OK]

Common Mistakes:

• Using incorrect bitwise operations
• Ignoring negative inputs
• Using floating point logs causing precision errors