How do you solve Minimum Flips to Make a OR b Equal to c on LeetCode?

Minimum Flips to Make a OR b Equal to c (LeetCode #1318) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Understanding bit manipulation is crucial for solving problems involving binary representations.

What is the time complexity of Minimum Flips to Make a OR b Equal to c?

The optimal solution for Minimum Flips to Make a OR b Equal to c runs in O(1) time and uses O(1) space. The optimal solution also runs in O(1) time as we are still iterating through a fixed number of bits (32).

What is the brute force solution for Minimum Flips to Make a OR b Equal to c?

The brute force approach for Minimum Flips to Make a OR b Equal to c is Brute Force, with O(1) time complexity and O(1) space complexity. The brute force approach involves checking each bit of a, b, and c to see if we need to flip any bits to make a OR b equal to c. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Minimum Flips to Make a OR b Equal to c?

Minimum Flips to Make a OR b Equal to c uses the following concepts: Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Greedy Algorithms.

What are the interview tips for Minimum Flips to Make a OR b Equal to c?

Always clarify the problem statement and constraints before jumping into coding. Think aloud while solving the problem to demonstrate your thought process. Practice bit manipulation problems to become more comfortable with binary operations.

What are common mistakes when solving Minimum Flips to Make a OR b Equal to c?

Forgetting to check all bits from 0 to 31, which can lead to incorrect results. Confusing the conditions for flipping bits when c is 0 versus when c is 1.

#1318

Minimum Flips to Make a OR b Equal to c

Medium

Bit ManipulationBit ManipulationGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(1)	O(1)
Space	O(1)	O(1)

💡

Intuition

Time O(1)Space O(1)

The optimal solution uses bit manipulation to directly assess the bits of a, b, and c without unnecessary iterations. It focuses on the conditions that dictate whether a flip is needed.

⚙️

Algorithm

4 steps

1Step 1: Initialize a flips counter to 0.
2Step 2: For each bit position from 0 to 31, extract the bits of a, b, and c.
3Step 3: If the c bit is 0, count the number of 1s in a and b at that position (both need to be flipped).
4Step 4: If the c bit is 1, check if both a and b are 0 (both need to be flipped to 1).

solution.py13 lines

1# Full working Python code
2
3def minFlips(a, b, c):
4    flips = 0
5    for i in range(32):
6        a_bit = (a >> i) & 1
7        b_bit = (b >> i) & 1
8        c_bit = (c >> i) & 1
9        if c_bit == 0:
10            flips += a_bit + b_bit
11        else:
12            flips += (1 - (a_bit | b_bit))
13    return flips

ℹ

Complexity note: The optimal solution also runs in O(1) time as we are still iterating through a fixed number of bits (32).

1Understanding bit manipulation is crucial for solving problems involving binary representations.
2Flipping bits can be visualized as toggling switches, where each switch represents a binary digit.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.