Leetcode 2220: Minimum Bit Flips to Convert Number

Input: You are given two integers, `start` and `goal`, representing the two numbers to be compared.

Example: start = 5, goal = 10

Constraints:

• 0 <= start, goal <= 10^9

Output: Return the minimum number of bit flips required to convert `start` into `goal`.

Example: Output: 3

Constraints:

• Both `start` and `goal` are non-negative integers.

Goal: The goal is to compute how many bits differ between `start` and `goal`.

Steps:

• XOR the two numbers `start` and `goal` to get the positions where the bits differ.

• Count the number of 1s in the result of the XOR operation, which represents the bits that differ.

Goal: The input integers `start` and `goal` are between 0 and 10^9. The solution must be efficient for large inputs.

Steps:

• 0 <= start, goal <= 10^9

Assumptions:

• The inputs are non-negative integers.

• The problem requires calculating the number of bit flips, which is based on bitwise operations.

• Input: Input: start = 5, goal = 10

• Explanation: The binary representation of 5 is 101, and the binary representation of 10 is 1010. To convert 5 to 10, flip the first, second, and fourth bits to get the number 1010, so the minimum number of bit flips is 3.

Approach: The approach involves using the XOR operation to identify which bits differ between `start` and `goal`, and then counting the number of differing bits.

Observations:

• The XOR operation will set a bit to 1 wherever the bits in `start` and `goal` differ.

• Counting the number of 1s in the result of XOR will give us the number of bit flips required.

• We need to perform the XOR operation between `start` and `goal` and count the number of bits set to 1 in the result.

Steps:

• Perform the XOR operation between `start` and `goal`.

• Count the number of 1s in the result using a function like `__builtin_popcount`.

Empty Inputs:

• The case where both `start` and `goal` are zero should return 0, as no bit flips are needed.

Large Inputs:

• The solution must handle large values for `start` and `goal`, up to 10^9.

Special Values:

• If `start` and `goal` are the same, the result should be 0, as no bit flips are needed.

Constraints:

• The solution must efficiently handle large input sizes.

int minBitFlips(int start, int goal) {
    return __builtin_popcount(start^goal);
}

1 : Function Declaration

int minBitFlips(int start, int goal) {

This is the function declaration for `minBitFlips`, which takes two integers, `start` and `goal`, and returns the minimum number of bit flips required to convert `start` into `goal`.

2 : Bitwise Operation and Count

    return __builtin_popcount(start^goal);

The bitwise XOR operation `start^goal` is performed to identify the differing bits between the two integers. The function `__builtin_popcount` is then used to count how many bits are set to 1, which represents the number of bit flips needed.

Best Case: O(1), when `start` and `goal` are the same.

Average Case: O(1), as bitwise operations and popcount are constant-time operations for fixed-size integers.

Worst Case: O(1), as the XOR operation and bit count are independent of the size of the integers within the 32-bit or 64-bit limit.

Description: The time complexity is constant for each operation.

Best Case: O(1), as no additional space is required.

Worst Case: O(1), as the solution uses constant space regardless of input size.

Description: The space complexity is constant, as no extra storage is needed beyond the input and output.

Leetcode 2220: Minimum Bit Flips to Convert Number

Explore →