Package g2201_2300.s2220_minimum_bit_flips_to_convert_number

Class Solution

java.lang.Object
g2201_2300.s2220_minimum_bit_flips_to_convert_number.Solution
public class Solution extends Object
2220 - Minimum Bit Flips to Convert Number\. Easy A **bit flip** of a number `x` is choosing a bit in the binary representation of `x` and **flipping** it from either `0` to `1` or `1` to `0`. * For example, for `x = 7`, the binary representation is `111` and we may choose any bit (including any leading zeros not shown) and flip it. We can flip the first bit from the right to get `110`, flip the second bit from the right to get `101`, flip the fifth bit from the right (a leading zero) to get `10111`, etc. Given two integers `start` and `goal`, return _the **minimum** number of **bit flips** to convert_ `start` _to_ `goal`. **Example 1:** **Input:** start = 10, goal = 7 **Output:** 3 **Explanation:** The binary representation of 10 and 7 are 1010 and 0111 respectively. We can convert 10 to 7 in 3 steps: - Flip the first bit from the right: 1010 -> 1011. - Flip the third bit from the right: 1011 -> 1111\. - Flip the fourth bit from the right: 1111 -> 0111\. It can be shown we cannot convert 10 to 7 in less than 3 steps. Hence, we return 3. **Example 2:** **Input:** start = 3, goal = 4 **Output:** 3 **Explanation:** The binary representation of 3 and 4 are 011 and 100 respectively. We can convert 3 to 4 in 3 steps: - Flip the first bit from the right: 011 -> 010. - Flip the second bit from the right: 010 -> 000\. - Flip the third bit from the right: 000 -> 100\. It can be shown we cannot convert 3 to 4 in less than 3 steps. Hence, we return 3. **Constraints:** * 0 <= start, goal <= 109

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • minBitFlips

      public int minBitFlips(int start, int goal)