Class Solution
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public final class Solution2220 - Minimum Bit Flips to Convert Number\.
Easy
A bit flip of a number
xis choosing a bit in the binary representation ofxand flipping it from either0to1or1to0.For example, for
x = 7, the binary representation is111and we may choose any bit (including any leading zeros not shown) and flip it. We can flip the first bit from the right to get110, flip the second bit from the right to get101, flip the fifth bit from the right (a leading zero) to get10111, etc.
Given two integers
startandgoal, return the minimum number of bit flips to convertstarttogoal.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:
<code>0 <= start, goal <= 10<sup>9</sup></code>
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
Modifier and Type Method Description final IntegerminBitFlips(Integer start, Integer goal)-
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Method Detail
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minBitFlips
final Integer minBitFlips(Integer start, Integer goal)
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