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# Solution1 bit operation

* t-complexity $O(C)$

    C is const with value 32
* s-complexity $O(1)$

iterate from low to high decimal

three cases of current bits:

* both bits are 1, current value depends on carry, and next carry = 1
* only one of the bits is 1
  * if carry is 1, then current value is 0, and next carry keeps 1
  * if carry is 0, then current value is 1, and next carry keeps 0

    these two cases can be summarized as res |= ((carry ^ 1) << i), and carry remain
* none of these bits are 0, current value is carry, next carry is 0