Super quick solution would be just count the no of 1 in binary form. If its 1. Return true.
In C++ just do __builtin_popcount(n) == 1
Yep, it one of the solutions, you're in right direction now combine the solution of the original question with this.
hint : if n&n-1 = 0, it means that the rest of the bit is guaranteed set to 0, so no need to parse them.
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Super quick solution would be just count the no of 1 in binary form. If its 1. Return true.
In C++ just do __builtin_popcount(n) == 1
Yep, it one of the solutions, you're in right direction now combine the solution of the original question with this.
hint : if n&n-1 = 0, it means that the rest of the bit is guaranteed set to 0, so no need to parse them.