Created
January 6, 2019 22:20
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Joshphus Problem using Bitwise Operation (Java)
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/* | |
* Solution of Joshphus Problem using Bitwise Operation | |
* Wikipedia: https://en.wikipedia.org/wiki/Josephus_problem | |
* YouTube: https://www.youtube.com/watch?v=uCsD3ZGzMgE | |
* | |
* ====================== EXPLANATION ====================== | |
* | |
* n (41) the number of people standing in the circle | |
* n = 101001 | |
* | |
* Left Shift n | |
* (n<<1) = 1010010 | |
* | |
* Flip the last bit | |
* ((n<<1) | 1) = 1010011 | |
* | |
* To Flip the First Set Bit | |
* n = 101001 | |
* | |
* n*2 | |
* Multiply n by 2 | |
* 41 x 2 = 82 | |
* 82 = 1010010 | |
* | |
* Integer.highestOneBit(n*2) | |
* Get highest set bit position | |
* 64 = 1000000 | |
* | |
* ~Integer.highestOneBit(n*2) | |
* Taking compliment | |
* ~64 = 0111111 | |
* | |
* ~Integer.highestOneBit(n*2) & ((n<<1) | 1) | |
* Bitwise And to copy bits exists in both operands. | |
* 1010011 | |
* & 0111111 | |
* ----------- | |
* 0010011 | |
* ----------- | |
*/ | |
public static int getSafePosition(int n) { | |
// int n1 = ~Integer.highestOneBit(n*2); | |
// int n2 = ((n<<1) | 1); | |
// return n1 & n2; | |
return ~Integer.highestOneBit(n*2) & ((n<<1) | 1); | |
} |
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