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/10/18 · Otherwise the bit is set in n ⊕ x as 0 ⊕ 1 = 1 Therefore for every set bit in n, we can have either a set bit or an unset bit in x However, we cannot have a set bit in x corresponding to an unset bit in n By this logic, the number of solutions comes out to be 2 raised to the power of the number of set bits in n The time complexity of this approach is O(log n) C // C. B O ¦ b \ q ¢151 ³ Ä w ¡ ¿ t z × M ö ¯ æ U ï ê Ö ` z Ë t x ¸ I \ ` o N ´ s w O s ù ± C b \ q U C ^ o M £ { y ï Ä w Á t T T c d ) p V { } f w w « y Ó å · Ø 0 ° ¼ g ¼ g 0 Å t w ù r s æ l h A L z N. â Ç Ê V þ { x G { ´ G Ê x Ç x { Ç x ?.
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