Find the missing no

#1

what is the thought process behind calculating setbit and doing bitwise & with each and every no…

#2

@ubaidshaikh9999,
It is basic XOR and AND operations.

See, when we XOR 2 numbers and if the numbers are same it will give 0 as output. Now in the question we are XORing 1,3,4,6 as giving in the array. so we will get Z = 1^3^4^6. Let this be z1

Now after that we are XORing Z with i where i =1 and i <=6. Z = 1^2^3^4^5^6 now let this be z2.

now Z = z1^z2 = (1^3^4^6) ^ (1^2^3^4^5^6). Now since XOR of same elements gives 0. 1,3,4,6 will cancel out since they occur twice and we are left with Z = 2^5 ( these are the missing numbers)

Similarly do a dry run of the setbit part yourself. Let me know if you are not able to understand anything.

#3

No No Bhaiya I already knew this I was asking about setbit,like what’s special in z&~(z-1) and how we came to know that setbit&arr[i] can give us setbit again

How it’s going to come in my mind suppose I encounter this question in a contest.

#4

@ubaidshaikh9999,
Set is basically the rightmost digit in the binary. z&~(z-1) in this say z is 7, then z-1 will be 6.
Binary of 7 is 111 and binary of 6 is 110. If I do ~6 then binary will be 001. Now and(&) of 111 and 001 will give us 1.

Another bitwise trick is:
x | (1 << n) : Returns the number x with the nth bit set. You can do a dry run of this on your own.

For bitwise you need to practice questions. You can participate in contests on codechef, codeforces.

#5

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