Given a number N, the task is to find the count of X such that N XOR X == N OR X, where 0<=X<=N
Input: N = 5
For N = 5,
5 XOR 2 == 5 OR 2
5 XOR 0 == 5 OR 0
Thus, count is 2.
Input: N = 7
For N = 7,
7 XOR 0 == 7 OR 0
Thus, count is 1.
Approach: The idea is to convert given number to binary and then count the unset bits in it. 2^count gives us the number of X such that N XOR X == N OR X.
Below is the implementation of the above approach:
- Count numbers whose sum with x is equal to XOR with x
- Count numbers whose difference with N is equal to XOR with N
- Count different numbers that can be generated such that there digits sum is equal to 'n'
- Count the numbers < N which have equal number of divisors as K
- Count pairs of natural numbers with GCD equal to given number
- Count numbers in a range having GCD of powers of prime factors equal to 1
- Maximum count of equal numbers in an array after performing given operations
- Count of numbers whose sum of increasing powers of digits is equal to the number itself
- Count of Numbers in Range where first digit is equal to last digit of the number
- Fill the missing numbers in the array of N natural numbers such that arr[i] not equal to i
- Count numbers < = N whose difference with the count of primes upto them is > = K
- Equal Sum and XOR of three Numbers
- Count of Multiples of A ,B or C less than or equal to N
- Numbers whose bitwise OR and sum with N are equal
- Count of subsets with sum equal to X using Recursion
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