Given a number N. The task is to count the all possible values of x such that nx is equal to (N-x), where denotes bitwise XOR operation.
Input: N = 3 Output: 4 The all possible values of x are respectively 0, 1, 2, 3. Input: N = 6 Output: 4 The all possible values of x are respectively 0, 2, 4, 6.
Approach: The XOR value of two bits will be 1 if both bits have opposite sign, and 0 when both bits are same. So on the basis of the property of XOR, we can say that n x is always greater than or equal to n-x. The only condition when its value is equal with n-x is bits of x form a subset of bits of n. Because if in the i’th position both x and n has set bits then after xor the value will decrease, and the decreased value will be , where i is 0-based position.
So the answer is the total count of subsets of bits of number n is , where k is the count of set bits in n.
Below is the implementation of above approach:
Time Complexity: O(k), where k is number of set bits in N.
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