Given an integer N, the task is to find the largest number M, where (M < N), such that N(XOR)M is equal to N(OR)M i.e. (N ^ M) = (N | M).
Input: N = 5
5 ^ 4 = 1 and 5 | 4 = 5. Therefore, XOR and OR between them are not equal.
5 ^ 3 = 6 and 5 | 3 = 7. Therefore, XOR and OR between them are not equal.
5 ^ 2 = 7 and 5 | 2 = 7. Therefore, XOR and OR between them are equal.
Input: N = 14
To get the required number M, traverse all the bits of N from its Least Significant Bit (LSB) to Most Significant Bit (MSB). Two cases arise here:
- If the ith bit of N is 1 then:
- If the ith bit of M is set to 1, then N^M will not be equal to N|M as (1^1 = 0) and (1|1 = 1).
- If the ith bit is set of M to 0, then N^M will be equal to N|M as (1^0 = 1) and (1|0 = 1).
- So if the ith bit of N is 1, set the ith bit of M to 0.
- If the ith bit of N is 0 then:
- If the ith bit of M is set to 1, then N^M will be equal to N|M as (0^1 = 1) and (0|1 = 1).
- If we set the ith bit of M to 0, then N^M will be equal to N|M as (0^0 = 0) and (0|0 = 0).
- So, if the ith bit of M is set to either 0 or 1, N^M will always be equal to N|M.
- As the largest value of M which is less than N has to be found out, always set the ith bit of M to 1.
- N = 5
- 32-bit representation of 5 = 00000000000000000000000000000101
- LSB index of 5 = 31
- MSB index of 5 = 29
- Traversing from LSB to MSB i.e. from 31 to 29:
- For index 31, N = 1. So M should be set to 0.
- For index 30, N = 0. So M should be set to 1.
- For index 29, N = 1. So M should be set to 0.
- Thus the 32-bit representation of M is 00000000000000000000000000000010, which is equal to 2 in decimal representation.
Below is the implementation of the above approach:
Time Complexity: O(log2 N)
Auxiliary Space: O(1)
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