Given a number maximize it by swapping bits at it’s extreme positions i.e. at first and last position, second and second last position and so on.

Examples:

Input : 4 (0000...0100) Output : 536870912 (0010...0000) In the above example swapped 3rd and 3rd last bit to maximize the given unsigned number. Input : 12 (00000...1100) Output : 805306368 (0011000...0000) In the above example 3rd and 3rd last bit and 4th and 4th last bit are swapped to maximize the given unsigned number.

**Naive Approach: **

**1.** Convert the number into it’s bit representation and store it’s bit representation in an array.

**2.** Traverse the array from both ends, if the less significant bit of the bit representation is greater than the more significant bit i.e. if less significant bit is 1 and more significant bit is 0 then swap them else take no action.

**3.** Convert the obtained binary representation back to the number.

**Efficient Approach: **

**1.** Create a copy of the original number because the original number would be modified, iteratively obtain the bits at the extreme positions.

**2.** If less significant bit is 1 and more significant bit is 0 then swap the bits in the bit from only, continue the process until less significant bit’s position is less than more significant bit’s position.

**3.** Display the maximized number.

// C++ program to find maximum number by // swapping extreme bits. #include <bits/stdc++.h> using namespace std; #define ull unsigned long long int ull findMax(ull num) { ull num_copy = num; /* Traverse bits from both extremes */ int j = sizeof(unsigned long long int) * 8 - 1; int i = 0; while (i < j) { // Obtaining i-th and j-th bits int m = (num_copy >> i) & 1; int n = (num_copy >> j) & 1; /* Swapping the bits if lesser significant is greater than higher significant bit and accordingly modifying the number */ if (m > n) { x = (1 << i | 1 << j); num = num ^ x; } i++; j--; } return num; } // Driver code int main() { ull num = 4; cout << findMax(num); return 0; }

Output:

536870912

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