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The key to this approach is analyzing each bit position of the numbers. We need to ensure that at each bit position i, (a[i] OR b[i]) matches c[i].
If c[i] is 0, both a[i] and b[i] must be 0. This may require flipping if either a[i] or b[i] is 1.
If c[i] is 1, we need at least one of a[i] or b[i] to be 1. If both are 0, we need one flip.
By iterating through each binary bit, we can count the necessary flips.
Time Complexity: O(1), since the number of bits is constant and we iterate a fixed number of times.
Space Complexity: O(1), as we use a constant amount of space.
1#include <iostream>
2using namespace std;
3
4int minFlips(int a, int b, int c) {
5 int flips = 0;
6 for(int i = 0; i < 32; ++i) {
7 int bitA = (a >> i) & 1;
8 int bitB = (b >> i) & 1;
9 int bitC = (c >> i) & 1;
10 if (bitC == 0) {
11 flips += bitA + bitB;
12 } else {
13 flips += !(bitA | bitB);
14 }
15 }
16 return flips;
17}
18
19int main() {
20 int a = 2, b = 6, c = 5;
21 cout << minFlips(a, b, c) << endl;
22 return 0;
23}This C++ solution follows the same logic as the C version. It utilizes bitwise operations to determine the state of each bit and count the necessary flips to make (a OR b) match c.
This alternative approach uses masks and logical reduction to simplify determining the minimum flips. Instead of checking each bit individually, create a mask that highlights the bits where a OR b does not match the bits of c.
For c's zeroed bits, count additions for every 1 found in corresponding bits in a or b.
For c's one bits, ensure at least one 1 is present in the corresponding a or b position. Adjust flips accordingly.
Time Complexity: O(1), consistently iterating fixed bits.
Space Complexity: O(1), as it uses primitive operations.
The C function uses bitwise operations to derive a differential mask of bits that need checking. It iteratively reduces a, b, and c by shifting and adjusts flips by analyzing each bit state.