Sponsored
Sponsored
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 <stdio.h>
2
3int minFlips(int a, int b, int c) {
4 int flips = 0;
5 for(int i = 0; i < 32; ++i) {
6 int bitA = (a >> i) & 1;
7 int bitB = (b >> i) & 1;
8 int bitC = (c >> i) & 1;
9 if (bitC == 0) {
10 flips += bitA + bitB;
11 } else {
12 flips += !(bitA | bitB);
13 }
14 }
15 return flips;
16}
17
18int main() {
19 int a = 2, b = 6, c = 5;
20 printf("%d\n", minFlips(a, b, c));
21 return 0;
22}The function minFlips calculates the number of flips required to make (a OR b) equal to c. The code iterates through 32 bits (considering an integer size of 32) and uses bitwise operations to check each individual bit position of a, b, and 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.
using namespace std;
int minFlips(int a, int b, int c) {
int flips = 0;
while (c > 0 || a > 0 || b > 0) {
if ((c & 1) == 0) {
flips += (a & 1) + (b & 1);
} else if ((a & 1) == 0 && (b & 1) == 0) {
flips++;
}
a >>= 1;
b >>= 1;
c >>= 1;
}
return flips;
}
int main() {
cout << minFlips(2, 6, 5) << endl;
return 0;
}In C++, the solution streamlines by reducing number sizes while maintaining the number of flips. Each subtraction (bit shift) curtails the evaluations for a, b, and c iteratively.