Description#
Given 3 positives numbers a
, b
and c
. Return the minimum flips required in some bits of a
and b
to make ( a
OR b
== c
). (bitwise OR operation).
Flip operation consists of change any single bit 1 to 0 or change the bit 0 to 1 in their binary representation.
Example 1:
![](https://spcdn.pages.dev/leetcode/problems/1318.Minimum%20Flips%20to%20Make%20a%20OR%20b%20Equal%20to%20c/images/sample_3_1676.png)
Input: a = 2, b = 6, c = 5
Output: 3
Explanation: After flips a = 1 , b = 4 , c = 5 such that (a
OR b
== c
)
Example 2:
Input: a = 4, b = 2, c = 7
Output: 1
Example 3:
Input: a = 1, b = 2, c = 3
Output: 0
Constraints:
1 <= a <= 10^9
1 <= b <= 10^9
1 <= c <= 10^9
Solutions#
Solution 1#
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| class Solution:
def minFlips(self, a: int, b: int, c: int) -> int:
ans = 0
for i in range(30):
x, y, z = a >> i & 1, b >> i & 1, c >> i & 1
if x | y != z:
ans += 2 if x == 1 and y == 1 else 1
return ans
|
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| class Solution {
public int minFlips(int a, int b, int c) {
int ans = 0;
for (int i = 0; i < 30; ++i) {
int x = a >> i & 1, y = b >> i & 1, z = c >> i & 1;
if ((x | y) != z) {
ans += x == 1 && y == 1 ? 2 : 1;
}
}
return ans;
}
}
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| class Solution {
public:
int minFlips(int a, int b, int c) {
int ans = 0;
for (int i = 0; i < 30; ++i) {
int x = a >> i & 1, y = b >> i & 1, z = c >> i & 1;
if ((x | y) != z) {
ans += x == 1 && y == 1 ? 2 : 1;
}
}
return ans;
}
};
|
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| func minFlips(a int, b int, c int) (ans int) {
for i := 0; i < 30; i++ {
x, y, z := a>>i&1, b>>i&1, c>>i&1
if (x | y) != z {
if x == 1 && y == 1 {
ans += 2
} else {
ans++
}
}
}
return
}
|