
Problem Statement
Given three positive integers a
, b
, and c
, the task is to determine the minimum number of flips required in some bits of a
and b
so that the result of the bitwise OR operation between a
and b
equals c
. Here, flipping a bit refers to changing it from 1
to 0
or from 0
to 1
. The goal is to align the binary representation of a OR b
to exactly match that of c
through the minimum possible number of bit flips.
Examples
Example 1
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
Approach and Intuition
To solve this problem, our primary goal is to align the result of a | b
(bitwise OR of a
and b
) to match c
. This can be visualized and tackled bit by bit. Here's a high-level approach:
- Convert numbers
a
,b
, andc
to their binary form. - Compare each bit position of
a
,b
, and the corresponding bit inc
. - Apply the following rules based on the bit values:
- If
c
has a bit set to1
, at least one among the corresponding bits ofa
orb
must be set to1
.- If both are
0
, one flip is needed.
- If both are
- If
c
has a bit set to0
, both corresponding bits ina
andb
should also be0
.- For any bit that is
1
in eithera
orb
, change it to0
, which might require one or two flips.
- For any bit that is
- If
- Count the total flips required and that would be the answer.
Examples Breakdown
Let's elaborate the approach with the provided examples:
- Example 1:
a = 2 (binary 010), b = 6 (binary 110), c = 5 (binary 101)
- For bit-0:
a
has0
,b
has0
, butc
needs1
. Flip eithera
orb
's bit. - For bit-1:
a
has1
,b
has1
, andc
has0
. Both bits ina
andb
need to be flipped to0
. - For bit-2:
a
has0
,b
has1
, andc
has1
. No flip needed as at least one bit matchesc
. - Total flips = 3.
- For bit-0:
- Example 2:
a = 4 (binary 100), b = 2 (binary 010), c = 7 (binary 111)
- For bit-0:
a
andb
are both0
whilec
is1
. Flip eithera
orb
's bit. - For bit-1 and bit-2: Already meeting the requirement as at least one bit is
1
. - Total flips = 1.
- For bit-0:
- Example 3:
a = 1 (binary 001), b = 2 (binary 010), c = 3 (binary 011)
- All bits already align with
c
's requirements. - Total flips = 0.
- All bits already align with
By following the above approach, we can determine the minimum flips necessary for any given values of a
, b
, and c
under the constraints.
Solutions
- C++
- Java
- Python
class Solution {
public:
int calculateMinFlips(int x, int y, int z) {
return __builtin_popcount((x | y) ^ z) + __builtin_popcount(x & y & ((x | y) ^ z));
}
};
In the given C++ solution to the problem "Minimum Flips to Make a OR b Equal to c", the function calculateMinFlips
computes the number of bit flips required to modify the bitwise OR of two integers, x and y, so that it equals a third integer, z.
Understand the approach by looking at the components of the function:
x | y
computes the bitwise OR of x and y.(x | y) ^ z
performs a bitwise XOR between the result ofx | y
and z. This XOR operation pinpoints the bits that are different betweenx | y
and z.__builtin_popcount((x | y) ^ z)
counts the number of bits set to 1 in the result of the XOR operation, representing the number of differing bits betweenx | y
and z which must be flipped.x & y & ((x | y) ^ z)
identifies the positions where both x and y have bits set that need to be flipped to make(x | y)
equal to z.__builtin_popcount(x & y & ((x | y) ^ z))
counts these specific positions which need correction.
By combining these two results, the function accurately calculates the number of flips required. This method is efficient and leverages bit manipulation functions available in C++ for quick computation.
class Solution {
public int calculateMinFlips(int x, int y, int z) {
return Integer.bitCount((x | y) ^ z) + Integer.bitCount(x & y & ((x | y) ^ z));
}
}
In this Java solution, you are given a method that calculates the minimum number of flips required to make the bitwise OR of two integers ( x ) and ( y ) equal to another integer ( z ). The implementation makes efficient use of bitwise operations.
- The
calculateMinFlips
method takes three integersx
,y
, andz
as inputs. - Utilize
Integer.bitCount()
method, which counts the number of one-bits in the binary representation of the passed integer. - Compute
(x | y) ^ z
to determine the bits that differ when comparing the bitwise OR ofx
andy
toz
. - Compute
x & y & ((x | y) ^ z)
to determine bits where bothx
andy
are set and yet differ fromz
. - The sum of the bit counts of these two operations gives the total number of flips needed.
This approach ensures optimal computation by directly using bitwise operators to identify the specific bits that require flipping. Adjust x
and y
such that their OR operation's outcome matches z
through minimal changes.
class Solution:
def calculateMinFlips(self, num1: int, num2: int, num3: int) -> int:
return bin((num1 | num2) ^ num3).count("1") + bin(num1 & num2 & ((num1 | num2) ^ num3)).count("1")
def calculateMinFlipsNew(self, num1: int, num2: int, num3: int) -> int:
return ((num1 | num2) ^ num3).bit_count() + (num1 & num2 & ((num1 | num2) ^ num3)).bit_count()
This solution involves implementing two methods in Python to determine the minimum number of bit flips required to make the bitwise OR of two numbers (num1
or num2
) equal to a third number (num3
). Here is an overview of how each method works:
calculateMinFlips
:- Calculate the XOR of the OR of
num1
andnum2
withnum3
. This XOR operation identifies positions where bits differ betweennum1 | num2
andnum3
. - Count the number of '1's in this XOR result using the
bin()
function andcount("1")
. This count gives the number of bits that need to be flipped. - Additionally, calculate the bitwise AND of
num1
andnum2
with the XOR result to find bits that are set in bothnum1
andnum2
and need to be flipped. Count the '1's in this result similarly. - Sum the counts from the XOR and AND operations to get the total number of flips.
- Calculate the XOR of the OR of
calculateMinFlipsNew
:- Similar to the first method but uses the
bit_count()
function introduced in Python 3.10 for determining the number of bits that are '1'. Thebit_count()
function is used directly on the XOR and AND results, making the code cleaner and possibly more efficient.
- Similar to the first method but uses the
Both methods provide an efficient way to compute the minimal flips needed by utilizing bitwise operations and ensure performance with in-built Python methods. The choice between the two might depend on the Python version in use due to the bit_count()
function's availability.
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