Given two binary numbers strings and of length . Find the number of ways of swapping two bits in s1(only s1 not s2) so that bit-wise OR of these two numbers s1 and s2 are changed.
Note: The length of both string must be equal, you can take leading zeros in case of different length.
Input: s1 = "01011", s2 = "11001" Output: 4 Explanation: You can swap the bit of s1 at indexed: (1, 4), (2, 3), (3, 4) and (3, 5) there are 4 ways possible. Input: s1 = "011000", s2 = "010011" Output: 6
Approach: Initialize a variable result as zero that stores number of ways to swap bits of s1 so that bitwise OR of s1 and s2 changes. Initialize four variables say , , , and as zero.
Traverse both strings from left side and check for the below conditions to increment values of the
variable declared above:
- If the current bit of both s1 and s2 are zero, increment c.
- If current bit of s2 is zero and s1 is one, increment d.
- If current bit of s2 is one and s1 is zero, increment a.
- If current bit of both s1 and s2 is one, increment b.
Update the result by (a*d) + (b*c) + (c*d) and return the result.
Below is the implementation of the above approach:
# Python 3 program to find no of ways
# to swap bits of s1 so that
# bitwise OR of s1 and s2 changes
# Function to find number of ways
def countWays(s1, s2, n):
a = b = c = d = 0
# intialise result that store
# No. of swaps required
result = 0;
# Traverse both strings and check
# the bits as explained
for i in range(0, n, 1):
if (s2[i] == ‘0’):
if (s1[i] == ‘0’):
c += 1;
d += 1
if (s1[i] == ‘0’):
a += 1
b += 1
# calculate result
result = a * d + b * c + c * d
# Driver code
if __name__ == ‘__main__’:
n = 5
s1 = “01011”
s2 = “11001”
print(countWays(s1, s2, n))
# This code is contributed by
- Leftover element after performing alternate Bitwise OR and Bitwise XOR operations on adjacent pairs
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- Count pairs with Bitwise OR as Even number
- Count pairs with Bitwise XOR as EVEN number
- Count pairs with Bitwise XOR as ODD number
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- Form the smallest number using at most one swap operation
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