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Problem Statement
Given two binary strings a
and b
, return a sum of a
and b
represented as binary strings.
Examples
-
Example 1:
- Input: a = "10110", b = "10101"
- Expected Output: "101011"
- Justification: The binary sum of "10110" (22 in decimal) and "10101" (21 in decimal) is "101011" (43 in decimal).
-
Example 2:
- Input: a = "111", b = "1"
- Expected Output: "1000"
- Justification: The binary sum of "111" (7 in decimal) and "1" (1 in decimal) is "1000" (8 in decimal).
-
Example 3:
- Input: a = "0", b = "0"
- Expected Output: "0"
- Justification: The binary sum of "0" and "0" remains "0".
Solution
To solve this problem, we'll iterate over the input strings from right to left, simulating the binary addition process. We choose this approach because it directly simulates how binary addition works, carrying over bits when necessary. Our solution needs to handle different lengths of binary strings and the carry that can occur after adding the last bits.
By processing the strings from the end towards the beginning, we can easily manage carries and ensure that we consider all bits in both strings. If one string is shorter, we'll treat missing bits as '0'. This method is effective because it mirrors manual binary addition, ensuring no bit is missed and carries are correctly applied, resulting in an accurate binary sum.
Step-by-Step Algorithm
-
Initialize Variables:
- Create a variable
result
to store the final binary sum as a string. - Initialize two index variables,
i
andj
, to point to the last characters of stringsa
andb
respectively. - Initialize a
carry
variable to 0, which will hold the carry-over value when adding two binary digits.
- Create a variable
-
Iterate Over Strings:
- Use a loop to iterate while either
i
orj
is non-negative, orcarry
is 1. This ensures that you continue addition until both strings are completely traversed and any remaining carry is processed.
- Use a loop to iterate while either
-
Calculate Sum:
- Within the loop, for each position starting from the end, add the corresponding binary digits of
a
andb
, along with thecarry
. - If
i
orj
has gone negative (indicating one string is shorter), treat the missing digits as 0.
- Within the loop, for each position starting from the end, add the corresponding binary digits of
-
Update Result and Carry:
- Calculate the binary digit to be appended to the result (
sum % 2
) and update thecarry
(sum / 2
). - Prepend the calculated digit to the
result
string.
- Calculate the binary digit to be appended to the result (
-
Handle Remaining Carry:
- After the loop, if there's a remaining
carry
, prepend it to theresult
.
- After the loop, if there's a remaining
-
Return the Result:
- Return the
result
string, which now contains the binary sum ofa
andb
.
- Return the
Algorithm Walkthrough
Let's consider the input: a = "10110", b = "10101"
-
Initialization:
result
= ""i
= 4 (pointing to the last character of "10110")j
= 4 (pointing to the last character of "10101")carry
= 0
-
First Iteration (rightmost digits):
a[i] = 0
,b[j] = 1
sum = 0 + 1 + 0 (carry) = 1
result
= sum%2 + result = 1%2 + "" = "1",carry
= sum/2 = 1/2 = 0i
= 3,j
= 3
-
Second Iteration:
a[i] = 1
,b[j] = 0
sum = 1 + 0 + 0 = 1
result
= sum%2 + result = 1%2 + "1" ="11",carry
= sum/2 = 1/2 = 0i
= 2,j
= 2
-
Third Iteration:
a[i] = 1
,b[j] = 1
sum = 1 + 1 + 0 = 2
result
= sum%2 + result = 2%2 + "11" = "011",carry
= sum/2 = 2/2 = 1i
= 1,j
= 1
-
Fourth Iteration:
a[i] = 0
,b[j] = 0
sum = 0 + 0 + 1 (carry) = 1
result
= sum%2 + result = 1%2 + "011" = "1011",carry
= sum/2 = 1/2 = 0i
= 0,j
= 0
-
Fifth Iteration:
a[i] = 1
,b[j] = 1
sum = 1 + 1 + 0 = 2
result
= sum%2 + result = 2%2 + "1011" = "01011",carry
= sum/2 = 2/2 = 1i
= -1,j
= -1
-
Handle Remaining Carry:
- Since
carry
= 1,result
= "101011"
- Since
-
Final Result:
- The final binary sum is "101011", which is the correct sum of "10110" and "10101".
Code
Complexity Analysis
Time Complexity
O(max(N, M)), where N and M are the lengths of the two input binary strings, respectively. This is because we iterate through each string once, from end to start, in the worst case. The time complexity is determined by the longer string since we perform the addition digit by digit, including handling of the carry.
Space Complexity
O(max(N, M)), which is required for storing the result. In the worst case, the length of the result string will be max(N, M) + 1
(due to a possible carry that adds an extra digit).
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