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Find Length of Longest Substring with 1s After One 0 Flip in Python
Suppose we have a binary string s. We are allowed to flip at most one "0" to "1", we have to find the length of the longest contiguous substring of 1s.
So, if the input is like s = "1010110001", then the output will be 4, as if we flip the zero present at index 3, then we get the string "1011110001", here length of the longest substring of 1s is 4.
To solve this, we will follow these steps −
- n := size of s
- ans := 0, ones := 0, left := 0, right := 0
- while right < n, do
- if s[right] is same as "1", then
- ones := ones + 1
- while right - left + 1 - ones > 1, do
- remove := s[left]
- if remove is same as "1", then
- ones := ones - 1
- left := left + 1
- ans := maximum of ans and (right - left + 1)
- right := right + 1
- if s[right] is same as "1", then
- return ans
Example
Let us see the following implementation to get better understanding −
def solve(s): n = len(s) ans = ones = left = right = 0 while right < n: if s[right] == "1": ones += 1 while right - left + 1 - ones > 1: remove = s[left] if remove == "1": ones -= 1 left += 1 ans = max(ans, right - left + 1) right += 1 return ans s = "1010110001" print(solve(s))
Input
"1010110001"
Output
4
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