DEV Community: JOY SINHA

Regular Expression Matching

JOY SINHA — Thu, 03 Feb 2022 05:37:19 +0000

Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:

'.' Matches any single character.
'*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).

Example 1:

Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
Example 2:

Input: s = "aa", p = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
Example 3:

Input: s = "ab", p = "."
Output: true
Explanation: "." means "zero or more (*) of any character (.)".

Constraints:

1 <= s.length <= 20
1 <= p.length <= 30
s contains only lowercase English letters.
p contains only lowercase English letters, '.', and ''.
It is guaranteed for each appearance of the character '', there will be a previous valid character to match.

*Code : - *

def isMatch(self, s: str, p: str) -> bool:
return self.isMat(s, p, len(s)-1, len(p)-1, False)

def isMat(self, s: str, p: str, si: int, pi: int, isAst):
    if pi < 0:
        return si < 0
    if p[pi] == '*':
        return self.isMat(s, p, si, pi-1, True)

    result = False
    if isAst:
        if si >= 0 and self.compare(s[si], p[pi]):
            result = result or self.isMat(s, p, si-1, pi, True) 
        result = result or self.isMat(s, p, si, pi-1, False)
    elif si >= 0 and self.compare(s[si], p[pi]):
        result = result or self.isMat(s, p, si-1, pi-1, False)

    return result

def compare(self, a, b):
    return a == b or a == '.' or b == '.'

11. Container With Most Water [Leetcode ]

JOY SINHA — Wed, 23 Jun 2021 16:39:47 +0000

Optimized Python 3 Solution for Container with most water problem.
I am using two pointer shifting technique to solve it so that we can reduce time complexity of the solution.

CODE:-

class Solution:
def maxArea(self, height: List[int]) -> int:
maxArea = 0
p1 = 0
p2 = len(height)-1
while(p1<=p2):
maxArea = max(maxArea,min(height[p1],height[p2])*(p2-p1))
if(height[p1]<height[p2]):
p1=p1+1
else:
p2=p2-1
return maxArea

1. Two Sum [From Leet code]

JOY SINHA — Wed, 23 Jun 2021 16:37:11 +0000

Brute force:

for i in range(len(nums)):
for j in range(i + 1, len(nums)):
if nums[i] + nums[j] == target:
return [i, j]
Runtime: 3584 ms, faster than 14.38% of Python online submissions for Two Sum.
Memory Usage: 14.5 MB, less than 14.64% of Python online submissions for Two Sum.

Two-pass Hash Table:

nums_hash = {}
for index, num in enumerate(nums):
nums_hash[num] = index

for index, num in enumerate(nums):
comp = target - num
if comp in nums_hash.keys() and index != nums_hash[comp]:
return [index, nums_hash[comp]]
Runtime: 1348 ms, faster than 24.52% of Python online submissions for Two Sum.
Memory Usage: 14.2 MB, less than 46.97% of Python online submissions for Two Sum.

One-pass Hash Table:

nums_hash = {}
for index, num in enumerate(nums):
comp = target - num
if comp in nums_hash.keys() and index != nums_hash[comp]:
return [index, nums_hash[comp]]
else:
nums_hash[num] = index
Runtime: 680 ms, faster than 27.87% of Python online submissions for Two Sum.
Memory Usage: 14.3 MB, less than 46.97% of Python online submissions for Two Sum.