Solving Leetcode Interviews in Seconds with AI: Subsequence With the Minimum Score
Introduction
In this blog post, we will explore how to solve the LeetCode problem "2565" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
Problem Statement
You are given two strings s and t. You are allowed to remove any number of characters from the string t. The score of the string is 0 if no characters are removed from the string t, otherwise: Let left be the minimum index among all removed characters. Let right be the maximum index among all removed characters. Then the score of the string is right - left + 1. Return the minimum possible score to make t a subsequence of s. A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of "abcde" while "aec" is not). Example 1: Input: s = "abacaba", t = "bzaa" Output: 1 Explanation: In this example, we remove the character "z" at index 1 (0-indexed). The string t becomes "baa" which is a subsequence of the string "abacaba" and the score is 1 - 1 + 1 = 1. It can be proven that 1 is the minimum score that we can achieve. Example 2: Input: s = "cde", t = "xyz" Output: 3 Explanation: In this example, we remove characters "x", "y" and "z" at indices 0, 1, and 2 (0-indexed). The string t becomes "" which is a subsequence of the string "cde" and the score is 2 - 0 + 1 = 3. It can be proven that 3 is the minimum score that we can achieve. Constraints: 1 <= s.length, t.length <= 105 s and t consist of only lowercase English letters.
Explanation
Here's the breakdown of the solution:
- Find Leftmost Matches: Iterate through
t. For each character int, find the earliest index inswhere it appears after matching the previous characters oft. Store these indices. - Find Rightmost Matches: Iterate through
tin reverse. For each character int, find the latest index inswhere it appears before matching the succeeding characters oft. Store these indices. Minimize the Score: Iterate through all possible removal ranges in
t. For each range, check if the remaining characters intform a subsequence ofsusing the precomputed leftmost and rightmost match indices. Calculate and minimize the score for each valid range.Runtime Complexity: O(m*n), where n is the length of s and m is the length of t.
- Storage Complexity: O(m), where m is the length of t.
def solve():
s = input()
t = input()
n = len(s)
m = len(t)
leftmost = [-1] * m
rightmost = [-1] * m
# Calculate leftmost matches
s_ptr = 0
for t_ptr in range(m):
while s_ptr < n:
if s[s_ptr] == t[t_ptr]:
leftmost[t_ptr] = s_ptr
s_ptr += 1
break
s_ptr += 1
else:
break # Could not find match
# Calculate rightmost matches
s_ptr = n - 1
for t_ptr in range(m - 1, -1, -1):
while s_ptr >= 0:
if s[s_ptr] == t[t_ptr]:
rightmost[t_ptr] = s_ptr
s_ptr -= 1
break
s_ptr -= 1
else:
break
ans = m
for left in range(m + 1):
for right in range(left, m + 1):
# Check if t[:left] + t[right:] is a subsequence of s
# Check left part
valid = True
if left > 0:
prev_s_index = -1
t_index = 0
while t_index < left:
found = False
for s_index in range(prev_s_index + 1, n):
if s[s_index] == t[t_index]:
prev_s_index = s_index
found = True
break
if not found:
valid = False
break
t_index += 1
if not valid:
continue
# Check right part
if right < m:
prev_s_index = -1
t_index = right
while t_index < m:
found = False
for s_index in range(prev_s_index + 1, n):
if s[s_index] == t[t_index]:
prev_s_index = s_index
found = True
break
if not found:
valid = False
break
t_index += 1
if not valid:
continue
if left == 0 and right == m:
if "" not in s:
continue
if left == right and left != m:
l = 0
r = 0
isValid = False
t_l = t[:left]
t_r = t[right:]
combined = t_l + t_r
if len(combined) == 0:
ans = min(ans, right - left + 1)
continue
s_ptr = 0
t_ptr = 0
while s_ptr < len(s) and t_ptr < len(combined):
if s[s_ptr] == combined[t_ptr]:
t_ptr += 1
s_ptr += 1
if t_ptr == len(combined):
ans = min(ans, right - left + 1)
elif left != right:
ans = min(ans, right - left)
print(ans)
# Code
```python
def solve(): s = input()
t = input()
n = len(s)
m = len(t)
leftmost = [-1] * m
rightmost = [-1] * m
# Calculate leftmost matches
s_ptr = 0
for t_ptr in range(m):
while s_ptr < n:
if s[s_ptr] == t[t_ptr]:
leftmost[t_ptr] = s_ptr
s_ptr += 1
break
s_ptr += 1
else:
break # Could not find match
# Calculate rightmost matches
s_ptr = n - 1
for t_ptr in range(m - 1, -1, -1):
while s_ptr >= 0:
if s[s_ptr] == t[t_ptr]:
rightmost[t_ptr] = s_ptr
s_ptr -= 1
break
s_ptr -= 1
else:
break
ans = m
for left in range(m + 1):
for right in range(left, m + 1):
# Check if t[:left] + t[right:] is a subsequence of s
# Check left part
valid = True
if left > 0:
prev_s_index = -1
t_index = 0
while t_index < left:
found = False
for s_index in range(prev_s_index + 1, n):
if s[s_index] == t[t_index]:
prev_s_index = s_index
found = True
break
if not found:
valid = False
break
t_index += 1
if not valid:
continue
# Check right part
if right < m:
prev_s_index = -1
t_index = right
while t_index < m:
found = False
for s_index in range(prev_s_index + 1, n):
if s[s_index] == t[t_index]:
prev_s_index = s_index
found = True
break
if not found:
valid = False
break
t_index += 1
if not valid:
continue
if left == 0 and right == m:
ans = min(ans, m)
continue
ans = min(ans, right - left)
print(ans)
solve()