Problem Description

You are given two strings s and t.

Your task is to find the length of the longest common prefix between s and t after removing at most one character from string s.

A prefix is a substring that starts from the beginning of a string. A common prefix between two strings is a prefix that both strings share.

The key constraint is that you can remove at most one character from string s (meaning you can remove 0 or 1 character). The removal, if done, can be from any position in s. After the optional removal, you need to find how long of a prefix from the modified s matches with the prefix of t.

For example:

• If s = "abc" and t = "ab", without removing any character, the common prefix is "ab" with length 2.
• If s = "xabc" and t = "abc", by removing the first character 'x' from s, we get "abc" which has a common prefix "abc" with t, giving us length 3.

The goal is to maximize the length of this common prefix by strategically choosing whether to remove a character from s and which character to remove if you do.

Intuition

When we think about finding the longest common prefix with at most one removal from s, we need to match characters from the beginning of both strings sequentially.

The key insight is that we're trying to match prefixes, which means we must start from index 0 of both strings and move forward. As we traverse:

1. When characters match (s[i] == t[j]), we can advance both pointers since this character contributes to our common prefix.

2. When characters don't match (s[i] != t[j]), we have a choice to make:

   • If we haven't removed any character yet, we can remove s[i] and continue. This is our one allowed removal.
   • If we've already used our removal, we must stop since we can't match any further.

The crucial observation is that when we encounter a mismatch, the only meaningful action is to skip the current character in s (effectively removing it). We don't skip characters in t because we're looking for a prefix of t, and prefixes must start from the beginning without gaps.

This naturally leads to a two-pointer approach where:

• Pointer i traverses string s
• Pointer j traverses string t
• A boolean flag rem tracks whether we've used our one removal

The length of the common prefix will be equal to how far we've advanced in t (the value of j), since that represents how many characters from t's prefix we've successfully matched.

Learn more about Two Pointers patterns.

Solution Approach

We implement the solution using a two-pointer technique with the following steps:

1. Initialize variables:

   • Record the lengths of strings s and t as n and m respectively
   • Set two pointers i = 0 and j = 0 to track positions in s and t
   • Initialize a boolean flag rem = False to track whether we've removed a character

2. Traverse both strings simultaneously:

   while i < n and j < m:

   We continue as long as we haven't reached the end of either string.

3. Handle character comparison:

   • Case 1: Characters don't match (s[i] != t[j]):
     • Check if we've already removed a character (if rem)
     • If yes, break the loop since we can't remove another character
     • If no, set rem = True to mark that we're using our removal
     • Only increment i (skip the character in s)
     • Keep j unchanged (we still need to match this character in t)
   • Case 2: Characters match (s[i] == t[j]):
     • Increment both pointers: i += 1 and j += 1
     • Move forward in both strings since we found a matching character

4. Return the result:

   • Return j, which represents the length of the longest common prefix
   • The value of j indicates how many characters from the beginning of t we successfully matched

The algorithm efficiently handles the removal decision in a single pass through the strings with O(n) time complexity where n is the length of string s, and O(1) space complexity since we only use a few variables.

Example Walkthrough

Let's walk through an example with s = "bxcd" and t = "bcd".

Initial State:

• s = "bxcd", n = 4
• t = "bcd", m = 3
• i = 0, j = 0, rem = False

Step 1: Compare s[0] with t[0]

• s[0] = 'b', t[0] = 'b'
• Characters match! Increment both pointers
• i = 1, j = 1, rem = False

Step 2: Compare s[1] with t[1]

• s[1] = 'x', t[1] = 'c'
• Characters don't match and rem = False
• We can use our removal: set rem = True
• Only increment i (skip 'x' in s)
• i = 2, j = 1, rem = True

Step 3: Compare s[2] with t[1]

• s[2] = 'c', t[1] = 'c'
• Characters match! Increment both pointers
• i = 3, j = 2, rem = True

Step 4: Compare s[3] with t[2]

• s[3] = 'd', t[2] = 'd'
• Characters match! Increment both pointers
• i = 4, j = 3, rem = True

Step 5: Loop terminates

• i = 4 equals n = 4, so we exit the loop
• Return j = 3

Result: The longest common prefix has length 3. By removing 'x' from position 1 in s, we transformed "bxcd" to "bcd", which shares the complete prefix "bcd" with t.

Solution Implementation

Time and Space Complexity

Time Complexity: O(min(n, m)) where n and m are the lengths of strings s and t respectively.

The while loop iterates through the strings with index i for string s and index j for string t. The loop continues while i < n and j < m. In each iteration, i is always incremented by 1, and j is incremented by 1 only when s[i] == t[j]. The loop terminates when either:

• i reaches n (processed all characters in s)
• j reaches m (processed all characters in t)
• A second mismatch is encountered (rem is already True and s[i] != t[j])

Since i is incremented in every iteration and the loop stops when i >= n or other conditions are met, the maximum number of iterations is min(n, m).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space for variables n, m, i, j, and rem, regardless of the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall: Misunderstanding When to Increment Pointers

The Problem: A common mistake is incorrectly handling pointer increments when characters don't match. Developers often mistakenly increment both pointers or only increment pointer_t when removing a character from s.

Incorrect Implementation:

if s[pointer_s] != t[pointer_t]:
    if removal_used:
        break
    removal_used = True
    pointer_s += 1
    pointer_t += 1  # WRONG! This skips a character in t

Why This Fails: When we remove a character from s, we're essentially skipping it to see if the next character in s matches the current character in t. We should NOT advance pointer_t because we still need to try matching the same character in t with the next character in s.

Example Where This Fails:

• s = "xabc", t = "abc"
• At position 0: s[0]='x' doesn't match t[0]='a'
• If we increment both pointers after removal, we'd be comparing s[1]='a' with t[1]='b' instead of t[0]='a'
• This would incorrectly return 0 instead of 3

Correct Solution: The original code correctly handles this by only incrementing pointer_t when characters match:

if s[pointer_s] != t[pointer_t]:
    if removal_used:
        break
    removal_used = True
    # Don't increment pointer_t here
else:
    pointer_t += 1  # Only increment when characters match

pointer_s += 1  # Always increment pointer_s

Alternative Pitfall: Not Considering the "No Removal" Case

The Problem: Some developers might assume they must always remove a character, leading to suboptimal solutions when no removal gives the best result.

Why This Matters: The problem states "at most one character", meaning 0 or 1 removal is allowed. Sometimes the best solution is to not remove any character at all.

Example:

• s = "abc", t = "abcd"
• Without removal: common prefix length = 3
• With any removal: common prefix length would be less than 3

The provided solution correctly handles this by using the removal_used flag as optional - it's only set to True when needed, not preemptively.