Longest Common Prefix After at Most One Removal - Problem

You are given two strings s and t. Return the length of the longest common prefix between s and t after removing at most one character from s.

Note: s can be left without any removal.

Input & Output

Example 1 — Basic Mismatch

$ Input: s = "abcde", t = "abfgh"

› Output: 2

💡 Note: The common prefix without removal is "ab" (length 2). Removing any character from s doesn't improve this, so the answer is 2.

Example 2 — Remove to Extend

$ Input: s = "axbc", t = "abc"

› Output: 3

💡 Note: Without removal: common prefix is "a" (length 1). Removing s[1]='x' gives "abc" vs "abc", common prefix "abc" (length 3).

Example 3 — No Improvement

$ Input: s = "abc", t = "def"

› Output: 0

💡 Note: No common prefix exists regardless of removal. The answer is 0.

Constraints

1 ≤ s.length, t.length ≤ 1000
s and t consist of only lowercase English letters

Visualization

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Asked in

M Microsoft 25 G Google 18

The key insight is to find where strings first differ, then try skipping that character in s. The two pointers approach efficiently handles this by avoiding string reconstruction. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(n)

For each possible removal position in string s (including no removal), calculate the common prefix length with string t. Return the maximum length found among all attempts.

Two Pointers Optimization

⏱️ Time: O(n) Space: O(1)

Start comparing characters from the beginning. When we find a mismatch, try skipping the current character in s and continue matching. This avoids generating new strings and reduces redundant work.

Brute Force — Algorithm Steps

Try with no character removed from s
Try removing character at each position i
For each attempt, calculate common prefix length with t
Return the maximum length found

Visualization

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Step-by-Step Walkthrough

No Removal

Compare original strings directly

Try Each Removal

Remove one character at a time from s

Find Maximum

Return the longest prefix found

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int commonPrefixLength(const char* str1, const char* str2) {
    int i = 0;
    while (str1[i] && str2[i] && str1[i] == str2[i]) {
        i++;
    }
    return i;
}

int solution(const char* s, const char* t) {
    int maxLength = 0;
    int sLen = strlen(s);
    
    // Try without removing any character
    int length = commonPrefixLength(s, t);
    if (length > maxLength) maxLength = length;
    
    // Try removing each character from s
    char* modifiedS = malloc(sLen);
    for (int i = 0; i < sLen; i++) {
        // Create modified string without character at position i
        strncpy(modifiedS, s, i);
        strcpy(modifiedS + i, s + i + 1);
        
        length = commonPrefixLength(modifiedS, t);
        if (length > maxLength) maxLength = length;
    }
    
    free(modifiedS);
    return maxLength;
}

int main() {
    char s[1001], t[1001];
    fgets(s, sizeof(s), stdin);
    fgets(t, sizeof(t), stdin);
    
    // Remove newline characters
    s[strcspn(s, "\n")] = 0;
    t[strcspn(t, "\n")] = 0;
    
    int result = solution(s, t);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n+1 attempts (no removal + n removals), we compare up to n characters

⚠ Quadratic Growth

Space Complexity

O(n)

Creating modified strings for each removal attempt

⚡ Linearithmic Space

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Longest Common Prefix After at Most One Removal - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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