Append Characters to String to Make Subsequence

Append Characters to String to Make Subsequence - Problem

You are given two strings s and t consisting of only lowercase English letters.

Return the minimum number of characters that need to be appended to the end of s so that t becomes a subsequence of s.

A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

Input & Output

Example 1 — Partial Match

$ Input: s = "coaching", t = "coding"

› Output: 4

💡 Note: We can match 'c' and 'o' from s, but need to append "ding" to complete the subsequence "coding"

Example 2 — Complete Match

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

› Output: 0

💡 Note: All characters 'a', 'c', 'e' from t are already present in s in correct order, so no characters need to be appended

Example 3 — No Match

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

› Output: 3

💡 Note: None of the characters in t are found in s, so we need to append all 3 characters "abc"

Constraints

1 ≤ s.length, t.length ≤ 10⁵
s and t consist only of lowercase English letters.

Visualization

Tap to expand

Understanding the Visualization

Input Strings

Given strings s and target t

Match Process

Find longest subsequence of t that exists in s

Calculate Result

Count remaining unmatched characters

Key Takeaway

🎯 Key Insight: Use two pointers to find the longest prefix of t that forms a subsequence in s, then append the remaining characters.

Asked in

f Meta 12 M Microsoft 8 G Google 6

The key insight is to use two pointers to match as many characters of t as possible from s in sequential order. The optimal approach uses two pointers with Time: O(n + m), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Greedy - Match First Available Character	O(n + m)	O(1)	Greedily match each character of t with the first available character in s
Two Pointers - Sequential Matching	O(n + m)	O(1)	Use two pointers to find how many characters of t can be matched in s
Brute Force - Check All Subsequences	O(2^n)	O(2^n)	Generate all possible subsequences of s and check if any matches t

Greedy - Match First Available Character — Algorithm Steps

For each character in t, find first matching character in remaining s
Mark matched characters as used
Count unmatched characters

Visualization

Tap to expand

Step-by-Step Walkthrough

Start Matching

For each character in t, find first match in remaining s

Mark Used

Move past matched characters to avoid reusing

Count Missing

Calculate how many characters couldn't be matched

Code -

solution.c — C

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

int solution(char* s, char* t) {
    int sIndex = 0;
    int matchedCount = 0;
    int sLen = strlen(s);
    int tLen = strlen(t);
    
    // For each character in t, try to find it in remaining s
    for (int i = 0; i < tLen; i++) {
        char ch = t[i];
        int found = 0;
        
        while (sIndex < sLen) {
            if (s[sIndex] == ch) {
                matchedCount++;
                sIndex++;
                found = 1;
                break;
            }
            sIndex++;
        }
        
        if (!found) {
            break;
        }
    }
    
    return tLen - matchedCount;
}

int main() {
    char s[1000], t[1000];
    fgets(s, sizeof(s), stdin);
    fgets(t, sizeof(t), stdin);
    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(n + m)

Single traversal of both strings where n is length of s and m is length of t

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables for tracking positions

✓ Linear Space

25.0K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy - Match First Available Character — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler