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

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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

✓ Brute Force - Check All Subsequences

⏱️ Time: O(2^n) Space: O(2^n)

This approach generates all possible subsequences of string s and checks if any of them equals string t. If found, we calculate how many characters need to be appended.

Greedy - Match First Available Character

⏱️ Time: O(n + m) Space: O(1)

This greedy approach works because we want to preserve as much of the original order as possible. We match each character in t with the first available matching character in s, ensuring we don't need to append more characters than necessary.

Two Pointers - Sequential Matching

⏱️ Time: O(n + m) Space: O(1)

This approach uses two pointers to traverse both strings simultaneously. We try to match as many characters of t as possible using characters from s in order, then calculate how many characters are left unmatched.

Brute Force - Check All Subsequences — Algorithm Steps

Generate all possible subsequences of s
Check if any subsequence equals t
Calculate remaining characters needed

Visualization

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Step-by-Step Walkthrough

Generate Subsequences

Create all 2^n possible subsequences of s

Check Matches

Find which subsequences are prefixes of t

Calculate Result

Return length of t minus longest match

Code -

solution.c — C

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

int appendCharacters(char* s, char* t) {
    int i = 0; // pointer for s
    int j = 0; // pointer for t
    int sLen = strlen(s);
    int tLen = strlen(t);
    
    // Try to match as many characters of t as possible using s
    while (i < sLen && j < tLen) {
        if (s[i] == t[j]) {
            j++; // found a match, move pointer in t
        }
        i++; // always move pointer in s
    }
    
    // Return the number of unmatched characters in t
    return tLen - j;
}

int main() {
    char s[101], t[101];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    fgets(t, sizeof(t), stdin);
    t[strcspn(t, "\n")] = 0;
    
    int result = appendCharacters(s, t);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

Generating all subsequences requires 2^n combinations where n is length of s

✓ Linear Growth

Space Complexity

O(2^n)

Storing all subsequences requires exponential space

✓ Linear Space

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Input & Output

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Related Problems

Common Approaches

Brute Force - Check All Subsequences — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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