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Formatted question description: https://leetcode.ca/all/2486.html

2486. Append Characters to String to Make Subsequence

  • Difficulty: Medium.
  • Related Topics: Two Pointers, String, Greedy.
  • Similar Questions: Is Subsequence, Minimum Operations to Make a 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.

  Example 1:

Input: s = "coaching", t = "coding"
Output: 4
Explanation: Append the characters "ding" to the end of s so that s = "coachingding".
Now, t is a subsequence of s ("coachingding").
It can be shown that appending any 3 characters to the end of s will never make t a subsequence.

Example 2:

Input: s = "abcde", t = "a"
Output: 0
Explanation: t is already a subsequence of s ("abcde").

Example 3:

Input: s = "z", t = "abcde"
Output: 5
Explanation: Append the characters "abcde" to the end of s so that s = "zabcde".
Now, t is a subsequence of s ("zabcde").
It can be shown that appending any 4 characters to the end of s will never make t a subsequence.

  Constraints:

  • 1 <= s.length, t.length <= 105

  • s and t consist only of lowercase English letters.

Solution (Java, C++, Python)

  • class Solution {
        public int appendCharacters(String s, String t) {
            int m = s.length(), n = t.length();
            for (int i = 0, j = 0; j < n; ++j) {
                while (i < m && s.charAt(i) != t.charAt(j)) {
                    ++i;
                }
                if (i++ == m) {
                    return n - j;
                }
            }
            return 0;
        }
    }
    
  • class Solution {
    public:
        int appendCharacters(string s, string t) {
            int m = s.size(), n = t.size();
            for (int i = 0, j = 0; j < n; ++j) {
                while (i < m && s[i] != t[j]) {
                    ++i;
                }
                if (i++ == m) {
                    return n - j;
                }
            }
            return 0;
        }
    };
    
  • class Solution:
        def appendCharacters(self, s: str, t: str) -> int:
            m, n = len(s), len(t)
            i = 0
            for j in range(n):
                while i < m and s[i] != t[j]:
                    i += 1
                if i == m:
                    return n - j
                i += 1
            return 0
    
    
  • func appendCharacters(s string, t string) int {
    	m, n := len(s), len(t)
    	for i, j := 0, 0; j < n; i, j = i+1, j+1 {
    		for i < m && s[i] != t[j] {
    			i++
    		}
    		if i == m {
    			return n - j
    		}
    	}
    	return 0
    }
    

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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