1003. Check If Word Is Valid After Substitutions

中文文档

Description

Given a string s, determine if it is valid.

A string s is valid if, starting with an empty string t = "", you can transform t into s after performing the following operation any number of times:

• Insert string "abc" into any position in t. More formally, t becomes tleft + "abc" + tright, where t == tleft + tright. Note that tleft and tright may be empty.

Return true if s is a valid string, otherwise, return false.

 

Example 1:

Input: s = "aabcbc"
Output: true
Explanation:
"" -> "abc" -> "aabcbc"
Thus, "aabcbc" is valid.

Example 2:

Input: s = "abcabcababcc"
Output: true
Explanation:
"" -> "abc" -> "abcabc" -> "abcabcabc" -> "abcabcababcc"
Thus, "abcabcababcc" is valid.

Example 3:

Input: s = "abccba"
Output: false
Explanation: It is impossible to get "abccba" using the operation.

 

Constraints:

• 1 <= s.length <= 2 * 104
• s consists of letters 'a', 'b', and 'c'

Solutions

Solution 1: Stack

We observe the operations in the problem and find that each time a string $\textit{"abc"}$ is inserted at any position in the string. Therefore, after each insertion operation, the length of the string increases by $3$. If the string $s$ is valid, its length must be a multiple of $3$. Thus, we first check the length of the string $s$. If it is not a multiple of $3$, then $s$ must be invalid, and we can directly return $\textit{false}$.

Next, we traverse each character $c$ in the string $s$. We first push the character $c$ onto the stack $t$. If the length of the stack $t$ is greater than or equal to $3$, and the top three elements of the stack form the string $\textit{"abc"}$, then we pop the top three elements from the stack. We then continue to traverse the next character in the string $s$.

After the traversal, if the stack $t$ is empty, it means the string $s$ is valid, and we return $\textit{true}$; otherwise, we return $\textit{false}$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.

Python3

class Solution:
    def isValid(self, s: str) -> bool:
        if len(s) % 3:
            return False
        t = []
        for c in s:
            t.append(c)
            if ''.join(t[-3:]) == 'abc':
                t[-3:] = []
        return not t

Java

class Solution {
    public boolean isValid(String s) {
        if (s.length() % 3 > 0) {
            return false;
        }
        StringBuilder t = new StringBuilder();
        for (char c : s.toCharArray()) {
            t.append(c);
            if (t.length() >= 3 && "abc".equals(t.substring(t.length() - 3))) {
                t.delete(t.length() - 3, t.length());
            }
        }
        return t.isEmpty();
    }
}

C++

class Solution {
public:
    bool isValid(string s) {
        if (s.size() % 3) {
            return false;
        }
        string t;
        for (char c : s) {
            t.push_back(c);
            if (t.size() >= 3 && t.substr(t.size() - 3, 3) == "abc") {
                t.erase(t.end() - 3, t.end());
            }
        }
        return t.empty();
    }
};

Go

func isValid(s string) bool {
	if len(s)%3 > 0 {
		return false
	}
	t := []byte{}
	for i := range s {
		t = append(t, s[i])
		if len(t) >= 3 && string(t[len(t)-3:]) == "abc" {
			t = t[:len(t)-3]
		}
	}
	return len(t) == 0
}

TypeScript

function isValid(s: string): boolean {
    if (s.length % 3 !== 0) {
        return false;
    }
    const t: string[] = [];
    for (const c of s) {
        t.push(c);
        if (t.slice(-3).join('') === 'abc') {
            t.splice(-3);
        }
    }
    return t.length === 0;
}