1003

1003. Check If Word Is Valid After Substitutions (Medium)

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.

Example 4:

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

 

Constraints:

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

Related Topics:
String, Stack

Similar Questions:

• Valid Parentheses (Easy)

Solution 1.

j is read pointer and i is write pointer. We always write s[j] to s[i].

If the last 3 characters in front of i is abc, we clean them by i -= 3.

In the end, return i == 0.

// OJ: https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    bool isValid(string s) {
        int i = 0, N = s.size();
        for (int j = 0; j < N; ++j) {
            s[i++] = s[j];
            if (i >= 3 && s[i - 3] == 'a' && s[i - 2] == 'b' && s[i - 1] == 'c') i -= 3;
        }
        return i == 0;
    }
};