1003. Check If Word Is Valid After Substitutions

Posted 2021-11-26 ruruozhenhao

We are given that the string "abc" is valid.

From any valid string V, we may split V into two pieces X and Y such that X + Y (X concatenated with Y) is equal to V.  (Xor Y may be empty.)  Then, X + "abc" + Y is also valid.

If for example S = "abc", then examples of valid strings are: "abc", "aabcbc", "abcabc", "abcabcababcc".  Examples of invalid strings are: "abccba", "ab", "cababc", "bac".

Return true if and only if the given string S is valid.

 

Example 1:

Input: "aabcbc"
Output: true
Explanation: 
We start with the valid string "abc".
Then we can insert another "abc" between "a" and "bc", resulting in "a" + "abc" + "bc" which is "aabcbc".

Example 2:

Input: "abcabcababcc"
Output: true
Explanation: 
"abcabcabc" is valid after consecutive insertings of "abc".
Then we can insert "abc" before the last letter, resulting in "abcabcab" + "abc" + "c" which is "abcabcababcc".

Example 3:

Input: "abccba"
Output: false

Example 4:

Input: "cababc"
Output: false

 

Note:

1. 1 <= S.length <= 20000
2. S[i] is ‘a‘, ‘b‘, or ‘c.

 

Approach #1: Stack. [Java]

class Solution {
    public boolean isValid(String S) {
        Stack<Character> stack = new Stack<>();
        for (int i = 0; i < S.length(); ++i) {
            if (S.charAt(i) == ‘c‘) {
                if (stack.empty() || stack.pop() != ‘b‘) return false;
                if (stack.empty() || stack.pop() != ‘a‘) return false;
            } else stack.push(S.charAt(i));
        }
        return stack.empty();
    }
}

　　

Analysis:

Keep a stack, whenever meet character of ‘c‘, pop ‘b‘ and ‘a‘ at the end of the stack. Otherwise, return false;

 

Reference:

https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/247626/JavaPythonC%2B%2B-Stack-Solution-O(N)