1003. Check If Word Is Valid After Substitutions

https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/

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'

----

Intuition

Brute force is replacing all "abc" with "" till no more substituions are possible

-----

Time - O(N)

Space - O(N)

----

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

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