Leetcode 1003. Check If Word Is Valid After Substitutions

Problem Explanation

The problem entails determining if a given string, S, can be made solely by a series of steps that involve repeatedly inserting the string "abc" in any valid string V. To insert "abc", split V into two parts, X and Y (either could be empty), and form a new string by concatenating X, "abc", and Y.

The strategy to solve this problem is to use a stack to keep track of each character in the string. The stack will simulate the process of inserting "abc" into the string. By looking at the top of the stack, we can understand whether the current configuration of the string could have been achieved by inserting "abc".


Let's take an example string 'abcabc'.

  1. Start by splitting abcabc into abc and abc.
  2. The first part matches the valid string, now we need to verify whether the second part 'abc' is valid.

This shows that 'abcabc' string is valid.

Now let's take another example string 'abcabcababcc'.

  1. Start by splitting subjects abcabcababcc into abc and abcababcc.
  2. The left part is valid. Let's split the remaining abcababcc into abc and ababcc.
  3. For the remaining part ababcc, we can split this into ab and abcc.
  4. Insert 'abc' between ab and abcc will get abcabcababcc.
  5. All parts are valid so the string is valid too.



3class Solution:
4    def isValid(self, S: str) -> bool:
5        # Stack for storing and matching the characters
6        stack = []
7        for c in S:
8            if c == 'c':
9                # Case where there aren't enough elements for matching "abc"
10                if len(stack) < 2:
11                    return False
12                b, a = stack.pop(), stack.pop()
13                # The last two elements don't match "b" and "a", invalid
14                if a != 'a' or b != 'b':
15                    return False
16            else:
17                # Push other characters into stack
18                stack.append(c)
20        # If there are unmatched characters left, the string is invalid
21        return len(stack) == 0


3class Solution {
4    public boolean isValid(String S) {
5        Stack<Character> stack = new Stack<>();
6        for (char c : S.toCharArray()) {
7            if (c == 'c') {
8                if (stack.isEmpty() || stack.pop() != 'b')
9                    return false;
10                if (stack.isEmpty() || stack.pop() != 'a')
11                    return false;
12            } 
13            else {
14                stack.push(c);
15            }
16        }
17        return stack.isEmpty();
18    }


3var isValid = function(S) {
4    let stack = [];
5    for(let char of S){
6        if(char === 'c'){
7            if(stack.length < 2) return false;
8            if(stack.pop() !== 'b') return false;
9            if(stack.pop() !== 'a') return false;
10        } 
11        else stack.push(char);
12    }  
13    return stack.length === 0;


3public class Solution {
4    public bool IsValid(string S) {
5        Stack<char> stack = new Stack<char>();
6        foreach(char c in S) {
7            if (c == 'c') {
8                if (stack.Count < 2 || stack.Pop() != 'b')
9                    return false;
10                if (stack.Count < 1 || stack.Pop() != 'a')
11                    return false;
12            } 
13            else {
14                stack.Push(c);
15            }
16        }
17        return stack.Count == 0;
18    }


3class Solution {
5    bool isValid(string s) {
6        stack<char> stack;
7        for (char c : s){
8            if(c == 'c'){
9                if(stack.size() < 2)
10                    return false;
11                if(stack.top() != 'b')
12                    return false;
13                stack.pop();
14                if(stack.top() != 'a')
15                    return false;
16                stack.pop();
17            } else {
18                stack.push(c);
19            }
20        }
21        return stack.empty();
22    }

In the solution, we maintain a stack that stores characters, simulating the process of inserting "abc". Whenever we encounter the character 'c', we check if there are at least two elements on the stack and if the last two characters were 'b' and 'a' in that order. If this is not the case, the string is invalid. Else, we pop the last two characters from the stack. For any non-'c' character, we push it onto the stack. If there are no unprocessed characters by the end of the string, the string is valid.# Complexity Analysis

The time complexity for this solution is O(n), where n is the length of the string. We iterate through the string only once, at each step performing a constant amount of work. In the worst case, we push and pop each character once, which are both O(1) operations.

The space complexity is also O(n), as in the worst case, we might end up storing all characters of the string in the stack before we start popping any of them.

In usage, this algorithm will be most efficient when dealing with strings of length in the thousands or tens of thousands. For particularly large strings, the space demands of the stack might become prohibitive.


The provided solution is already quite efficient, but there are a few possible optimizations:

  1. Early Exit: If the length of the string isn't a multiple of 3, we can return false immediately as there's no way to form the string with "abc" sections.

  2. Embed the Pattern Matching: Instead of stacking all characters and then verifying if they match the "b" and "a" pattern when a "c" appears, alternately we could directly check during the iteration if the current character and the two prior ones match the "abc" pattern.

These changes wouldn't alter the complexity of the solution but could provide some small speed improvements.


The problem of checking if a string is made by repeating "abc" can be effectively solved by using a stack to keep track of string characters and match them with "abc". This solution is both easy to implement and understand, and works efficiently for large strings. Considering the simplicity and effectiveness of this approach, it serves as an excellent demonstration of how stacks can effectively be used in string manipulation tasks.

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