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Formatted question description: https://leetcode.ca/all/1003.html

# 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:

## 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/
// 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;
}
};

• class Solution {
public boolean isValid(String S) {
Stack<Character> stack = new Stack<Character>();
int length = S.length();
for (int i = 0; i < length; i++) {
char c = S.charAt(i);
if (c == 'a' || c == 'b')
stack.push(c);
else {
if (stack.size() < 2)
return false;
char c2 = stack.pop();
char c1 = stack.pop();
if (c2 != 'b' || c1 != 'a')
return false;
}
}
return stack.isEmpty();
}
}

############

class Solution {
public boolean isValid(String s) {
if (s.length() % 3 > 0) {
return false;
}
StringBuilder stk = new StringBuilder();
for (char c : s.toCharArray()) {
int n = stk.length();
if (c == 'c' && n > 1 && stk.charAt(n - 2) == 'a' && stk.charAt(n - 1) == 'b') {
stk.deleteCharAt(n - 1);
stk.deleteCharAt(n - 2);
} else {
stk.append(c);
}
}
return stk.length() == 0;
}
}

• // OJ: https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/
// 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;
}
};

• class Solution:
def isValid(self, s: str) -> bool:
if len(s) % 3:
return False
stk = []
for c in s:
if c == 'c' and len(stk) > 1 and stk[-2] == 'a' and stk[-1] == 'b':
stk = stk[:-2]
else:
stk.append(c)
return not stk

############

class Solution(object):
def isValid(self, S):
"""
:type S: str
:rtype: bool
"""
while "abc" in S:
S = S.replace("abc", "")
return not S

• func isValid(s string) bool {
if len(s)%3 > 0 {
return false
}
stk := []rune{}
for _, c := range s {
n := len(stk)
if c == 'c' && n > 1 && stk[n-2] == 'a' && stk[n-1] == 'b' {
stk = stk[:n-2]
} else {
stk = append(stk, c)
}
}
return len(stk) == 0
}

• 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;
}