Check If Word Is Valid After Substitutions

Check If Word Is Valid After Substitutions - Problem

String Stack 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 t_left + "abc" + t_right, where t == t_left + t_right. Note that t_left and t_right may be empty.

Return true if s is a valid string, otherwise, return false.

Input & Output

Example 1 — Valid String

$ Input: s = "aabcbc"

› Output: true

💡 Note: Start with empty string, insert "abc" to get "abc", then insert "abc" at beginning to get "abcabc", but wait - we can think of it as: "aabcbc" → remove "abc" from middle → "abc" → remove "abc" → empty string.

Example 2 — Invalid String

$ Input: s = "abacbc"

› Output: false

💡 Note: Cannot form this by inserting only "abc" patterns. The 'a' in the middle breaks the pattern.

Example 3 — Simple Valid

$ Input: s = "abcabcabc"

› Output: true

💡 Note: Three "abc" patterns in sequence can be built by inserting "abc" three times.

Constraints

1 ≤ s.length ≤ 2 × 10⁴
s consists only of letters 'a', 'b', and 'c'

Visualization

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Asked in

G Google 15 f Facebook 12

The key insight is that valid strings can always be reduced to empty by removing "abc" patterns. Best approach uses a stack to detect and remove patterns in O(n) time. Time: O(n), Space: O(n).

Common Approaches

✓ One Pass Hash

⏱️ Time: N/A Space: N/A

Brute Force String Replacement

⏱️ Time: O(n²) Space: O(n)

Keep searching for "abc" patterns in the string and remove them. If we can reduce the string to empty, it's valid. Otherwise, it's invalid.

Stack-Based Parsing

⏱️ Time: O(n) Space: O(n)

Push characters onto a stack. When we see 'c' and the top two characters are 'b' and 'a', we can remove the "abc" pattern. If stack is empty at the end, the string is valid.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(char* s) {
    int len = strlen(s);
    char* pos;
    
    // Keep removing "abc" until no more found
    while ((pos = strstr(s, "abc")) != NULL) {
        // Shift remaining string left to remove "abc"
        memmove(pos, pos + 3, strlen(pos + 3) + 1);
        len -= 3;
    }
    
    // Return true if string is now empty
    return len == 0;
}

int main() {
    char buffer[100000];
    fgets(buffer, sizeof(buffer), stdin);
    
    // Remove newline if present
    int len = strlen(buffer);
    if (len > 0 && buffer[len-1] == '\n') {
        buffer[len-1] = '\0';
    }
    
    // Call solution and print result
    int result = solution(buffer);
    printf("%s\n", result ? "true" : "false");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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