Check if All A's Appears Before All B's

Check if All A's Appears Before All B's - Problem

String Easy

Given a string s consisting of only the characters 'a' and 'b', return true if every 'a' appears before every 'b' in the string. Otherwise, return false.

In other words, the string should have all 'a' characters grouped together at the beginning (or none at all), followed by all 'b' characters grouped together at the end (or none at all).

Input & Output

Example 1 — All a's before all b's

$ Input: s = "aaabbb"

› Output: true

💡 Note: All 'a' characters (positions 0,1,2) appear before all 'b' characters (positions 3,4,5)

Example 2 — Mixed pattern

$ Input: s = "abab"

› Output: false

💡 Note: The 'a' at position 2 appears after the 'b' at position 1, violating the pattern

Example 3 — Only one character type

$ Input: s = "aaa"

› Output: true

💡 Note: String contains only 'a' characters, so the pattern is valid

Constraints

1 ≤ s.length ≤ 100
s[i] is either 'a' or 'b'

Visualization

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Asked in

🍎 Apple 15 Ⓜ Microsoft 12

The key insight is that once we encounter a 'b' character, no 'a' should appear afterward. The optimal approach uses a single pass with a boolean flag to track if we've seen 'b', returning false immediately if 'a' appears after 'b' is encountered. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check All Pairs

⏱️ Time: O(n²) Space: O(1)

For each 'a' in the string, verify that it appears before every 'b' by comparing indices. This ensures no 'b' appears before any 'a'.

Single Pass - Early Detection

⏱️ Time: O(n) Space: O(1)

Traverse the string once from left to right. Track if we've seen any 'b' character. If we encounter an 'a' after seeing a 'b', the pattern is invalid.

Two Pointers - Find Critical Positions

⏱️ Time: O(n) Space: O(1)

Use two passes to find the position of the last 'a' and the first 'b'. If the last 'a' appears after the first 'b', the pattern is invalid.

Brute Force - Check All Pairs — Algorithm Steps

Find all positions of 'a' characters
Find all positions of 'b' characters
Check every 'a' position is less than every 'b' position

Visualization

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Step-by-Step Walkthrough

Find 'a'

Locate each 'a' character

Check backwards

Look for any 'b' before this 'a'

Validate

If 'b' found before 'a', return false

Code -

solution.c — C

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

bool solution(char* s) {
    int len = strlen(s);
    // Check each 'a' to ensure no 'b' appears before it
    for (int i = 0; i < len; i++) {
        if (s[i] == 'a') {
            // Check if any 'b' appears before this 'a'
            for (int j = 0; j < i; j++) {
                if (s[j] == 'b') {
                    return false;
                }
            }
        }
    }
    return true;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    // Remove newline character
    s[strcspn(s, "\n")] = 0;
    
    bool result = solution(s);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each 'a' we might check against all 'b's, worst case n/2 * n/2 comparisons

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

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Medium Frequency

~8 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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