Class Solution
java.lang.Object
g2101_2200.s2124_check_if_all_as_appears_before_all_bs.Solution
2124 - Check if All A’s Appears Before All B’s.<p>Easy</p>
<p>Given a string <code>s</code> consisting of <strong>only</strong> the characters <code>'a'</code> and <code>'b'</code>, return <code>true</code> <em>if <strong>every</strong></em> <code>'a'</code> <em>appears before <strong>every</strong></em> <code>'b'</code> <em>in the string</em>. Otherwise, return <code>false</code>.</p>
<p><strong>Example 1:</strong></p>
<p><strong>Input:</strong> s = “aaabbb”</p>
<p><strong>Output:</strong> true</p>
<p><strong>Explanation:</strong></p>
<p>The ’a’s are at indices 0, 1, and 2, while the ’b’s are at indices 3, 4, and 5.</p>
<p>Hence, every ‘a’ appears before every ‘b’ and we return true.</p>
<p><strong>Example 2:</strong></p>
<p><strong>Input:</strong> s = “abab”</p>
<p><strong>Output:</strong> false</p>
<p><strong>Explanation:</strong></p>
<p>There is an ‘a’ at index 2 and a ‘b’ at index 1.</p>
<p>Hence, not every ‘a’ appears before every ‘b’ and we return false.</p>
<p><strong>Example 3:</strong></p>
<p><strong>Input:</strong> s = “bbb”</p>
<p><strong>Output:</strong> true</p>
<p><strong>Explanation:</strong> There are no ’a’s, hence, every ‘a’ appears before every ‘b’ and we return true.</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 <= s.length <= 100</code></li>
<li><code>s[i]</code> is either <code>'a'</code> or <code>'b'</code>.</li>
</ul>
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Method Summary
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Constructor Details
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Solution
public Solution()
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Method Details
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checkString
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