existence-of-a-substring-in-a-string-and-its-reverse

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Algorithms

Given a string <code>s</code>, find any substring of length <code>2</code> which is also present in the reverse of <code>s</code>.

Return <code>true</code> if such a substring exists, and <code>false</code> otherwise.

&nbsp;
Example 1:

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
Input: s = &quot;leetcode&quot;

Output: true

Explanation: Substring <code>&quot;ee&quot;</code> is of length <code>2</code> which is also present in <code>reverse(s) == &quot;edocteel&quot;</code>.
</div>

Example 2:

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
Input: s = &quot;abcba&quot;

Output: true

Explanation: All of the substrings of length <code>2</code> <code>&quot;ab&quot;</code>, <code>&quot;bc&quot;</code>, <code>&quot;cb&quot;</code>, <code>&quot;ba&quot;</code> are also present in <code>reverse(s) == &quot;abcba&quot;</code>.
</div>

Example 3:

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
Input: s = &quot;abcd&quot;

Output: false

Explanation: There is no substring of length <code>2</code> in <code>s</code>, which is also present in the reverse of <code>s</code>.
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> consists only of lowercase English letters.</li>
</ul>

isSubstringPresent

Existence of a Substring in a String and Its Reverse

Hash Table

String

给你一个字符串 <code>s</code> ，请你判断字符串 <code>s</code> 是否存在一个长度为 <code>2</code> 的子字符串，在 <code>s</code> 反转后的字符串中也出现。

如果存在这样的子字符串，返回 <code>true</code>；如果不存在，返回 <code>false</code> 。

&nbsp;

示例 1：

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
输入：s = "leetcode"

输出：true

解释：子字符串 <code>"ee"</code> 的长度为 <code>2</code>，它也出现在 <code>reverse(s) == "edocteel"</code> 中。
</div>

示例 2：

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
输入：s = "abcba"

输出：true

解释：所有长度为 <code>2</code> 的子字符串 <code>"ab"</code>、<code>"bc"</code>、<code>"cb"</code>、<code>"ba"</code> 也都出现在 <code>reverse(s) == "abcba"</code> 中。
</div>

示例 3：

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
输入：s = "abcd"

输出：false

解释：字符串 <code>s</code> 中不存在满足「在其反转后的字符串中也出现」且长度为 <code>2</code> 的子字符串。
</div>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li>字符串 <code>s</code> 仅由小写英文字母组成。</li>
</ul>