<p>Given two strings <code>s1</code> and <code>s2</code>, return <code>true</code><em> if </em><code>s2</code><em> contains a permutation of </em><code>s1</code><em>, or </em><code>false</code><em> otherwise</em>.</p>

<p>In other words, return <code>true</code> if one of <code>s1</code>&#39;s permutations is the substring of <code>s2</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;ab&quot;, s2 = &quot;eidbaooo&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> s2 contains one permutation of s1 (&quot;ba&quot;).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;ab&quot;, s2 = &quot;eidboaoo&quot;
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s1.length, s2.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s1</code> and <code>s2</code> consist of lowercase English letters.</li>
</ul>

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        int a = s1.size();
        int b = s2.size();

        if (a > b) return false;
        std::unordered_map<char, int> s1_map;
        for(char c : s1) {
            s1_map.emplace(c, 0);
            ++s1_map[c];
        }
        std::unordered_map<char, int> s2_map;
        for(int i = 0; i < a; ++i) {
            s2_map.emplace(s2[i], 0);
            ++s2_map[s2[i]];
        }
        for(int i = a; i < b; ++i) {

            if (s2_map == s1_map) return true;
            char prev_char = s2[i-a];
            char curr_char = s2[i];
            s2_map[prev_char]-=1;
            if (s2_map[prev_char] == 0) {
                s2_map.erase(prev_char);
            }
            s2_map.emplace(curr_char, 0);
            ++s2_map[curr_char];
        }
        return s2_map == s1_map;
    }
};
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