Minimum Window Substring
<p>Given two strings <code>s</code> and <code>t</code> of lengths <code>m</code> and <code>n</code> respectively, return <em>the <strong>minimum window</strong></em> <span data-keyword="substring-nonempty"><strong><em>substring</em></strong></span><em> of </em><code>s</code><em> such that every character in </em><code>t</code><em> (<strong>including duplicates</strong>) is included in the window</em>. If there is no such substring, return <em>the empty string </em><code>&quot;&quot;</code>.</p>

<p>The testcases will be generated such that the answer is <strong>unique</strong>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;ADOBECODEBANC&quot;, t = &quot;ABC&quot;
<strong>Output:</strong> &quot;BANC&quot;
<strong>Explanation:</strong> The minimum window substring &quot;BANC&quot; includes &#39;A&#39;, &#39;B&#39;, and &#39;C&#39; from string t.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;a&quot;, t = &quot;a&quot;
<strong>Output:</strong> &quot;a&quot;
<strong>Explanation:</strong> The entire string s is the minimum window.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;a&quot;, t = &quot;aa&quot;
<strong>Output:</strong> &quot;&quot;
<strong>Explanation:</strong> Both &#39;a&#39;s from t must be included in the window.
Since the largest window of s only has one &#39;a&#39;, return empty string.
</pre>

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

<ul>
	<li><code>m == s.length</code></li>
	<li><code>n == t.length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> and <code>t</code> consist of uppercase and lowercase English letters.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you find an algorithm that runs in <code>O(m + n)</code> time?</p>
Hash Table; String; Sliding Window
use std::collections::HashMap;

impl Solution {
    pub fn min_window(s: String, t: String) -> String {
        let mut need: HashMap<char, usize> = HashMap::new();
        let mut window: HashMap<char, usize> = HashMap::new();
        for c in t.chars() {
            *need.entry(c).or_insert(0) += 1;
        }
        let m = s.len();
        let n = t.len();
        let mut k = -1;
        let mut mi = m + 1;
        let mut cnt = 0;

        let s_bytes = s.as_bytes();
        let mut l = 0;
        for r in 0..m {
            let c = s_bytes[r] as char;
            *window.entry(c).or_insert(0) += 1;
            if window[&c] <= *need.get(&c).unwrap_or(&0) {
                cnt += 1;
            }
            while cnt == n {
                if r - l + 1 < mi {
                    mi = r - l + 1;
                    k = l as i32;
                }

                let c = s_bytes[l] as char;
                if window[&c] <= *need.get(&c).unwrap_or(&0) {
                    cnt -= 1;
                }
                *window.entry(c).or_insert(0) -= 1;
                l += 1;
            }
        }
        if k < 0 {
            return String::new();
        }
        s[k as usize..(k as usize + mi)].to_string()
    }
}
