Sync LeetCode submission Runtime - 65 ms (75.61%), Memory - 18.2 MB (39.66%)

Author: jimit105

0076-minimum-window-substring/README.md

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+<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>

0076-minimum-window-substring/solution.py

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+# Approach: Sliding Window
+
+# n = len(s), m = len(t)
+# Time: O(m + n)
+# Space: O(m + n)
+
+from collections import Counter
+
+class Solution:
+    def minWindow(self, s: str, t: str) -> str:
+        if not t or not s:
+            return ''
+
+        t_count = Counter(t)
+        required = len(t_count) # No. of unique chars in t
+
+        # formed is used to keep track of how many unique characters in t are present in the current window in its desired frequency.
+        formed = 0
+        window_counts = {}
+
+        left, right = 0, 0
+        min_length = float('inf')
+        min_start = 0
+
+        while right < len(s):
+            char = s[right]
+            window_counts[char] = window_counts.get(char, 0) + 1
+
+            if char in t_count and window_counts[char] == t_count[char]:
+                formed += 1
+
+            # If all required characters are found in the window
+            while left <= right and formed == required:
+                char = s[left]
+
+                # If the current window size is smaller than the min window size
+                if min_length > right - left + 1:
+                    min_length = right - left + 1
+                    min_start = left
+
+                window_counts[char] -= 1
+
+                if char in t_count and window_counts[char] < t_count[char]:
+                    # If the character is in t and its count in the window is less than its count in t
+                    formed -= 1 
+                left += 1 # shrink window
+
+            right += 1 # expand window
+
+        if min_length > len(s):
+            return ''
+
+        return s[min_start : min_start + min_length]
+