Sync LeetCode submission Runtime - 884 ms (69.74%), Memory - 38 MB (50.00%)

Author: jimit105

2003-distinct-numbers-in-each-subarray/README.md

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+<p>Given an integer array <code>nums</code> and an integer <code>k</code>, you are asked to construct the array <code>ans</code> of size <code>n-k+1</code> where <code>ans[i]</code> is the number of <strong>distinct</strong> numbers in the subarray <code>nums[i:i+k-1] = [nums[i], nums[i+1], ..., nums[i+k-1]]</code>.</p>
+
+<p>Return <em>the array </em><code>ans</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,3,2,2,1,3], k = 3
+<strong>Output:</strong> [3,2,2,2,3]
+<strong>Explanation: </strong>The number of distinct elements in each subarray goes as follows:
+- nums[0:2] = [1,2,3] so ans[0] = 3
+- nums[1:3] = [2,3,2] so ans[1] = 2
+- nums[2:4] = [3,2,2] so ans[2] = 2
+- nums[3:5] = [2,2,1] so ans[3] = 2
+- nums[4:6] = [2,1,3] so ans[4] = 3
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,1,1,1,2,3,4], k = 4
+<strong>Output:</strong> [1,2,3,4]
+<strong>Explanation: </strong>The number of distinct elements in each subarray goes as follows:
+- nums[0:3] = [1,1,1,1] so ans[0] = 1
+- nums[1:4] = [1,1,1,2] so ans[1] = 2
+- nums[2:5] = [1,1,2,3] so ans[2] = 3
+- nums[3:6] = [1,2,3,4] so ans[3] = 4
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
+</ul>

2003-distinct-numbers-in-each-subarray/solution.py

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+# Approach 1: Hash Map
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def distinctNumbers(self, nums: List[int], k: int) -> List[int]:
+        n = len(nums)
+        answer = [0] * (n - k + 1)
+
+        freq = {}
+
+        # Process first window
+        for num in nums[:k]:
+            freq[num] = freq.get(num, 0) + 1
+        answer[0] = len(freq)
+
+        # Slide window and update freq
+        for pos in range(k, n):
+            left = nums[pos - k]
+            freq[left] -= 1
+            if freq[left] == 0:
+                del freq[left]
+
+            right = nums[pos]
+            freq[right] = freq.get(right, 0) + 1
+
+            answer[pos - k + 1] = len(freq)
+
+        return answer
+