Sync LeetCode submission Runtime - 111 ms (6.27%), Memory - 17.5 MB (15.29%)

Author: jimit105

2724-convert-an-array-into-a-2d-array-with-conditions/README.md

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+<p>You are given an integer array <code>nums</code>. You need to create a 2D array from <code>nums</code> satisfying the following conditions:</p>
+
+<ul>
+	<li>The 2D array should contain <strong>only</strong> the elements of the array <code>nums</code>.</li>
+	<li>Each row in the 2D array contains <strong>distinct</strong> integers.</li>
+	<li>The number of rows in the 2D array should be <strong>minimal</strong>.</li>
+</ul>
+
+<p>Return <em>the resulting array</em>. If there are multiple answers, return any of them.</p>
+
+<p><strong>Note</strong> that the 2D array can have a different number of elements on each row.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,3,4,1,2,3,1]
+<strong>Output:</strong> [[1,3,4,2],[1,3],[1]]
+<strong>Explanation:</strong> We can create a 2D array that contains the following rows:
+- 1,3,4,2
+- 1,3
+- 1
+All elements of nums were used, and each row of the 2D array contains distinct integers, so it is a valid answer.
+It can be shown that we cannot have less than 3 rows in a valid array.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,3,4]
+<strong>Output:</strong> [[4,3,2,1]]
+<strong>Explanation:</strong> All elements of the array are distinct, so we can keep all of them in the first row of the 2D array.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 200</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
+</ul>

2724-convert-an-array-into-a-2d-array-with-conditions/solution.py

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+# Approach 1: Frequency Counter
+
+# Time: O(N)
+# Space: O(N)
+
+class Solution:
+    def findMatrix(self, nums: List[int]) -> List[List[int]]:
+        freq = [0] * (len(nums) + 1)
+        res = []
+
+        for i in nums:
+            if freq[i] >= len(res):
+                res.append([])
+            res[freq[i]].append(i)
+            freq[i] += 1
+
+        return res