Sync LeetCode submission Runtime - 8 ms (64.55%), Memory - 16.7 MB (23.35%)

Author: jimit105

3291-find-if-array-can-be-sorted/README.md

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+<p>You are given a <strong>0-indexed</strong> array of <strong>positive</strong> integers <code>nums</code>.</p>
+
+<p>In one <strong>operation</strong>, you can swap any two <strong>adjacent</strong> elements if they have the <strong>same</strong> number of <span data-keyword="set-bit">set bits</span>. You are allowed to do this operation <strong>any</strong> number of times (<strong>including zero</strong>).</p>
+
+<p>Return <code>true</code> <em>if you can sort the array, else return </em><code>false</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [8,4,2,30,15]
+<strong>Output:</strong> true
+<strong>Explanation:</strong> Let&#39;s look at the binary representation of every element. The numbers 2, 4, and 8 have one set bit each with binary representation &quot;10&quot;, &quot;100&quot;, and &quot;1000&quot; respectively. The numbers 15 and 30 have four set bits each with binary representation &quot;1111&quot; and &quot;11110&quot;.
+We can sort the array using 4 operations:
+- Swap nums[0] with nums[1]. This operation is valid because 8 and 4 have one set bit each. The array becomes [4,8,2,30,15].
+- Swap nums[1] with nums[2]. This operation is valid because 8 and 2 have one set bit each. The array becomes [4,2,8,30,15].
+- Swap nums[0] with nums[1]. This operation is valid because 4 and 2 have one set bit each. The array becomes [2,4,8,30,15].
+- Swap nums[3] with nums[4]. This operation is valid because 30 and 15 have four set bits each. The array becomes [2,4,8,15,30].
+The array has become sorted, hence we return true.
+Note that there may be other sequences of operations which also sort the array.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,3,4,5]
+<strong>Output:</strong> true
+<strong>Explanation:</strong> The array is already sorted, hence we return true.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,16,8,4,2]
+<strong>Output:</strong> false
+<strong>Explanation:</strong> It can be shown that it is not possible to sort the input array using any number of operations.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 2<sup>8</sup></code></li>
+</ul>

3291-find-if-array-can-be-sorted/solution.py

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+# Approach 2: Sortable Segments
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def canSortArray(self, nums: List[int]) -> bool:
+        num_of_set_bits = bin(nums[0]).count('1')
+        max_of_segment = nums[0]
+        min_of_segment = nums[0]
+
+        max_of_prev_segment = float('-inf')
+
+        for i in range(len(nums)):
+
+            # element belongs to same segment
+            if bin(nums[i]).count('1') == num_of_set_bits:
+                # update min and max of segment
+                max_of_segment = max(max_of_segment, nums[i])
+                min_of_segment = min(min_of_segment, nums[i])
+
+            # element belongs to new segment   
+            else:
+                # check if segments are arranged properly
+                if min_of_segment < max_of_prev_segment:
+                    return False
+
+                # update max of previous segment
+                max_of_prev_segment = max_of_segment
+
+                # start new segment with current element
+                max_of_segment = nums[i]
+                min_of_segment = nums[i]
+                num_of_set_bits = bin(nums[i]).count('1')
+
+        # final check for proper segment arrangement
+        if min_of_segment < max_of_prev_segment:
+            return False
+
+        return True
+