Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 17.3 MB (30.05%)

Author: jimit105

3555-final-array-state-after-k-multiplication-operations-i/README.md

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+<p>You are given an integer array <code>nums</code>, an integer <code>k</code>, and an integer <code>multiplier</code>.</p>
+
+<p>You need to perform <code>k</code> operations on <code>nums</code>. In each operation:</p>
+
+<ul>
+	<li>Find the <strong>minimum</strong> value <code>x</code> in <code>nums</code>. If there are multiple occurrences of the minimum value, select the one that appears <strong>first</strong>.</li>
+	<li>Replace the selected minimum value <code>x</code> with <code>x * multiplier</code>.</li>
+</ul>
+
+<p>Return an integer array denoting the <em>final state</em> of <code>nums</code> after performing all <code>k</code> operations.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [2,1,3,5,6], k = 5, multiplier = 2</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[8,4,6,5,6]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<table>
+	<tbody>
+		<tr>
+			<th>Operation</th>
+			<th>Result</th>
+		</tr>
+		<tr>
+			<td>After operation 1</td>
+			<td>[2, 2, 3, 5, 6]</td>
+		</tr>
+		<tr>
+			<td>After operation 2</td>
+			<td>[4, 2, 3, 5, 6]</td>
+		</tr>
+		<tr>
+			<td>After operation 3</td>
+			<td>[4, 4, 3, 5, 6]</td>
+		</tr>
+		<tr>
+			<td>After operation 4</td>
+			<td>[4, 4, 6, 5, 6]</td>
+		</tr>
+		<tr>
+			<td>After operation 5</td>
+			<td>[8, 4, 6, 5, 6]</td>
+		</tr>
+	</tbody>
+</table>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,2], k = 3, multiplier = 4</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[16,8]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<table>
+	<tbody>
+		<tr>
+			<th>Operation</th>
+			<th>Result</th>
+		</tr>
+		<tr>
+			<td>After operation 1</td>
+			<td>[4, 2]</td>
+		</tr>
+		<tr>
+			<td>After operation 2</td>
+			<td>[4, 8]</td>
+		</tr>
+		<tr>
+			<td>After operation 3</td>
+			<td>[16, 8]</td>
+		</tr>
+	</tbody>
+</table>
+</div>
+
+<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;= 100</code></li>
+	<li><code>1 &lt;= k &lt;= 10</code></li>
+	<li><code>1 &lt;= multiplier &lt;= 5</code></li>
+</ul>

3555-final-array-state-after-k-multiplication-operations-i/solution.py

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+# Approach 2: Heap
+
+# Time: O(n + k log n)
+# Space: O(n)
+
+import heapq
+
+class Solution:
+    def getFinalState(self, nums: List[int], k: int, multiplier: int) -> List[int]:
+
+        pq = [(val, i) for i, val in enumerate(nums)]
+        heapq.heapify(pq)
+
+        for _ in range(k):
+            _, i = heapq.heappop(pq)
+            nums[i] *= multiplier
+            heapq.heappush(pq, (nums[i], i))
+
+        return nums
+