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0025-reverse-nodes-in-k-group/0025-reverse-nodes-in-k-group.py

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+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
+        if k == 1:
+            return head
+
+        dummyHead = ListNode(-1, head)
+
+        beforeSeg = dummyHead
+        firstInSeg = head
+        lastInSeg = None
+        afterSeg = None
+
+        # Get initial values for lastInSeg and afterSeg
+        curr = head
+        for i in range(k):
+            if i == k - 1:
+                lastInSeg = curr
+                afterSeg = curr.next
+            elif curr is None:
+                return head
+
+            curr = curr.next
+        
+
+        # hold one before segment
+        # hold last in segment
+        # hold one after segment
+
+        while lastInSeg is not None:
+            # in segment, flip all links
+            prev = None
+            curr = beforeSeg.next
+            next = beforeSeg.next.next
+
+            for i in range(k):
+                newPrev = curr
+                newCurr = next
+                newNext = next.next if next is not None else None
+
+                curr.next = prev
+
+                prev, curr, next = newPrev, newCurr, newNext
+
+            
+            beforeSeg.next = lastInSeg
+            firstInSeg.next = afterSeg
+
+            beforeSeg = firstInSeg
+            firstInSeg = afterSeg
+
+            curr = firstInSeg
+            lastInSeg = None
+            afterSeg = None
+            for i in range(k):
+                if curr is None:
+                    break
+                elif i == k - 1:
+                    lastInSeg = curr
+                    afterSeg = curr.next
+
+                curr = curr.next
+
+        return dummyHead.next

0025-reverse-nodes-in-k-group/README.md

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+<h2><a href="https://leetcode.com/problems/reverse-nodes-in-k-group">25. Reverse Nodes in k-Group</a></h2><h3>Hard</h3><hr><p>Given the <code>head</code> of a linked list, reverse the nodes of the list <code>k</code> at a time, and return <em>the modified list</em>.</p>
+
+<p><code>k</code> is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of <code>k</code> then left-out nodes, in the end, should remain as it is.</p>
+
+<p>You may not alter the values in the list&#39;s nodes, only nodes themselves may be changed.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex1.jpg" style="width: 542px; height: 222px;" />
+<pre>
+<strong>Input:</strong> head = [1,2,3,4,5], k = 2
+<strong>Output:</strong> [2,1,4,3,5]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg" style="width: 542px; height: 222px;" />
+<pre>
+<strong>Input:</strong> head = [1,2,3,4,5], k = 3
+<strong>Output:</strong> [3,2,1,4,5]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the list is <code>n</code>.</li>
+	<li><code>1 &lt;= k &lt;= n &lt;= 5000</code></li>
+	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Follow-up:</strong> Can you solve the problem in <code>O(1)</code> extra memory space?</p>

0133-clone-graph/0133-clone-graph.py

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+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val = 0, neighbors = None):
+        self.val = val
+        self.neighbors = neighbors if neighbors is not None else []
+"""
+
+from typing import Optional
+from collections import deque
+class Solution:
+    def cloneGraph(self, node: Optional['Node']) -> Optional['Node']:
+        if not node:
+            return node
+
+        newHead = Node(node.val)
+
+        frontier = deque([ node ])
+        nodeMapping = { node.val: newHead }
+        while len(frontier) > 0:
+            current = frontier.popleft()
+
+            for n in current.neighbors:
+                if n.val not in nodeMapping:
+                    newNode = Node(n.val)
+                    nodeMapping[newNode.val] = newNode
+                    frontier.append(n)
+                nodeMapping[current.val].neighbors.append(nodeMapping[n.val])
+
+        return newHead
+        

0133-clone-graph/README.md

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+<h2><a href="https://leetcode.com/problems/clone-graph">133. Clone Graph</a></h2><h3>Medium</h3><hr><p>Given a reference of a node in a <strong><a href="https://en.wikipedia.org/wiki/Connectivity_(graph_theory)#Connected_graph" target="_blank">connected</a></strong> undirected graph.</p>
+
+<p>Return a <a href="https://en.wikipedia.org/wiki/Object_copying#Deep_copy" target="_blank"><strong>deep copy</strong></a> (clone) of the graph.</p>
+
+<p>Each node in the graph contains a value (<code>int</code>) and a list (<code>List[Node]</code>) of its neighbors.</p>
+
+<pre>
+class Node {
+    public int val;
+    public List&lt;Node&gt; neighbors;
+}
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>Test case format:</strong></p>
+
+<p>For simplicity, each node&#39;s value is the same as the node&#39;s index (1-indexed). For example, the first node with <code>val == 1</code>, the second node with <code>val == 2</code>, and so on. The graph is represented in the test case using an adjacency list.</p>
+
+<p><b>An adjacency list</b> is a collection of unordered <b>lists</b> used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.</p>
+
+<p>The given node will always be the first node with <code>val = 1</code>. You must return the <strong>copy of the given node</strong> as a reference to the cloned graph.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2019/11/04/133_clone_graph_question.png" style="width: 454px; height: 500px;" />
+<pre>
+<strong>Input:</strong> adjList = [[2,4],[1,3],[2,4],[1,3]]
+<strong>Output:</strong> [[2,4],[1,3],[2,4],[1,3]]
+<strong>Explanation:</strong> There are 4 nodes in the graph.
+1st node (val = 1)&#39;s neighbors are 2nd node (val = 2) and 4th node (val = 4).
+2nd node (val = 2)&#39;s neighbors are 1st node (val = 1) and 3rd node (val = 3).
+3rd node (val = 3)&#39;s neighbors are 2nd node (val = 2) and 4th node (val = 4).
+4th node (val = 4)&#39;s neighbors are 1st node (val = 1) and 3rd node (val = 3).
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/01/07/graph.png" style="width: 163px; height: 148px;" />
+<pre>
+<strong>Input:</strong> adjList = [[]]
+<strong>Output:</strong> [[]]
+<strong>Explanation:</strong> Note that the input contains one empty list. The graph consists of only one node with val = 1 and it does not have any neighbors.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> adjList = []
+<strong>Output:</strong> []
+<strong>Explanation:</strong> This an empty graph, it does not have any nodes.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the graph is in the range <code>[0, 100]</code>.</li>
+	<li><code>1 &lt;= Node.val &lt;= 100</code></li>
+	<li><code>Node.val</code> is unique for each node.</li>
+	<li>There are no repeated edges and no self-loops in the graph.</li>
+	<li>The Graph is connected and all nodes can be visited starting from the given node.</li>
+</ul>