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Merge pull request #547 from TonyKim9401/main
[Tony] WEEK 11 Solutions
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// TC: O(n) | ||
// visit all nodes to find maximum path | ||
// SC: O(h) | ||
// h means the high of binary tree | ||
class Solution { | ||
private int output = Integer.MIN_VALUE; | ||
public int maxPathSum(TreeNode root) { | ||
max(root); | ||
return output; | ||
} | ||
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private int max(TreeNode node) { | ||
if (node == null) return 0; | ||
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int leftSum = Math.max(max(node.left), 0); | ||
int rightSum = Math.max(max(node.right), 0); | ||
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int currentMax = node.val + leftSum + rightSum; | ||
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output = Math.max(output, currentMax); | ||
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return node.val + Math.max(leftSum, rightSum); | ||
} | ||
} |
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// TC: O(n * m) | ||
// n = the length of edges, m = the length of each element list of edges | ||
// SC: O(n) | ||
// n = recursively visit everywhere | ||
public class Solution { | ||
public boolean validTree(int n, int[][] edges) { | ||
if (edges.length != n - 1) return false; | ||
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List<List<Integer>> graph = new ArrayList<>(); | ||
for (int i = 0; i < n; i++) { | ||
graph.add(new ArrayList<>()); | ||
} | ||
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for (int[] edge : edges) { | ||
graph.get(edge[0]).add(edge[1]); | ||
graph.get(edge[1]).add(edge[0]); | ||
} | ||
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boolean[] visited = new boolean[n]; | ||
if (!dfs(0, -1, graph, visited)) return false; | ||
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for (boolean v : visited) if(!v) return false; | ||
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return true; | ||
} | ||
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private boolean dfs(int node, int parent, List<List<Integer>> graph, boolean[] visited) { | ||
visited[node] = true; | ||
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for (int neighbor : graph.get(node)) { | ||
if (neighbor == parent) continue; | ||
if (visited[neighbor]) return false; | ||
if (!dfs(neighbor, node, graph, visited)) return false; | ||
} | ||
return true; | ||
} | ||
} |
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// TC: O(n) | ||
// must retrieve all elements | ||
// SC: O(n) | ||
// need the same size of memory space in the worst case | ||
class Solution { | ||
public int[][] insert(int[][] intervals, int[] newInterval) { | ||
List<int[]> output = new ArrayList<>(); | ||
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int i = 0; | ||
int n = intervals.length; | ||
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while (i < n && intervals[i][1] < newInterval[0]) { | ||
output.add(intervals[i]); | ||
i += 1; | ||
} | ||
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while (i < n && intervals[i][0] <= newInterval[1]) { | ||
newInterval[0] = Math.min(newInterval[0], intervals[i][0]); | ||
newInterval[1] = Math.max(newInterval[1], intervals[i][1]); | ||
i += 1; | ||
} | ||
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output.add(newInterval); | ||
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while (i < n) { | ||
output.add(intervals[i]); | ||
i += 1; | ||
} | ||
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return output.toArray(new int[output.size()][]); | ||
} | ||
} |
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// TC: O(n) | ||
// visiting all nodes | ||
// SC: O(1) | ||
// constant space complexity | ||
class Solution { | ||
public int maxDepth(TreeNode root) { | ||
return getMax(root, 0); | ||
} | ||
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private int getMax(TreeNode node, int depth) { | ||
if (node == null) return depth; | ||
depth += 1; | ||
return Math.max(getMax(node.left, depth), getMax(node.right, depth)); | ||
} | ||
} |
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// TC: O(n) | ||
// -> find middle, reverse, merge -> max O(n) | ||
// SC: O(1) | ||
class Solution { | ||
public void reorderList(ListNode head) { | ||
if (head == null || head.next == null) return; | ||
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ListNode slow = head, fast = head; | ||
while (fast != null && fast.next != null) { | ||
slow = slow.next; | ||
fast = fast.next.next; | ||
} | ||
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ListNode backSide = null; | ||
ListNode curr = slow.next; | ||
slow.next = null; | ||
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while (curr != null) { | ||
ListNode temp = curr.next; | ||
curr.next = backSide; | ||
backSide = curr; | ||
curr = temp; | ||
} | ||
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ListNode first = head; | ||
ListNode second = backSide; | ||
while (second != null) { | ||
ListNode temp = first.next; | ||
first.next = second; | ||
first = second; | ||
second = temp; | ||
} | ||
} | ||
} |