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[Tony] WEEK 12 Solutions #564

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Nov 3, 2024
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30 changes: 30 additions & 0 deletions merge-intervals/TonyKim9401.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,30 @@
// TC: O(n log n)
// It takes n log n to sort array, visit all elements in O(n) time, total O(n log n)
// SC: O(n)
// Needs max O(n) space to save intervals
class Solution {
public int[][] merge(int[][] intervals) {
int n = intervals.length;
if (n == 1) return intervals;

Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
List<int[]> output = new ArrayList<>();
output.add(intervals[0]);

int[] currentInterval = intervals[0];

for (int[] interval : intervals) {
int currentEnd = currentInterval[1];
int nextStart = interval[0];
int nextEnd = interval[1];
if (currentEnd >= nextStart) {
currentInterval[1] = Math.max(currentEnd, nextEnd);
Comment on lines +20 to +21
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@bky373 bky373 Nov 3, 2024

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(사소) 이 조건이면 max 로 하지 않고, currentInterval[1] = currentEnd 로 해도 괜찮아 보입니다!

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아앗 nextStartnextEnd 를 잘못 봤군요 위 내용은 넘어가시면 될 것 같습니다!

} else {
currentInterval = interval;
output.add(currentInterval);
}
}

return output.toArray(new int[output.size()][]);
}
}
Original file line number Diff line number Diff line change
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// TC: O(n + m)
// n = the number of nodes, m = the number of edges
// SC: O(n + m)
// n, m are the each size of the 2 demension array 'edges'
public class Solution {
public int countComponents(int n, int[][] edges) {
List<List<Integer>> graph = new ArrayList<>();

for (int i = 0; i < n; i++) graph.add(new ArrayList<>());

for (int[] edge : edges) {
graph.get(edge[0]).add(edge[1]);
graph.get(edge[1]).add(edge[0]);
}

boolean[] visit = new boolean[n];
int count = 0;

for (int i = 0; i < n; i++) {
if (!visit[i]) {
count += 1;
dfs(i, graph, visit);
}
}
return count;
}

private void dfs(int node, List<List<Integer>> graph, boolean[] visit) {
visit[node] = true;
for (int neighbor : graph.get(node)) {
if (!visit[neighbor]) dfs(neighbor, graph, visit);
}
}
}
21 changes: 21 additions & 0 deletions remove-nth-node-from-end-of-list/TonyKim9401.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,21 @@
// TC: O(n)
// Visit all elements in the worst case
// SC: O(1)
// Keep using ready assigned variables only
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode output = new ListNode(0, head);
ListNode dummy = output;

for (int i = 0; i < n; i++) head = head.next;

while (head != null) {
head = head.next;
dummy = dummy.next;
}

dummy.next = dummy.next.next;

return output.next;
}
}
17 changes: 17 additions & 0 deletions same-tree/TonyKim9401.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,17 @@
// TC: O(n)
// retreive all given nodes
// SC: O(1)
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dfs 스택을 사용하는 만큼 O(N) 이지 않을까요~?

// doesn't require additional space
class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if (p == null || q == null) {
if (p == null && q == null) return true;
return false;
Comment on lines +8 to +9
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😉

Suggested change
if (p == null && q == null) return true;
return false;
return p == null && q == null;

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@bky373 bky373 Nov 3, 2024

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L7 ~ L10 을 조금 더 줄여서 아래처럼 가능할 것 같습니다!

if (p == null || q == null) {
    return p == q;
}

}

if (p.val != q.val) return false;

return isSameTree(p.left, q.left) &&
isSameTree(p.right, q.right);
}
}