[Tony] WEEK 13 Solutions

house-robber/TonyKim9401.java

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+// TC: O(n)
+// always need to check every case
+// SC: O(n)
+// the length of the result int list is same with the length of the given nums int list
+class Solution {
+    public int rob(int[] nums) {
+        int[] result = new int[nums.length];
+
+        if (nums.length < 2) return nums[0];
+        if (nums.length < 3) return Math.max(nums[0], nums[1]);
+
+        result[0] = nums[0];
+        result[1] = Math.max(nums[0], nums[1]);
+
+        for (int i = 2; i < nums.length; i++) {
+            result[i] = Math.max(result[i - 1], result[i - 2] + nums[i]);
+        }
+
+        return result[nums.length-1];
+    }
+}

lowest-common-ancestor-of-a-binary-search-tree/TonyKim9401.java

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+// TC: O(h)
+// h = the high of binary search tree
+// SC: O(1)
+// Doesn't require additional space
+class Solution {
+    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+        if (p.val < root.val && q.val < root.val)
+            return lowestCommonAncestor(root.left, p, q);
+        if (p.val > root.val && q.val > root.val)
+            return lowestCommonAncestor(root.right, p, q);
+        return root;
+    }
+}

meeting-rooms/TonyKim9401.java

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+// TC: O(n log n)
+// list sort
+// SC: O(n)
+// list sort max use O(n)
+public class Solution {
+    public boolean canAttendMeetings(List<Interval> intervals) {
+        intervals.sort(Comparator.comparingInt(o -> o.start));
+
+        for (int i = 0; i < intervals.size() - 1; i++) {
+            Interval preInterval = intervals.get(i);
+            Interval postInterval = intervals.get(i+1);
+
+            if (preInterval.end > postInterval.start) return false;
+        }
+        return true;
+    }
+}

non-overlapping-intervals/TonyKim9401.java

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+// TC: O(n log n)
+// in order to order the given intervals array
+// SC: O(1)
+// only constant space necessary
+class Solution {
+    public int eraseOverlapIntervals(int[][] intervals) {
+        Arrays.sort(intervals, (o1, o2) -> Integer.compare(o1[1], o2[1]));
+
+        int output = 0;
+        int end = intervals[0][1];
+
+        for (int i = 1; i < intervals.length; i++) {
+            int[] currentInterval = intervals[i];
+
+            if (currentInterval[0] < end) output += 1;
+            else end = currentInterval[1];
+        }
+
+        return output;
+    }
+}