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[jaejeong1] Week 13 Solutions #583

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Nov 10, 2024
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20 changes: 20 additions & 0 deletions house-robber/jaejeong1.java
Original file line number Diff line number Diff line change
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class Solution {
public int rob(int[] nums) {
// 인접한 경우는 제외한 최대 합
// 풀이: dp로 풀이한다. dp[i] = max(dp[i-1], dp[i-2] + nums[i])
// TC: O(N), SC: O(N)
if (nums.length == 1) {
return nums[0];
}

var dp = new int[nums.length];
dp[0] = nums[0];
dp[1] = Math.max(nums[0], nums[1]);

for (int i=2; i<nums.length; i++) {
dp[i] = Math.max(dp[i-1], dp[i-2] + nums[i]);
}

return dp[nums.length-1];
}
}
33 changes: 33 additions & 0 deletions lowest-common-ancestor-of-a-binary-search-tree/jaejeong1.java
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// Definition for a binary tree node.
class TreeNode {
int val;
TreeNode left;
TreeNode right;

TreeNode(int x) {
val = x;
}
}


class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
// 플이: p,q 둘다 root 보다 값이 적으면 왼쪽 서브트리를, 둘다 크면 오른쪽 서브트리를 탐색해야한다.
// 그렇지 않으면 해당 root가 최저 공통 조상이 된다.
// TC: O(H), H: 트리의 높이
// SC: O(1)
var node = root;
while (node != null) {
if (p.val < node.val && q.val < node.val) {
node = node.left;
} else if (p.val > node.val && q.val > node.val) {
node = node.right;
} else {
break;
}
}

return node;
}
}
32 changes: 32 additions & 0 deletions meeting-rooms/jaejeong1.java
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import java.util.List;

// Definition of Interval:
class Interval {
int start, end;
Interval(int start, int end) {
this.start = start;
this.end = end;
}
}


public class Solution {
/**
* @param intervals: an array of meeting time intervals
* @return: if a person could attend all meetings
*/
public boolean canAttendMeetings(List<Interval> intervals) {
// 풀이: 정렬 후 양옆을 비교해가며 조건에 맞지 않으면 false 를 반환한다.
// TC: O(N)
// SC: O(1)

var sortedIntervals = intervals.stream().sorted().toList();

for (int i=0; i<sortedIntervals.size()-1; i++) {
if (sortedIntervals.get(i).end > sortedIntervals.get(i+1).start) {
return false;
}
}
return true;
}
}
27 changes: 27 additions & 0 deletions non-overlapping-intervals/jaejeong1.java
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import java.util.Arrays;
import java.util.Comparator;

class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
// 풀이: 정렬 후 각 구간의 끝 값을 저장해가며 양 옆을 비교, 제거 대상일때 카운트를 증가시켜 반환한다.
// TC: O(N)
// SC: O(1)

// intervals를 각 구간의 끝 값을 기준으로 정렬
Arrays.sort(intervals, Comparator.comparingInt(a -> a[1]));

int answer = 0;
int end = intervals[0][1]; // 첫 번째 구간의 끝 값

// 두 번째 구간부터 순회하며 겹치는지 확인
for (int i = 1; i < intervals.length; i++) {
if (intervals[i][0] < end) { // 현재 구간이 이전 구간과 겹치면
answer++; // 제거 횟수를 증가
} else {
end = intervals[i][1]; // 겹치지 않으면 현재 구간의 끝 값을 갱신
}
}

return answer;
}
}