For example, Meanwhile, the security system for these houses remain the same as for those in the previous street.great solution! LeetCode – Best Meeting Point (Java) A group of two or more people wants to meet and minimize the total travel distance. Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), find the minimum number of conference rooms required. 14 VIEWS. Copyright © 2008 - 2020Program Creekpublic int rob(int[] nums) {if(nums==null || nums.length==0)return 0;if(nums.length==1)return nums[0];int max1 = robHelper(nums, 0, nums.length-2);int max2 = robHelper(nums, 1, nums.length-1);return Math.max(max1, max2);}public int robHelper(int[] nums, int i, int j){if(i==j){return nums[i];}int[] dp = new int[nums.length];dp[i]=nums[i];dp[i+1]=Math.max(nums[i+1], dp[i]);for(int k=i+2; k<=j; k++){dp[k]=Math.max(dp[k-1], dp[k-2]+nums[k]);}return dp[j];}After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. [Leetcode] Meeting Rooms, Solution Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (s i < e i ), determine if a person could attend all meetings.
The distance is calculated using Manhattan Distance, where distance(p1, p2) = |p2.x - p1.x| + |p2.y - p1.y|.This problem is converted to find the median value on x-axis and y-axis.Copyright © 2008 - 2020Program CreekWhat if question ask to find co-ordinates. Solutions of LeetCode Online Judge. Output is one meeting room.
like if there is an overlap then inc count then no extra space is requiredWhat would be the runtime (big O) of your approach?I don’t think your solution is right. LeetCode – Meeting Rooms (Java) Given an array of meeting time intervals consisting of start and end times [s1, e1], [s2, e2], ... , determine if a person could attend all meetings. Last Edit: February 18, 2020 5:25 AM . Thanks!public int minTotalDistance(int[][] grid) {int m=grid.length;int n=grid[0].length;ArrayList<Integer> cols = new ArrayList<Integer>();ArrayList<Integer> rows = new ArrayList<Integer>();for(int i=0; i<m; i++){for(int j=0; j<n; j++){if(grid[i][j]==1){cols.add(j);rows.add(i);}}}int sum=0;for(Integer i: rows){sum += Math.abs(i - rows.get(rows.size()/2));}Collections.sort(cols);for(Integer i: cols){sum+= Math.abs(i-cols.get(cols.size()/2));}return sum;}The point (0,2) is an ideal meeting point, as the total travel distance of 2+2+2=6 is minimal.