Implement a function meetingPlanner that given the availability, slotsA and slotsB, of two people and a meeting duration dur, returns the earliest time slot that works for both of them and is of duration dur. If there is no common time slot that satisfies the duration requirement, return null.

Examples:  input:  slotsA = [[10, 50], [60, 120], [140, 210]]         slotsB = [[0, 15], [60, 70]]         dur = 8 output: [60, 68]  input:  slotsA = [[10, 50], [60, 120], [140, 210]]         slotsB = [[0, 15], [60, 70]]         dur = 12 output: null # since there is no common slot whose duration is 12 

This is my approach to the solution:

    import java.io.*;     import java.util.*;     import java.lang.*;      class Solution {        static int[] meetingPlanner(int[][] slotsA, int[][] slotsB, int dur) {         // your code goes here         int i = 0,j = 0;         int maxStartDur,minEndDur, overlap;         int [] optTime = new int[2];         int [] nullArr = new int[0];         for( i = 0; i < slotsA.length; i++)         {           for(j = 0; j < slotsB.length; j++ )           {             //Calculate the overlap between corresponding times               maxStartDur = Math.max(slotsA[i][0],slotsB[j][0]);               minEndDur = Math.min(slotsA[i][1], slotsB[j][1]);               overlap = minEndDur - maxStartDur;                if( overlap >= dur )               {                 optTime[0] = maxStartDur;                 optTime[1] = maxStartDur + dur;                 return optTime;               }              /* minDur = Math.min((slotsA[i][1] - slotsA[i][0]),(slotsB[j][1] - slotsB[j][0]));               if( minDur >= dur)               {                 optTime[0] = slotsA[i][0];                 optTime[1] = slotsA[i][0] + dur;               }               else                 break;             }             */           }        }          return nullArr;       } // Time complexity:  O(slotsA.length*slotsB.length) //Space complexity: O(array of size [2])       public static void main(String[] args) {        }      } 

These are some of the questions regarding this code:

1) Can I further optimize this code to have an improved time and space complexity?

2) Is there a smarter approach to solve this question?

3) Are there better data structures available which I can use to solve this question more appropriately?

Reference