1024. Video Stitching

Problem Link : https://leetcode.com/problems/video-stitching/
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Complexity : Time-O(NlogN) Memory-O(N), TimeTaken-<2hr
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Logic :
First we sort the array then we traverse the array.
For each iteration till some 'i' we iterate till 'i' till
maxt is greater/ equal and also store the maxi for remembering the index
whose end is the most far then we send the remaining array for next traverse.
Base cond is that whenever we reach the end of array it means we haven't found the 
end time hence return -1 and whenver we find the end time we return 1 and for a final base cond
whenever we see a clip with disjoint interval it means that we cant form the req ans clip 
hence return -1

class Solution {
    public int videoStitching(int[][] clips, int time) {
       Integer[][] arr=new Integer[clips.length][2];
        for(int i=0;i<clips.length;i++){
            arr[i][0]=clips[i][0];
            arr[i][1]=clips[i][1];
        }
        Arrays.sort(arr, (a, b)->{
           if(a[0]==b[0])return a[1]-b[1];
            return a[0]-b[0];
        });
        for(Integer[] i:arr)
            System.out.println(i[0]+" , "+i[1]);
        return traverse(0, arr, 0, time);
    }
    private int traverse(int i, Integer[][] arr, int maxt, int time){
        if(i>=arr.length)
            return -1;
        System.out.println(i+" "+maxt);
        int ans=-1;
        int maxi=i;
        boolean found=false;
        while(i<arr.length && arr[i][0]<=maxt){
            if(time<=arr[i][1])return 1;
            if(arr[maxi][1]<=arr[i][1]){
                maxi=i;
                found=true;
            }
            ++i;
        }
        if(found)
            ans=traverse(maxi+1, arr, arr[maxi][1], time);
        if(ans==-1)
            return -1;
        return ans+1;
    }
}