binary search code -:

 void insertInterval(List<int[]> intervals, int left, int right) {

    int lo = 0, hi = intervals.size();

    while (lo < hi) {

        int mid = (lo + hi) / 2;

        if (intervals.get(mid)[0] < left) {

            lo = mid + 1;

        } else {

            hi = mid;

        }

    }

    intervals.add(lo, new int[]{left, right});

}

import java.util.*;

class Solution{
    public int bi(int []arr,int tar){
        int low=0;
        int high=arr.length-1;
        while (low<=high){
            int mid=(low+high)/2;
            if(arr[mid]<tar){
                low=mid+1;
            }else{
                high=mid-1;
            }
        }
        return low;
    }



    public static void main(String[] args) {
        Solution sol=new Solution();
        int []st={1,1,4,4,5};
        System.out.println(sol.bi(st,3));


    }
} 

so ye upar wala if duplicate 1,2,2,2  and target=2 so isme 1 return and if maan le 1,2,4,5 aisa kuch toh if target =3 toh 2 return