Day 35.2.txt

1200. Minimum Absolute Difference

Given an array of distinct integers arr, find all pairs of elements with the minimum absolute difference of any two elements. 

Return a list of pairs in ascending order(with respect to pairs), each pair [a, b] follows

a, b are from arr
a < b
b - a equals to the minimum absolute difference of any two elements in arr
 

Example 1:

Input: arr = [4,2,1,3]
Output: [[1,2],[2,3],[3,4]]
Explanation: The minimum absolute difference is 1. List all pairs with difference equal to 1 in ascending order.
Example 2:

Input: arr = [1,3,6,10,15]
Output: [[1,3]]
Example 3:

Input: arr = [3,8,-10,23,19,-4,-14,27]
Output: [[-14,-10],[19,23],[23,27]]
 

Constraints:

2 <= arr.length <= 10^5
-10^6 <= arr[i] <= 10^6



class Solution {
    static int a[];
    public List<List<Integer>> minimumAbsDifference(int[] ar) {
        List<List<Integer>> nm=new ArrayList<List<Integer>>();
        a=new int[ar.length];
        int i,k=Integer.MAX_VALUE;
        for(i=0;i<a.length;i++)
        {
            a[i]=ar[i];
        }
        task(a,0,ar.length-1);
        for(i=0;i<a.length-1;i++)
        {
            if((Math.abs(a[i]-a[i+1]))<k)
            {
                k=Math.abs(a[i]-a[i+1]);
            }
        }
        ArrayList<Integer> yy=new ArrayList<>();
        for(i=0;i<a.length-1;i++)
        {
            if((Math.abs(a[i]-a[i+1]))==k)
            {
                yy.add(a[i]);
                yy.add(a[i+1]);
                nm.add(new ArrayList (yy));
                yy.clear();
            }
        }
        return nm;
    }
    void task(int n[] , int i , int j)
    {
        if(i<j)
        {
            int m=(i+j)/2;
            task(n,i,m);
            task(n,m+1,j);
            task1(n,i,m,j);
        }
    }
    void task1(int n[] , int s , int mid , int e)
    {
        int i=s,j=mid+1,k=0;
        int b[]=new int[e-s+1];
        while(i<=mid&&j<=e)
        {
            if(n[i]<=n[j])
            {
                b[k++]=n[i++];
            }
            else
            {
                b[k++]=n[j++];
            }
        }
        while(i<=mid)
        {
            b[k++]=n[i++];
        }
        while(j<=e)
        {
            b[k++]=n[j++];
        }
        for(i=s;i<=e;i++)
        {
            a[i]=b[i-s];
        }
    }
}