Found_HashMap_DistinctPoints

Question

 Given N 2D Points, Calculate no. of distinct points among them.

 Ex: x[5] = {2, 1, 3, 2, 2}
 y[5] = {3, 1, 2, 3, 4}

 The first array represents the x co-ordinates, the second array represents the y co-ordinate.       Acoording to above examples the points are 
(x[0],y[0])->(2,3)
(x[1],y[1])->(1,1)
(x[2],y[2])->(3,2)
(x[3],y[3])->(2,3)
(x[4],y[4])->(2,4)

Total number of distinct points are 4.

I wrote the implementation for the above problem by creating two HashMaps and storing the index as keys and elements as values. However, the code gave wrong output for ex: for the above test case the output of my code was "5 " as against "4". The implementation is as below:

      HashMap<Integer,Integer> mpp1 = new HashMap<>();
        HashMap<Integer,Integer> mpp2 = new HashMap<>();
        for(int i =0;i<n;i++){
            mpp1.put(i,x[i]);
        }
        for(int i =0;i<n;i++){
            mpp2.put(i,y[i]);
        }
        int count=0;
        boolean b=true;
        for(int i=0;i<n;i++){
            if(mpp1.get(i).equals(mpp2.get(i))==false){
                    b=false;
                   count++;
            }
          //  System.out.print(mpp1.get(i)+" ");
        }
       // if(b==true){
        //System.out.print("-1");
        //}
        
        System.out.print(count);

Don't compare objects using == or !=, but using equals(). — RealSkeptic
– RealSkeptic, Commented Oct 29, 2023 at 10:00
user_program: can you please make those changes in the code? Seriously? You can't make those yourself? — WJS
– WJS, Commented Oct 29, 2023 at 19:16
for(int i=0;i<n;i++){ if(mpp1.get(i).equals(mpp2.get(i))==false){ count++; } } System.out.print(count); — user_program
– user_program, Commented Oct 30, 2023 at 4:04

Ankit Singh · Accepted Answer · 2024-06-23 17:27:17Z

0

    Set<Integer> set = new HashSet();
    int count=0;
    
    for(int i=0;i<n;i++){
        int x = arr1[i]-arr2[i];
        
        if(!set.contains(x)){
            count++;
        }
        set.add(x);
    }
    
    System.out.println(count);

answered Jun 23, 2024 at 17:27

Ankit Singh

1

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Comments

user10717742 · Accepted Answer · 2023-10-29 19:15:54Z

Your implementation gave me the answer 4 for the stated problem.

    int n=5;
    int[] x = new int[] {2, 1, 3, 2, 2};
    int[] y = new int[] {3, 1, 2, 3, 4};

So it works, since it gives the expected answer. But I think it is the wrong approach, since you are comparing the x, y coordinates of a point, where x=1, y=1 is the only point that matches x with y and as such is not counted, which results in 4.

But you are looking for distinct 2D points, a 2D point is defined by x, y coordinates. So point-0 is x[0] and y[0], point-1 x[1] and y[1] etc.. So going by this you would need to first find your points, then compare their respective coordinates.

(not optimised) An approach would be to first find matching x coordinates, then check if y also matches if so, then this is your none distinct point. So in your example it would be point-0 (x=2, y=3) and point-3 (x=2, y=3).

The code could look like this:

    int[] x = new int[] {2, 1, 3, 2, 2};
    int[] y = new int[] {3, 1, 2, 3, 4};
    int noneUnique = 0;

    for(int i = x.length - 1; i > 0; i--) {
        for(int j = i - 1; j > -1; j--) {
            if (x[i] == x[j] && y[i] == y[j]) {
                noneUnique++;
            }
        }
    }

    System.out.format("found %d distinct points", x.length - noneUnique);

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