Pairing numbers (a,b) in an array such a way that a*2 >=b

Question

I am trying to solve pairing numbers (a,b) in an array such a way that a*2 >=b. Where a and b are numbers from input array.

Examples:

input: a[] = {1,2,3,4,5}

output: 2

explanation:

• we can pair 1 with 3
• 2 with 4 or 5

input: a[] = {4,3,2,1,5}

output: 2

explanation:

• we can pair 1 with 3
• 2 with 4 or 5

input: a[] = {4,3,2,1,5,6}

output: 3

explanation:

• we can pair 5 with 1
• 2 with 4
• 3 with 6

I tried to solve the problem using recursion like below, but this does not give any right results. Any help would be appreciated.

• Sort the input array
• if a[start] * 2 >= [end] found then add 1 to result recur for start +1 and end - 1
• else recur for (start + 1, end), (start, end - 1) and (start + 1, end - 1)

Idea is matching a[start] with remaining elements in the array and get max result.

    public static int countPairs(int[] a){
       Arrays.sort(a);
       return countPairs(a,a.length,0,a.length-1);
    }

    public static int countPairs(int[] a, int n, int start, int end){


        if(end == start){
            return 0;
        }
        if(start >= n || end < 0){
            return 0;
        }

         System.out.print("matching start: "+start + " and end "+end+"   ");System.out.println("matching "+a[start] + " and "+a[end]);

        if(a[start] < a[end] && a[start] * 2 >= a[end]  ){

            int res = countPairs(a,n,start+1,end-1) +1;
             //System.out.print("inside if matching start: "+start + " and end "+end+"   ");System.out.println("matching "+a[start] + " and "+a[end] + " count is "+res);
            return res;
        }
        else{

            return max(countPairs(a,n,start+1,end) ,
                    countPairs(a,n,start,end - 1),countPairs(a,n,start+1,end - 1));
        }

    }

tests:

import org.junit.Test;

import java.util.Arrays;
import java.util.Random;


public class CountingPairsTest {

    static int countPairs(int[] a){
        return PairingNumbers.countPairs(a);
    }

    @Test
     public void test1(){
        int[] a = {1,2,3,4,5};
        System.out.println("****************************************\n" + Arrays.toString(a));
        int count = countPairs(a);
        System.out.println("count "+count);
    }

    @Test public void test2(){
        int[] a = {1,2,3,4,5,6};
        System.out.println("****************************************\n" + Arrays.toString(a));
        int count = countPairs(a);
        System.out.println("count "+count);
    }

    @Test public void test5(){
        int[] a = {1,2,3,7,4,5,6};
        System.out.println("****************************************\n" + Arrays.toString(a));
        int count = countPairs(a);
        System.out.println("count "+count);
    }

    @Test public void test6(){
        int[] a = {9,8,20,15,21};

        System.out.println("****************************************\n" + Arrays.toString(a));
        int count = countPairs(a);
        System.out.println("count "+count);
    }

    @Test public  void test3(){
        int[] a = {5,4,3,2,1};
        System.out.println("****************************************\n" + Arrays.toString(a));
        int count = countPairs(a);
        System.out.println("count "+count);
    }

    @Test public void test4(){
        int[] a = {2,4,5,3,1};

        System.out.println("****************************************\n" + Arrays.toString(a));
        int count = countPairs(a);
        System.out.println("count "+count);
    }

    @Test public void test7(){
        int[] a = new Random().ints(10,1,100).toArray();// IntStream.range(1,100).toArray();


        System.out.println("****************************************\n" + Arrays.toString(a));
        int count = countPairs(a);
        System.out.println("count "+count);
    }
    @Test public void test8(){
        int[] a = new Random().ints(10,1,10).toArray();// IntStream.range(1,100).toArray();


        System.out.println("****************************************\n" + Arrays.toString(a));
        int count = countPairs(a);
        System.out.println("count "+count);
    }
}

Answer 1

I suggest that the answer is a.length / 2. Half of the length of the array (rounded down if the length was odd). You can pair the numbers any way you like. For non-negative a and b if a * 2 < b, just swap a and b and you will have a * 2 >= b. So since it takes two numbers to make a pair, you can always form precisely as many pairs as half of the length of the array.

Example (from the comments): [1, 2, 2, 2]. Length is 4, half of the length is 2, so there should be 2 pairs. Let’s check: (1, 2) is a nice pair because 1 * 2 >= 2. (2, 2) is another nice pair since 2 * 2 >= 2. In this case we didn’t even need any swapping (on the other hand the same pairs would have worked with swapping too: 2 * 2 >= 1 and 2 * 2 >= 2).

It will not always work if the array may contain negative numbers. So you may want to add a validation of the array that checks that it doesn’t.

What went wrong in your solution?

Your recursive algorithm is wrong. Say the array is [2, 3, 7, 9]. Clearly (2, 3) and (7, 9) are nice pairs, so there are two pairs here. The way you describe your algorithm, since (2, 9) is not a valid pair, you discard at least one of 2 and 9, leaving no chance of forming two pairs from the remaining numbers.