CODEWITHSUNDEEP
javaalgorithmsearchoptimization
Sahil Gupta
Sahil Gupta

Reputation: 53

Best way to find the number of pairs in an array whose difference is k

I was solving the problem on hackerrank. I had two approaches in my mind :

input : unsorted array(a) and k

First Approach :

1) Sort the array

2) for each array element a[i] ,find the element a[i]+K using binary search.If found increament the count and break the inner loop.

Second Approach :

1) Sort the array

2) for each array element a[i] ,find the element a[i]+K using linearsearch.If found increament the count and break the inner loop.

I found the First approach to be better as it will solve the problem in n(logn). But when multiple test cases are on the solutions the approach 2 takes lesser time . Can some one please explain why ?

Below are the code for the two approaches :

First Approach Code : 

static int pairs(int[] a,int k) {
  /* Complete this function */
    int temp;
    int len=a.length;
    int count=0;
    int beg;
    int mid;
    int end;
    int midVal;
    Arrays.sort(a);
    for(int i=0;i<len-1;i++){
    temp=a[i]+k;
    beg=i+1;  
    end=len-1;
        for(int l=beg;l<len;l++){
            mid=(beg+end)/2;
            midVal=a[mid];
            if(midVal==temp){
                count++;
                break;
            }
            else if(midVal>temp){
                end=mid-1;
            }
            else{
                beg=mid+1;
            }
        }

    }
    return count;
}

Second Approach Code :

static int pairs(int[] a,int k) {
  /* Complete this function */
    int temp;
    int len=a.length;
    int count=0;
    Arrays.sort(a);
    for(int i=0;i<len;i++){
    temp=a[i];
        for(int j=i+1;j<len;j++){
            if(temp-a[j]==-k){
                count++;
                break;
            }
        }
    }
    return count;
}

Upvotes: 4

Views: 24368

Answers (6)

Wei-Hsuan Chou
Wei-Hsuan Chou

Reputation: 51

My solution

System.out.println("findPair:"+findPair(new int[]{4,7,1,2,5,3,0,7,8,5,2}, 3));

public static int findPair(int[] nums, int target){ 
    Arrays.sort(nums);

    int count = 0;
    String pairText = "";
    for(int i =0; i < nums.length;i++){
        for(int j =i+1; j < nums.length;j++){

            if(nums[j]-nums[i] == target && (!pairText.equals(nums[i]+", "+nums[j]))){

                //System.out.println(nums[i]+", "+nums[j]);
                pairText = nums[i]+", "+nums[j];
                count++;
            }
        }

    }  


    return count;


}

Upvotes: 0

Deepak Maddeshiya
Deepak Maddeshiya

Reputation: 7

static boolean checkDuplicatesWithinK(int arr[], int k)
    {
        // Creates an empty hashset
        HashSet<Integer> set = new HashSet<>();

        // Traverse the input array
        for (int i=0; i<arr.length; i++)
        {
            // If already present n hash, then we found 
            // a duplicate within k distance
            if (set.contains(arr[i]))
               return true;

            // Add this item to hashset
            set.add(arr[i]);

            // Remove the k+1 distant item
            if (i >= k)
              set.remove(arr[i-k]);
        }
        return false;
    }

public static void main (String[] args)
    {
        int arr[] = {10, 5, 3, 4, 3, 5, 6};
        if (checkDuplicatesWithinK(arr, 3))
           System.out.println("Yes");
        else
           System.out.println("No");
    }

Upvotes: -1

Saif ali Karedia
Saif ali Karedia

Reputation: 912

private static int CountDistinctPairs(int[] nums, int k) {
    // TODO Auto-generated method stub
    HashMap<Integer, Boolean> map = new HashMap<Integer, Boolean>();
    HashSet<String> set = new HashSet<String>();
    for(int i =0;i < nums.length;i++){
        map.put(nums[i],true);
    }
    for (int i = 0 ; i < nums.length; i++){
        if(map.containsKey(nums[i]+k)){
            String a = "";
            if(nums[i]<nums[i]+k){
                a = "("+nums[i]+","+(nums[i]+k)+")";
            }
            else{
                a = "("+(nums[i] + k)+","+nums[i]+")";
            }
            set.add(a);
        }
    }
    System.out.println(set);
    return set.size();
}

Upvotes: 0

fatih tekin
fatih tekin

Reputation: 989

You can also use set no need to use hashmap 

public static void main(String[] args) throws Exception {

    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

    int[] firstLine = Stream.of(in.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
    int count = firstLine[0];
    int diff = firstLine[1];           
    Set<Integer> numbers = new HashSet<Integer>(Stream.of(in.readLine().split(" ")).map(Integer::valueOf).collect(Collectors.toList()));

    int pairCount = 0;
    for (Integer number : numbers) {
        pairCount += (numbers.contains(number+diff) ? 1:0); 
    }
    System.out.println(pairCount);                
}

Upvotes: 0

Abhishek Bansal
Abhishek Bansal

Reputation: 12705

In your 1st code, your inner loop termination condition is incorrect. It should terminate if the beg and end pointers cross each other.

Change it to while (beg <= end) or for (; beg <= end; )

Also, for the second approach there is no need for sorting the array.

Upvotes: 0

Vikram Bhat
Vikram Bhat

Reputation: 6246

The first approach is the best here among the two but there is better approach than both:-

Here a pseudo code for a better approach:-

for(i=0;i<Arr.length;i++) {
  Hashmap.add(Arr[i])
}
count=0;
for(j=0;j<Arr.length;j++) {

  if(Hashmap.containsKey(Arr[j]+k))
     count++;

}

Time complexity: O(N) whereas your approach = O(NlogN)

Edit:-

Note:- My approach has extra space complexity of O(N) for Hash table whereas suggested approach is inplace.

Upvotes: 7

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