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;
}

```