# how to reduce time complexity of nested loop or replace nested loop

Hi experts,

I tried to resolve the issue here, but my solution exceeded 3s for 3 cases. Can you all please suggest?

**HERE IS THE PROBLEM:**

Given an array a of positive integers, find the number of pairs of integers for which the difference between the two numbers is equal to a given number k. Since the number of pairs could be quite large, take it modulo 109 + 7 for your output.

Example

For k = 3 and a = [1, 6, 8, 2, 4, 9, 12], the output should be

solution(k, a) = 3.

There are three pairs of integers whose difference is equal to 3: (1, 4), (6, 9) and (9, 12).

Input/Output

[execution time limit] 3 seconds (java)

[input] integer k

The specified difference between two numbers.

Guaranteed constraints:

1 ≤ k ≤ 1000.

[input] array.integer a

Guaranteed constraints:

1 ≤ a.length ≤ 105,

1 ≤ a[i] ≤ 1000.

[output] integer

The number of pairs of integers with difference k modulo 109 + 7.

[Java] Syntax Tips

// Prints help message to the console

// Returns a string

//

// Globals declared here will cause a compilation error,

// declare variables inside the function instead!

String helloWorld(String name) {

System.out.println("This prints to the console when you Run Tests");

return "Hello, " + name;

}

===================================

**Here's my solution:**

int solution(int k, int[] a) {

int number = 0;

for(int i=0; i<a.length; i++)

{

for(int j=i+1; j<a.length; j++)

{

if(Math.abs(a[i]-a[j])==k)

number++;

}

}

return number;

}

## Comments

I have no idea how I landed on this page while looking for writing assistance. I don’t know how students manage to find Chicago Reference Generator and writing service providers so easily. I have been looking for a reliable website for two hours now and still have not found what I’ve been looking for.