Two solutions for finding pairs of numbers in a list with a given difference. See https://www.hackerrank.com/challenges/pairs for more test cases.

one-pointer.js

function onePointer (input) {
  const lines = input.split('\n')
  const k = Number(lines[0].split(' ')[1])
  let nums = lines[1].split(' ').map(x => Number(x))
  let count = 0

  // Sort the numbers in order to limit the number of required comparisons
  nums.sort((a, b) => a - b)

  // For each element, calculate the difference between it and subsequent elements up to
  // the point at which their difference is equal to K. O(n^2)
  for (let i = 0; i < nums.length; i++) {
    for (let j = i + 1; j < nums.length; j++) {
      // Since nums is sorted, we can always subtract the smaller number from the larger number,
      // so diff is always positive
      let diff = nums[j] - nums[i]

      if (diff === k) 
        count ++
        break
      }
      // If diff is greater than K, we know that diff will greater than K for all subsequent elements
      else if (diff > k) {
          break
      }
    }
    return count
}

pairs.md

Given the input

5 2
1 5 3 4 2

where 5 is the number of elements in the list consider, 2 is k, and 1 5 3 4 2 is the list to consider, the expected output is 3.

set.js

function set (input) {
  const lines = input.split('\n')
  const k = Number(lines[0].split(' ')[1])
  let nums = lines[1].split(' ').map(x => Number(x))
  let count = 0
  
  const nums_set = new Set(nums)
  
  // We only need to loop through the set once, so this runs in linear time (O(n)). Woohoo!
  nums_set.forEach(num => {
    // Find the difference between k and num. If the difference is in the set,
    // then there exists a pair with difference k.
    if (nums_set.has(num - k)) {
      count ++ 
    }
  })

  return count
}