Determine overlapping numbers in ranges

You will be given an array with 5 numbers. The first 2 numbers represent a range, and the next two numbers represent another range. The final number in the array is X. The goal of your program is to determine if both ranges overlap by at least X numbers. For example, in the array [4, 10, 2, 6, 3] the ranges 4 to 10 and 2 to 6 overlap by at least 3 numbers (4, 5, 6), so your program should return true. Solve with and without looping.

If the array is [10, 20, 4, 14, 4] then the ranges are:

10 11 12 13 14 15 16 17 18 19 20
4 5 6 7 8 9 10 11 12 13 14

These ranges overlap by at least 4 numbers, namely: 10, 11, 12, 13, 14 so your program should return true.

With loop:

function OverlappingRanges(arr) {

  // keep a count of how many numbers overlap
  var counter = 0;
  
  // loop through one of the ranges
  for (var i = arr[0]; i < arr[1]; i++) {

    // check if a number within the first range exists
    // in the second range
    if (i >= arr[2] && i <= arr[3]) { 
      counter += 1;
    }

  }
 
  // check if the numbers that overlap is equal to or greater
  // than the last number in the array
  return (counter >= arr[4]) ? true : false;
}

OverlappingRanges([4, 10, 2, 6, 3]); 

Without loop:

function overlapping(input){
  var nums1 = listOfNums(input[0], input[1]);
  var nums2 = listOfNums(input[2], input[3]);
  var overlappingNum = 0;

  if(nums1[0] >= nums2[0] && nums1[0] <= nums2[1]){
    overlappingNum =  nums2[1] - nums1[0] + 1;
  } else {
    overlappingNum =  nums1[1] - nums2[0] + 1;
  }
  if(overlappingNum >= input[4]){
    return true;
  }
}

function listOfNums(a, b){
  var start = a;
  var end = b;
  if(a > b){
    start = b;
    end = a;
  }

  return [a, b];
}

var a = [4, 10, 2, 6, 3];
overlapping(a)