Daily Challenge #215 - Difference of 2

The objective is to implement a function that will return all pairs of integers from a given array of integers that have a difference of 2. The result array should be sorted in ascending order of values.

Assume there are no duplicate integers in the array. The order of the integers in the input array should not matter.

Examples

[1, 2, 3, 4]      -->  [[1, 3], [2, 4]]
[4, 1, 2, 3]      -->  [[1, 3], [2, 4]]
[1, 23, 3, 4, 7]  -->  [[1, 3]]
[4, 3, 1, 5, 6]   -->  [[1, 3], [3, 5], [4, 6]]

Tests

pairDifference([1,2,3,4])
pairDifference([1,3,4,6])

Happy coding!

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Comments

[Vidit Sarkar]

Python solution

pairDifference = lambda lst: sorted([(num,num+2) for num in lst if (num+2) in lst])

print(pairDifference([1, 2, 3, 4])) # output [(1, 3), (2, 4)]
print(pairDifference([4, 1, 2, 3])) # output [(1, 3), (2, 4)]
print(pairDifference([1, 23, 3, 4, 7])) # output [(1, 3)]
print(pairDifference([4, 3, 1, 5, 6])) # output [(1, 3), (3, 5), (4, 6)]
print(pairDifference([1,3,4,6])) # output [(1, 3), (4, 6)]

[SavagePixie]

Whoa, that reads almost as if it were English!

[Vidit Sarkar]

Thanks!

[Vidit Sarkar]

C++ solution

vector<pair<int, int>> pairDifference(vector<int> arr){
    // store all the numbers in a set for ease of searching
    // cpp set stores all numbers in ascending order by default
    set<int> numbers(arr.begin(), arr.end());

    // stores the pairs which have difference of two
    vector<pair<int, int>> numPairs;

    for(int num : numbers){
        // for any number num if num+2 is also in the set
        // then include it in the vector
        if(numbers.find(num+2) != numbers.end()){
            numPairs.push_back(make_pair(num,num+2));
        }
    }
    return numPairs;
}

[SavagePixie]

Something like this should do it in JavaScript. I'm not entirely sure if this sort is okay, though. I can never remember which way it's ascendent and which descendent.

const pairDifference = list => list
    .sort()
    .filter(x => list.includes(x + 2))
    .map(x => [ x, x + 2 ])

[Usman Khalil]

This really is Savage

[Antonio J.]

Ruby solution

def pair_difference(lst)
  lst.sort # sort array
    .combination(2) # get all posible combinations of pairs
    .select{|arr| (arr.max - arr.min) == 2 } # only combinations that have difference of 2
    .sort_by{|arr| arr[0] } # sort by the first number
end

[Khatri-Ran-Bahadur]

 <script>
      var myFunction = (list) =>
        list
          .sort((num1, num2) => num1 - num2)
          .reduce((returnList, eachVal, i, assendingArray) => {
            for (let j = i + 1; j < assendingArray.length; j++) {
              if (Math.abs(eachVal - assendingArray[j]) === 2) {
                returnList.push([eachVal, assendingArray[j]]);
              }
            }
            return returnList;
          }, []);
      console.log(myFunction([3, 4, 7, 6, 9, 8]));
      console.log(myFunction([12, 4, 6, 7, 9, 70]));
      console.log(myFunction([23, 40, 7, 16, 9, 38]));
      console.log(myFunction([23, 25, 27, 29, 30, 32]));
    </script>

[Matt Lott]

JS solution

const pairDifference = (nums) => {
  const results = []
  if (!nums || nums.length < 2) return results
  nums.sort()
  for (let i = 0; i < nums.length - 1; i++) {
    // 2nd num can only be 1 or 2 positions away when sorted and no duplicates
    if (nums[i + 1] - nums[i] === 2) {
      results.push([nums[i], nums[i + 1]])
    }
    else if (i < nums.length - 2 && nums[i + 2] - nums[i] === 2) {
      results.push([nums[i], nums[i + 2]])
    }
  }
  return results
}

const pairsEqual = (p1, p2) => p1[0] === p2[0] && p1[1] === p2[1]

const test = (nums, expected) => {
  const results = pairDifference(nums)
  const passed = results.length === expected.length &&
    !results.find((p, i) => !pairsEqual(p, expected[i]))
  console.log(`${passed? 'Pass' : 'FAIL'} - input: ${JSON.stringify(nums)} ` +
  `expected: ${JSON.stringify(expected)} results: ${JSON.stringify(results)}`)
}

test([1, 2, 3, 4], [[1, 3], [2, 4]])
test([1, 3, 4, 6], [[1, 3], [4, 6]])
test([4, 1, 2, 3], [[1, 3], [2, 4]])
test([1, 23, 3, 4, 7], [[1, 3]])
test([4, 3, 1, 5, 6], [[1, 3], [3, 5], [4, 6]])
test(undefined, [])
test([], [])
test([1], [])
test([2, -1, 0, 1], [[-1, 1], [0, 2]])
test([2000, -100, -98, 1999], [[-100, -98]])

[Valts Liepiņš]

Took some iterations to get to this solution, but here it is.
Haskell:

import Data.Maybe (catMaybes)
import Data.List (sort)

pairDifference :: [Int] -> [(Int, Int)]
pairDifference xs = catMaybes $ f <$> sorted <*> sorted
  where
    sorted = sort xs
    f x y | y - x == 2 = Just (x, y)
          | otherwise  = Nothing

[kesprit]

My Swift solution :

func pairDifference(tab: [Int]) -> [[Int]]{
    tab.sorted().reduce(into: [[Int]]()) { (result, intValue) in
        if let value = tab.first(where: { $0 == intValue + 2 }) {
            result.append([intValue, value])
        }
    }
}

pairDifference(tab: [1, 2, 3, 4]) // [[1, 3], [2, 4]]
pairDifference(tab: [4, 1, 2, 3]) // [[1, 3], [2, 4]]
pairDifference(tab: [1, 23, 3, 4, 7]) // [[1, 3]]
pairDifference(tab: [4, 3, 1, 5, 6]) // [[1, 3], [3, 5], [4, 6]]

[Sabin Pandelovitch]

JS solution with reduce

const pairDifference = arr =>
  arr.sort((a,b)=>a-b).reduce((acc, val, i, oa) => {
    for (let j = i + 1; j < oa.length; j++) {
      if (Math.abs(val - oa[j]) === 2) {
        acc.push([val, oa[j]]);
      }
    }
    return acc;
  }, []);

[mushtaqasif]

JS Solution

function pairDifference(data) {
    var pairs = [];
    var added = [];

    data.sort(function(x, y) {
        return x - y;
    });

    for (let i = 0; i < data.length; i++) {
        var x = data[i];

        for (let j = 0; j < data.length; j++) {
            var y = data[j];

            if (added.indexOf(x) < 0 && added.indexOf(y) < 0
                && Math.abs(x - y) == 2
            ) {
                pairs.push([x, y]);
                added.push(x);
                added.push(y);
            }
        }
    }

    return pairs;
}

var testData = [
    [1, 2, 3, 4],
    [4, 1, 2, 3],
    [1, 23, 3, 4, 7],
    [4, 3, 1, 5, 6],
    [1,3,4,6]
];

for (let k = 0; k < testData.length; k++) {
    var data = testData[k];

    console.log(JSON.stringify(data) + ' --> ' 
        + JSON.stringify(pairDifference(data)))
}

output:

[1,2,3,4] --> [[1,3],[2,4]]
[4,1,2,3] --> [[1,3],[2,4]]
[1,23,3,4,7] --> [[1,3]]
[4,3,1,5,6] --> [[1,3],[4,6]]
[1,3,4,6] --> [[1,3],[4,6]]