We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Sorting
  4. Closest Numbers
  5. Discussions

Closest Numbers

Sort by

recency

|

632 Discussions

|

  • shiaramoon
     + 0 comments

    python 3

    def closestNumbers(arr):
    arr.sort()
    mindif = arr[-1] - arr[0]
    results = []
    
    for i in range(len(arr) - 1):
        diff = arr[i + 1] - arr[i]
        if diff < mindif:
            mindif = diff + 0
            results = [arr[i], arr[i + 1]]
        elif diff == mindif:
            results.append(arr[i])
            results.append(arr[i + 1])
    
    return results

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here: https://youtu.be/-sX3IgdQ6Wg

    vector<int> closestNumbers(vector<int> arr) {
    sort(arr.begin(), arr.end());
    int diff = INT_MAX;
    vector<int> result;
    for(int i = 1; i < arr.size(); i++){
        if(arr[i] - arr[i-1] < diff){
            diff = arr[i] - arr[i-1];
            result = {arr[i-1], arr[i]};
        }
        else if(arr[i] - arr[i-1] == diff){
            result.insert(result.end(), {arr[i-1], arr[i]});
        }
    }
    return result;
}

  • emhhacker
     + 0 comments

    My Java solution:

    public static List<Integer> closestNumbers(List<Integer> arr) {
        if(arr.size() == 2) return Arrays.asList(arr.get(0), arr.get(1));
        //goal: find pair with smallest abs diff
        //sort arr
        Collections.sort(arr);
        //init new min
        int min = arr.get(1) - arr.get(0);
        List<Integer> minIdices = new ArrayList<>();
        minIdices.add(arr.get(0));
        minIdices.add(arr.get(1));
        //compare adj elements
        for(int i = 2; i < arr.size(); i++){
            int diff = arr.get(i) - arr.get(i - 1);
            //if new min is found clear ans list
            if(diff < min){
                min = diff;
                minIdices.clear();
                minIdices.add(arr.get(i - 1));
                minIdices.add(arr.get(i));
            }
            else if(diff == min){
                minIdices.add(arr.get(i - 1));
                minIdices.add(arr.get(i));
            }
        }
        return minIdices;
    }

  • instanderapk8
     + 0 comments

    This information is really helpfull for thoese who are interested in custom ecommerce development

  • muktadeer
     + 0 comments

    public static List closestNumbers(List arr) { // Write your code here

        // 5,4,3,2
    List<List<Integer>> list = new ArrayList<>(arr.size() - 1);
    for (int i = 0; i < arr.size() - 1; i++) {
        list.add(new ArrayList<>());
    }
    arr = arr.stream().sorted().collect(Collectors.toList());

    for (int i = 0; i < arr.size() - 1; i++) {
        list.get(i).add(arr.get(i));
        list.get(i).add(arr.get(i + 1));
    }

    List<Integer> ind = new ArrayList<>();

    for (int i = 0; i < list.size(); i++) {
        ind.add(list.get(i).get(1) - list.get(i).get(0));
    }

    int smaller = ind.stream().min(Comparator.naturalOrder()).orElse(0);

    List<Integer> res = new ArrayList<>();
    for (int i = 0; i < ind.size(); i++) {
        if (ind.get(i) == smaller) {
            list.get(i).forEach(f -> res.add(f));
        }
    }
    // System.out.println(res);
    return res;
}