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. Search
  4. Missing Numbers
  5. Discussions

Missing Numbers

Sort by

recency

|

1254 Discussions

|

  • leonardofa
     + 0 comments

    JS solution using reduce. EZ

    function missingNumbers(arr, brr) {
    
    const groupA = arr.reduce((acc, n) => {
        acc[n] = (acc[n] || 0) + 1;
        return acc;
    }, {});
    
    const groupB = brr.reduce((acc, n) => {
        acc[n] = (acc[n] || 0) + 1;
        return acc;
    }, {});
    
    var missing = [];
    for(const nu of brr){
        if(groupA[nu] !== groupB[nu] && missing.indexOf(nu) < 0){
            missing.push(nu);
        }
    }
   
    return missing.sort(function(a, b) {
        return a - b;
    });
}

  • mr_waite
     + 0 comments

    Counting Sort is your friend, and the answer to many problems of this nature.

    O(m + n) solution is the goal.

  • E1201
     + 0 comments

    Most simple solution that i could think of: def missingNumbers(arr, brr): for item in arr: if brr.contains(item): brr.remove(item) brr = sorted(set(brr),key = int) return brr

  • afzal_saiyed931
     + 0 comments
    def missingNumbers(arr, brr):
    # Write your code here
    
    arr_freq = Counter(arr)
    brr_freq = Counter(brr)
    
    missing = []
    
    for key in brr_freq:
        if brr_freq[key] != arr_freq[key]:
            missing.append(key)
            
    return sorted(missing)

  • awadallayossef
     + 0 comments

    Time Complexity = 

    def missingNumbers(arr:list, brr:list):
    sol = []
    
    arr_frequency = Frequency(arr)
    brr_frequency = Frequency(brr)

    for number , freq in brr_frequency.items():
        if number not in arr_frequency:
            sol.append(number)
        elif number in arr_frequency:
            if arr_frequency[number] != freq: 
                sol.append(number)
    return sorted(sol)

def Frequency(elements:list):
    temp = {}
    for element in elements:
        if element not in temp:
            temp[element] = 1
        else:
            temp[element] += 1
    return temp