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Missing Numbers

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1241 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/6AXUA5_XwUw

    vector<int> missingNumbers(vector<int> arr, vector<int> brr) {
    map<int, int> mp;
    for(int i = 0; i < brr.size(); i++) mp[brr[i]]++;
    for(int i = 0; i < arr.size(); i++) mp[arr[i]]--;
    map<int, int>::iterator it;
    vector<int> result;
    for(it = mp.begin(); it != mp.end(); it++)
        if(it->second > 0) result.push_back(it->first);
    
    return result;
}

  • vidaureanto2002
     + 0 comments

    Dont judge me i am just starting 

    public static List<Integer> missingNumbers2(List<Integer> arr, List<Integer> brr) {
    if (arr.isEmpty() || brr.isEmpty() || arr.size() > brr.size()) return null;
    HashMap<Integer, Integer> frequency = new HashMap<>();
    HashMap<Integer, Integer> frequency2 = new HashMap<>();

    List<Integer> list = new ArrayList<>();
    for (int eArr : arr) {
        frequency2.put(eArr, frequency2.getOrDefault(eArr, 0) + 1);
    }
    for (int eBrr : brr) {
        frequency.put(eBrr, frequency.getOrDefault(eBrr, 0) + 1);
    }
    for (var entry : frequency.entrySet()){
        int key = entry.getKey();
        int value = entry.getValue();
        int value2 = frequency2.getOrDefault(key,0);
        if (value != value2){
            list.add(key);
        }
    }

    return list;
}

  • david1194
     + 0 comments

    Typescript:

    function missingNumbers(arr: number[], brr: number[]): number[] {
    // convert all of arr to arrMap
    const arrMap = new Map();
    let shifted;
    while(arr.length){
        shifted = arr.shift();
        arrMap.set(shifted, (arrMap.get(shifted) || 0) + 1);
    }
    const diff = new Set<number>();
    
    // remove any brr that matches in arrMap,
    // then remove from arrMap and brr
    // any non matches, save in diff
    while(brr.length){
        shifted = brr.shift();
     if (arrMap.has(shifted)) {
        arrMap.set(shifted, arrMap.get(shifted) - 1);
        if (arrMap.get(shifted) === 0) {
            arrMap.delete(shifted);
        }
     } else {
        diff.add(shifted);
     }
    }
    // diff is prefiltered and just needs to be sorted
    return Array.from(diff).sort((a, b) => a - b);
}

  • highsoft_soi
     + 0 comments

    Perl:

    sub missingNumbers {
    my $arr = shift;
    my $brr = shift;
    
    my (%h1, %h2);
    my @res;

    grep { $h1{$_} += 1; } @$arr;
    grep { $h2{$_} += 1; } @$brr;

    foreach my $e (keys %h2) {
        if (!exists($h1{$e})) {
            push(@res, $e);
        } elsif (exists($h1{$e}) && $h1{$e} != $h2{$e}) {
            push(@res, $e);
        }
    }

    return sort {$a <=> $b} @res;
}

  • pub_ashishk28
     + 1 comment

    Python3

    def missingNumbers(arr, brr):
    # Write your code here
   
    diff = Counter(brr) - Counter(arr)
    return sorted(diff.keys())