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Missing Numbers

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  • brewjunk
    3 days ago+ 0 comments
    from collections import Counter
    
    
    def missingNumbers(arr, brr):
        count_a = Counter(arr)
        count_b = Counter(brr)
        result = []
        for i in arr:
            if i in brr and count_a[i] == count_b[i]:
                pass
            else:
                if i not in result:
                    result.append(i)
        for i in brr:
            if i not in arr:
                result.append(i)
                    
        return sorted(result, reverse = False)
    
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  • augustoribeiro_1
    1 week ago+ 0 comments
    public static List<Integer> missingNumbers(List<Integer> arr, List<Integer> brr) {
        
            Collections.sort(brr);
            
            for(int i = 0; i < arr.size(); i++) {
                brr.remove(arr.get(i));
            }
            
            // delete duplicates
            for(int i = 0; i < brr.size() - 1; i++) {
                int current = brr.get(i);
                int next = brr.get(i + 1);
                
                if(current == next) {
                    brr.remove(i);
                }
            }
    
            return brr;
        }
    
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  • grbz_kerem
    3 weeks ago+ 0 comments

    JavaScript Solution

    function missingNumbers(arr, brr) {
      const missing = [];
      const frequencyArr = Array.from({
        length: Math.max(...brr) + 1,
      }).fill(0);
      const frequencyBrr = [...frequencyArr];
      for (let i = 0; i < brr.length; i++) {
        if (arr[i]) frequencyArr[arr[i]]++;
        frequencyBrr[brr[i]]++;
      }
      for (let i = 0; i < frequencyBrr.length; i++) {
        if (frequencyArr[i] !== frequencyBrr[i]) missing.push(i);
      }
      return missing;
    }
    
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  • JimmyDev
    4 weeks ago+ 0 comments

    JS Solution

    function missingNumbers(arr, brr) {
        // Write your code here
        var oArr = {};
        var oBrr = {};
        for (var i=0; i<arr.length; i++) {
            if (!oArr[arr[i]]) oArr[arr[i]] = 0;
            oArr[arr[i]]++;
        }
        for (var i=0; i<brr.length; i++) {
            if (!oBrr[brr[i]]) oBrr[brr[i]] = 0;
            oBrr[brr[i]]++;
        }
        var lMissingNumber = [];
        for (var key in oBrr) {
            if (!oBrr.hasOwnProperty(key)) continue;
            if (oBrr[key] != oArr[key]) {
                lMissingNumber.push(key);
            }
        }
        return lMissingNumber;
    }
    
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  • job_av
    1 month ago+ 0 comments

    my JS solution:

    let s_arr = arr.sort((a,b)=>a-b);
    let s_brr = brr.sort((a,b)=>a-b);
    
    let missing = new Set;
    
    let i=0, j=0;
    while(i<s_brr.length){
        if (s_brr[i]!=s_arr[j]){
            missing.add(s_brr[i++]);
        } else {
            i++;
            j++;
        }
    }
    
    return Array.from(missing);
    
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