Discussion on Missing Numbers Challenge

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2 days ago+ 0 comments

Hash map and reallocate array

Time Complexity: O(n)
Space Complexity: O(n)

int find_min(int* arr, int arr_count){
    int min_num = arr[0];
    for (int i=1; i<arr_count; i++){
        if(arr[i] < min_num){
            min_num = arr[i];
        }
    }
    return min_num;
}

int* append(int* arr, int* size, int number){
    int new_size = *size+1;
    arr = (int* )realloc(arr, new_size*sizeof(int));
    arr[*size] = number;
    *size = new_size;
    return arr;
}

int* missingNumbers(int arr_count, int* arr, int brr_count, int* brr, int* result_count) {
    // Initialize NULL pointer
    int* res_arr = NULL;
    *result_count = 0;
    
    // find brr minimum element
    int brr_min = find_min(brr, brr_count);
    // initialize the array with zeros
    int* brr_freq = (int* )calloc(100, sizeof(int));
    
    for (int i = 0; i<brr_count; i++){
        brr_freq[brr[i]-brr_min] ++;
    }    
    for (int i = 0; i< arr_count; i++){
        brr_freq[arr[i]-brr_min] --;
    }
    for (int i = 0; i<100; i++){
        if (brr_freq[i] != 0){
            res_arr = append(res_arr, result_count, (i+brr_min));
        }
    }
    return res_arr;
}

3 weeks ago+ 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;
}

4 weeks ago+ 0 comments

RUST:

fn missingNumbers(arr: &[i32], brr: &[i32]) -> Vec<i32> {
    let mut count_arr: HashMap<i32,i32> = HashMap::new();
    let mut count_brr: HashMap<i32,i32> = HashMap::new();
    let mut result: Vec<i32> = Vec::new();

    for num in arr {
        count_arr.entry(*num).and_modify(|c| *c+=1).or_insert(1);
    }
    for num in brr {
        count_brr.entry(*num).and_modify(|c| *c+=1).or_insert(1);
    }
    for (num, _) in &count_brr {
        if count_brr.get(&num).unwrap_or(&0) > count_arr.get(&num).unwrap_or(&0) {
            result.push(*num);
        }
    }
    result.sort();
    result
}

1 month ago+ 1 comment

Here is my Python solution!

def missingNumbers(arr, brr):
    arr = Counter(arr)
    brr = Counter(brr)
    missing = []
    for num in brr.keys():
        if arr[num] < brr[num]:
            missing.append(num)
    return sorted(list(set(missing)))

1 month ago+ 1 comment

I think, you can just do the difference between the Counter objects. def missingNumbers(arr, brr): arr = Counter(arr) brr = Counter(brr) missing = brr - arr return sorted(list(missing))

1 month ago+ 0 comments

I'm not very familiar with Counter objects so I didn't know this existed.

2 months ago+ 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;
}

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