Count the Specials | Practice | GeeksforGeeks

Given an array A (may contain duplicates) of N elements and a positive integer K. The task is to count the number of elements which occurs exactly floor(N/K) times in the array.
Example:
Input:N = 5K = 2A[] = 1 4 1 2 4Output:2Explanation:In the given array, 1 and 4 occurs floor(5/2) = 2 times. So count is 2.
Your Task:Your task is to complete the function countSpecials() which should count the elements which occurs exactly floor(N/K) times.
Constrains:1 <= N <= 1031 <= Ai <= 1031 <= K <= 103
0
hgaur7011 month ago
TIME COMPLEXITY : O(n+m) , TIME TAKEN : 0.0
BY USING MAP :-
int countSpecials(int arr[], int sizeof_array, int K){
    
    int f = floor(sizeof_array/K), count = 0;
    
    int cot = 0;
    
    map<int, int> ob;
    map<int, int>::iterator it;

    for (int i = 0; i < sizeof_array; i++)
    {
        ob[arr[i]]++;
    }

    for (auto &&i : ob)
    {
        if (i.second == f)
        {
            count++;
        }
    }

    
    return count;
    
}
0
sumitd12991 month ago
SIMPLEST CODE WITH LEAST TIME COMPLEXITY YOU WILL GET HERE IS :
 
int countSpecials(int arr[], int sizeof_array, int K){      int f = floor(sizeof_array/K), count = 0;      int arr2[1000]={0};   // Your code here   for(int i=0;i<sizeof_array;i++){       arr2[arr[i]]++;   }   for(int i=0;i<1000;i++){       if(arr2[i]==f){           count++;       }   }
return count;
}
+1
benklomp1 month ago
int countSpecials(int arr[], int n, int K)
{
    int f = floor(n/K), count = 0;
    
    // Your code here
    
    // simple bubble sort
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < n; ++j)
            if(arr[i] > arr[j])
            {
                int T = arr[i];
                arr[i] = arr[j];
                arr[j] = T;
            }
            
    int counter = 0, p = -1;
    for(int i = 0; i < n; i += 1)
    {
        if(arr[i] == p) continue;
        
        int counter = 1, j = i;
        while(++j <  n && arr[i] == arr[j])
                ++counter;
                
        if(counter == f)
            ++count;
            
        p = arr[i];
    }
            
    return count;
0
sumitrodge272 months ago
i dont know how map works, someone help!
int countSpecials(int arr[], int sizeof_array, int K){
    
    int f = floor(sizeof_array/K), count = 0;
    int t=0;
    for(int i=0;i<sizeof_array;i++)
    {
        for(int j=0;j<sizeof_array;j++)
        {
            if(arr[i]==arr[j])
            {
                t++;
            }
        }
        if(t==f)
        {
            count++;
        }
        t=0;
    }
    
    count=ceil(count/f);
    
    return count;
    
}
0
sumitrodge273 months ago
someone please tell me what's wrong with my code 
 
int countSpecials(int arr[], int sizeof_array, int K){   int f = floor(sizeof_array/K), count = 0;   int p=0;   for(int i=0;i<sizeof_array;i++)   {              for(int j=0;j<sizeof_array;j++)       {           if(arr[i]==arr[j])           {              p+=1;            }       }       if(p==f)       {           count+=1;       }       p=0;   }   return count;   }
 
-2
dipawalimandaokar3213 months ago
// { Driver Code Starts//Initial Template for C++#include <bits/stdc++.h>using namespace std;
// Function Prototypeint countSpecials(int[], int, int);
// } Driver Code Ends//User function Template for C++
/*Function to count number of elements with occurrence* exactly equal to floor(sizeof_array/K)* arr : input array* sizeof_array : number of array elements*/int countSpecials(int arr[], int sizeof_array, int K){      int f = floor(sizeof_array/K),p=0;      for(int i=0;i<sizeof_array;i++)   {        int count;       for(int j=0;j<sizeof_array;j++)
    {   if(arr[i]==arr[j])       count++ ;    }         if(count==f)        p++;                 }   // Your code here            return p;   }
// { Driver Code Starts.
// Driver code to rest countSpecials functionint main() {// Testcase inputint testcase;cin >> testcase;while(testcase--){        // sizeof_array : number of array elements    // K : as given in statement    int sizeof_array, K;    cin >> sizeof_array >> K;        // array of size sizeof_array    int arr[sizeof_array];        for(int i = 0;i<sizeof_array;i++){        cin >> arr[i];    }        // calling function and printing count the occurrence    cout << countSpecials(arr, sizeof_array, K) << endl;    }return 0;}  // } Driver Code Ends
Whats the problem in countspecial function..am unable to understand.
0
hitentandon3 months ago
JS Sol:
        let f = Math.floor(N/K), count = 0, cc=1;
       arr.sort();
       for(let i = 1; i < N; i++)
           if(arr[i]==arr[i-1])
               cc++;
           else if(cc===f)
           {
               count++;
               cc=1;
           }
           else
               cc=1;
       if(cc===f)
           count++;
       return count;
C++ Sol:
   int f = floor(N/K), count = 0, cc=1;
   sort(arr, arr+N);
   for(int i = 1; i < N; i++)
       if(arr[i]==arr[i-1])
           cc++;
       else if(cc==f)
       {
           count++;
           cc=1;
       }
       else
           cc=1;
   if(cc==f)
       count++;
   return count;
+1
sourabhsingh191220014 months ago
Easy C++ Solution 
int countSpecials(int arr[], int sizeof_array, int K){
    
    int f = floor(sizeof_array/K), cnt = 0;
    unordered_map<int,int>mp;
    
    // Your code here
    for(int i=0;i<sizeof_array;i++){
        if(mp.find(arr[i]) == mp.end()){
            mp[arr[i]]=1;
        }else
       mp[arr[i]]++;
    }
    for(auto &i:mp){
        if(i.second==f)
        cnt++;
    }
    
    return cnt;
    
}
0
harshitd8914 months ago
int countSpecials(int arr[], int N, int K){   int f = floor(N/K), count = 0;int array[1000];for(int i=0;i<1000;i++){array[i]=0;   }
for(int i=0;i<1000;i++){   if(i<N){   int p=arr[i];   array[p]=array[p]+1;       }   else{       break;   }}for(int i=0;i<1000;i++){   if(array[i]==f){       count++;   }}   return count;   } 
0
ayushnautiyal11106 months ago
int countSpecials(int arr[], int sizeof_array, int K){      int f = floor(sizeof_array/K), count = 0;   unordered_map<int,int>mp;   for(int i=0;i<sizeof_array;i++){       mp[arr[i]]++;   }   for(auto it =mp.begin();it!=mp.end();it++){       if(it->second==f){           count++;       }   }   return count;}
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