C++ Program to find all the elements that appear more than ” n/k ” times.

October 13, 2022

Elements that appear more than “n/k” times in C++

Here, in this page we will discuss the program to find all the elements that appear more than “n/k” times in C++ . We are given with an array of size n, find all elements in array that appear more than n/k times.

Example :

  • Input : arr[ ] : {3, 1, 2, 2, 1, 2, 3, 3} and k = 4,
  • Output : 2, 3
  • Explanation : As, size of array i.e, n = 8 so, n/k => 2 and in the given input array 2 and 3 occurs 3 and 3 times respectively. 
Elements that appear more than " n/k " times in C++

Method 1 (Brute force Approach) :

In this approach we will discuss the naive approach to solve this problem.

  • Iterate over the array and count its frequency.
  • If count is more than n/k times then print that element.
  • Now, here we need to handle the duplicate printing, so for that make every j-th element to -1 (Here, we are assuming that array will contain only positive elements).

Time and Space complexity of above approach :

  • Time – complexity : O(n^2)
  • Space-complexity – O(1)

Code to find all Elements that appear more than " n/k " times in C++

#include<bits/stdc++.h>
using namespace std;

int main(){

   int n;
   cin>>n;

   int arr[n];
   for(int i=0; i<n; i++)
     cin>>arr[i];

   int k;
   cin>>k;

   int x = n/k;

   for(int i=0; i<n; i++){

     int count = 1;

     if(arr[i] != -1){
         for(int j=i+1; j<n; j++){

            if(arr[i]==arr[j]){
                 count++;
                 arr[j] = -1; //set that visited element to -1 to avoid repetition
            }
         }
     }
     if(count > x) cout<<arr[i]<<" ";
    }

}
Input :
9 
1 2 2 6 6 6 6 7 9
4

Output :
6

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Minimum no. of Jumps to reach the end of an array 

Find Largest sum contiguous Subarray

Minimize the maximum difference between heights 

Method 2 :

In this approach we will discuss the efficient approach than the above approach to solve this problem .

  • Sort the array.
  • Iterate over the array and  count the frequency of every distinct array element by checking if adjacent elements are equal or not.
  • If frequency of the array element is found to be greater than n / k, then print the array element.

Time and Space complexity of above approach :

  • Time – complexity : O(n logn)
  • Space-complexity – O(1)

Code to find all Elements that appear more than " n/k " times in C++

#include<bits/stdc++.h>
using namespace std;

int main(){

  int n;
  cin>>n;

  int arr[n];
  for(int i=0; i<n; i++)
    cin>>arr[i];

  int k;
  cin>>k;

  int x = n/k;

  sort(arr, arr + n);

  for(int i = 0; i < n;) {

    int count = 1;

    while ((i + 1) < n && arr[i] == arr[i + 1]) {
      count++;
      i++;
     }

    if (count > x) {
      cout << arr[i] << " ";
    }
    i++;
  }

}
Input :
9 
1 2 2 6 6 6 6 7 9
4

Output :
6

Method 3 (Using Hash-map) :

In this approach we will discuss the efficient approach . We will use map to store the frequency of the element.

  • Declare a hash map.
  • Iterate over the array and increase that value in map by 1.
  • Now, iterate over the map and check if frequency is found to be greater than n / k, then print that key value.

Time and Space complexity of above approach :

  • Time – complexity : O(n )
  • Space-complexity – O(n)
Elements that occur more than n/k times 
#include<bits/stdc++.h>
using namespace std;

int main(){

    int n;
    cin>>n;

    int arr[n];
    for(int i=0; i<n; i++)
      cin>>arr[i];

    int k;
    cin>>k;

    int x = n/k;
    unordered_map<int, int>mp;

    for (int i = 0; i < n; i++) {
     mp[arr[i]]++;
    }

    for(auto it=mp.begin(); it!=mp.end(); it++){
      if(it->second>x)
        cout<first<<" ";
    }

}
Input :
9 
1 2 2 6 6 6 6 7 9
4

Output :
6

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