Print all array elements appearing more than N / K times

Difficulty Level : Medium
Last Updated : 22 Jun, 2021

Given an array arr[] of size N and an integer K, the task is to find all the array elements that appear more than (N / K) times.

Examples:

Input: arr[] = { 1, 2, 6, 6, 6, 6, 6, 10 }, K = 4
Output: 6
Explanation:
The frequency of 6 in the array is greater than N / K(= 2). Therefore, the required output is 6.
Input: arr[] = { 3, 4, 4, 5, 5, 5, 5 }, K = 4
Output: 4 5
Explanation:
The frequency of 4 in the array is greater than N / K(= 1).
The frequency of 5 in the array is greater than N / K(= 1).
Therefore, the required output is 4 5.

Naive Approach: The simplest approach to solve this problem is to traverse the array and for every distinct array element, count its frequency and check if exceeds N / K or not. If found to be true, then print the array element.

Time Complexity: O(N²)
Auxiliary Space: O(1)

Sorting-based Approach: The idea is to sort the array followed by traversal of the array to 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.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to print all array elements
// whose frequency is greater than N / K
void NDivKWithFreq(int arr[], int N, int K)
{
    // Sort the array, arr[]
    sort(arr, arr + N);
 
    // Traverse the array
    for (int i = 0; i < N;) {
 
        // Stores frequency of arr[i]
        int cnt = 1;
 
        // Traverse array elements which
        // is equal to arr[i]
        while ((i + 1) < N
               && arr[i] == arr[i + 1]) {
 
            // Update cnt
            cnt++;
 
            // Update i
            i++;
        }
 
        // If frequency of arr[i] is
        // greater than (N / K)
        if (cnt > (N / K)) {
 
            cout << arr[i] << " ";
        }
        i++;
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 2, 6, 6, 6, 6, 7, 10 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 4;
 
    NDivKWithFreq(arr, N, K);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to print all array elements
// whose frequency is greater than N / K
static void NDivKWithFreq(int arr[], int N, int K)
{
    // Sort the array, arr[]
    Arrays.sort(arr);
 
    // Traverse the array
    for (int i = 0; i < N;) {
 
        // Stores frequency of arr[i]
        int cnt = 1;
 
        // Traverse array elements which
        // is equal to arr[i]
        while ((i + 1) < N
               && arr[i] == arr[i + 1]) {
 
            // Update cnt
            cnt++;
 
            // Update i
            i++;
        }
 
        // If frequency of arr[i] is
        // greater than (N / K)
        if (cnt > (N / K)) {
 
            System.out.print(arr[i]+ " ");
        }
        i++;
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 2, 6, 6, 6, 6, 7, 10 };
    int N = arr.length;
    int K = 4;
 
    NDivKWithFreq(arr, N, K);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement
# the above approach
 
# Function to prall array elements
# whose frequency is greater than N / K
def NDivKWithFreq(arr, N, K):
     
    # Sort the array, arr[]
    arr = sorted(arr)
 
    # Traverse the array
    i = 0
     
    while i < N:
         
        # Stores frequency of arr[i]
        cnt = 1
 
        # Traverse array elements which
        # is equal to arr[i]
        while ((i + 1) < N and
               arr[i] == arr[i + 1]):
 
            # Update cnt
            cnt += 1
 
            # Update i
            i += 1
 
        # If frequency of arr[i] is
        # greater than (N / K)
        if (cnt > (N // K)):
            print(arr[i], end = " ")
             
        i += 1
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, 2, 2, 6, 6, 6, 6, 7, 10 ]
    N = len(arr)
    K = 4
 
    NDivKWithFreq(arr, N, K)
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach 
using System;
    
class GFG{
    
// Function to print all array elements
// whose frequency is greater than N / K
static void NDivKWithFreq(int[] arr, int N,
                          int K)
{
     
    // Sort the array, arr[]
    Array.Sort(arr);
  
    // Traverse the array
    for(int i = 0; i < N;)
    {
         
        // Stores frequency of arr[i]
        int cnt = 1;
  
        // Traverse array elements which
        // is equal to arr[i]
        while ((i + 1) < N &&
               arr[i] == arr[i + 1])
        {
             
            // Update cnt
            cnt++;
  
            // Update i
            i++;
        }
  
        // If frequency of arr[i] is
        // greater than (N / K)
        if (cnt > (N / K))
        {
            Console.Write(arr[i] + " ");
        }
        i++;
    }
}
    
// Driver Code
public static void Main()
{
    int[] arr = { 1, 2, 2, 6, 6, 6, 6, 7, 10 };
    int N = arr.Length;
    int K = 4;
  
    NDivKWithFreq(arr, N, K);
}
}
 
// This code is contributed by code_hunt

Javascript

<script>
 
// JavaScript program for above approach
 
// Function to prlet all array elements
// whose frequency is greater than N / K
function NDivKWithFreq(arr, N, K)
{
    // Sort the array, arr[]
    arr.sort();
  
    // Traverse the array
    for (let i = 0; i < N;) {
  
        // Stores frequency of arr[i]
        let cnt = 1;
  
        // Traverse array elements which
        // is equal to arr[i]
        while ((i + 1) < N
               && arr[i] == arr[i + 1]) {
  
            // Update cnt
            cnt++;
  
            // Update i
            i++;
        }
  
        // If frequency of arr[i] is
        // greater than (N / K)
        if (cnt > (N / K)) {
  
            document.write(arr[i]+ " ");
        }
        i++;
    }
}
 
 
// Driver Code
 
    let arr = [1, 2, 2, 6, 6, 6, 6, 7, 10 ];
    let N = arr.length;
    let K = 4;
  
    NDivKWithFreq(arr, N, K);
           
</script>

Output:

Time Complexity: O(N * log₂N)
Auxiliary Space: O(1)

Binary Search-Based Approach: The problem can be solved using Binary Search technique. The idea is to traverse the array and count the frequency of every distinct array element by calculating the upper bound of array elements. Finally, check if the frequency of the array element is greater than N / K or not. If found to be true, then print the array element. Follow the steps below to solve the problem:

Sort the array, arr[].
Traverse the array using the variable i and find the upper_bound of arr[i] say, X and check if (x – i) is greater than N / K or not. If found to be true then print the arr[i].
Finally, update i = X.

C++

// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to+ find the upper_bound of
// an array element
int upperBound(int arr[], int N, int K)
{
 
    // Stores minimum index
    // in which K lies
    int l = 0;
 
    // Stores maximum index
    // in which K lies
    int r = N;
 
    // Calculate the upper
    // bound of K
    while (l < r) {
 
        // Stores mid element
        // of l and r
        int mid = (l + r) / 2;
 
        // If arr[mid] is less
        // than or equal to K
        if (arr[mid] <= K) {
 
            // Right subarray
            l = mid + 1;
        }
 
        else {
 
            // Left subarray
            r = mid;
        }
    }
    return l;
}
 
// Function to print all array elements
// whose frequency is greater than N / K
void NDivKWithFreq(int arr[], int N, int K)
{
 
    // Sort the array arr[]
    sort(arr, arr + N);
 
    // Stores index of
    // an array element
    int i = 0;
 
    // Traverse the array
    while (i < N) {
 
        // Stores upper bound of arr[i]
        int X = upperBound(arr, N, arr[i]);
 
        // If frequency of arr[i] is
        // greater than N / 4
        if ((X - i) > N / 4) {
 
            cout << arr[i] << " ";
        }
 
        // Update i
        i = X;
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 2, 6, 6, 6, 6, 7, 10 };
 
    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    int K = 4;
 
    // Function Call
    NDivKWithFreq(arr, N, K);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function to+ find the upper_bound of
// an array element
static int upperBound(int arr[], int N, int K)
{
 
    // Stores minimum index
    // in which K lies
    int l = 0;
 
    // Stores maximum index
    // in which K lies
    int r = N;
 
    // Calculate the upper
    // bound of K
    while (l < r)
    {
 
        // Stores mid element
        // of l and r
        int mid = (l + r) / 2;
 
        // If arr[mid] is less
        // than or equal to K
        if (arr[mid] <= K)
        {
 
            // Right subarray
            l = mid + 1;
        }
 
        else
        {
 
            // Left subarray
            r = mid;
        }
    }
    return l;
}
 
// Function to print all array elements
// whose frequency is greater than N / K
static void NDivKWithFreq(int arr[], int N, int K)
{
 
    // Sort the array arr[]
    Arrays.sort(arr);
 
    // Stores index of
    // an array element
    int i = 0;
 
    // Traverse the array
    while (i < N)
    {
 
        // Stores upper bound of arr[i]
        int X = upperBound(arr, N, arr[i]);
 
        // If frequency of arr[i] is
        // greater than N / 4
        if ((X - i) > N / 4)
        {
 
            System.out.print(arr[i] + " ");
        }
 
        // Update i
        i = X;
    }
}
 
// Driver Code
public static void main(String[] args)
{
    // Given array arr[]
    int arr[] = { 1, 2, 2, 6, 6, 6, 6, 7, 10 };
 
    // Size of array
    int N = arr.length;
    int K = 4;
 
    // Function Call
    NDivKWithFreq(arr, N, K);
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python program to implement
# the above approach
 
# Function to+ find the upper_bound of
# an array element
def upperBound(arr, N, K):
 
  # Stores minimum index
    # in which K lies
    l = 0;
 
    # Stores maximum index
    # in which K lies
    r = N;
 
    # Calculate the upper
    # bound of K
    while (l < r):
 
        # Stores mid element
        # of l and r
        mid = (l + r) // 2;
 
        # If arr[mid] is less
        # than or equal to K
        if (arr[mid] <= K):
 
            # Right subarray
            l = mid + 1;
        else:
 
            # Left subarray
            r = mid;
    return l;
 
# Function to prall array elements
# whose frequency is greater than N / K
def NDivKWithFreq(arr, N, K):
   
  # Sort the array arr
    arr.sort();
 
    # Stores index of
    # an array element
    i = 0;
 
    # Traverse the array
    while (i < N):
 
        # Stores upper bound of arr[i]
        X = upperBound(arr, N, arr[i]);
 
        # If frequency of arr[i] is
        # greater than N / 4
        if ((X - i) > N // 4):
            print(arr[i], end="");
 
        # Update i
        i = X;
 
# Driver Code
if __name__ == '__main__':
   
    # Given array arr
    arr = [1, 2, 2, 6, 6, 6, 6, 7, 10];
 
    # Size of array
    N = len(arr);
    K = 4;
 
    # Function Call
    NDivKWithFreq(arr, N, K);
 
    # This code is contributed by 29AjayKumar

C#

// C# program to implement
// the above approach
using System;
class GFG
{
 
  // Function to+ find the upper_bound of
  // an array element
  static int upperBound(int []arr, int N, int K)
  {
 
    // Stores minimum index
    // in which K lies
    int l = 0;
 
    // Stores maximum index
    // in which K lies
    int r = N;
 
    // Calculate the upper
    // bound of K
    while (l < r)
    {
 
      // Stores mid element
      // of l and r
      int mid = (l + r) / 2;
 
      // If arr[mid] is less
      // than or equal to K
      if (arr[mid] <= K)
      {
 
        // Right subarray
        l = mid + 1;
      }
 
      else
      {
 
        // Left subarray
        r = mid;
      }
    }
    return l;
  }
 
  // Function to print all array elements
  // whose frequency is greater than N / K
  static void NDivKWithFreq(int []arr, int N, int K)
  {
 
    // Sort the array arr[]
    Array.Sort(arr);
 
    // Stores index of
    // an array element
    int i = 0;
 
    // Traverse the array
    while (i < N)
    {
 
      // Stores upper bound of arr[i]
      int X = upperBound(arr, N, arr[i]);
 
      // If frequency of arr[i] is
      // greater than N / 4
      if ((X - i) > N / 4)
      {
 
        Console.Write(arr[i] + " ");
      }
 
      // Update i
      i = X;
    }
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
 
    // Given array arr[]
    int []arr = { 1, 2, 2, 6, 6, 6, 6, 7, 10 };
 
    // Size of array
    int N = arr.Length;
    int K = 4;
 
    // Function Call
    NDivKWithFreq(arr, N, K);
  }
}
 
// This code is contributed by AnkThon

Javascript

<script>
// Javascript program to implement
// the above approach
 
//function to sort the array
function sortt(number)
{
var n= number.length;
    for (i = 0; i < n; ++i)
        {
             for (j = i + 1; j < n; ++j)
            {
                 if (number[i] > number[j])
                {
                     a =  number[i];
                    number[i] = number[j];
                    number[j] = a;
                 }
             }
         }
}
 
// Function to+ find the upper_bound of
// an array element
function upperBound( arr, N, K)
{
 
    // Stores minimum index
    // in which K lies
    var l = 0;
 
    // Stores maximum index
    // in which K lies
    var r = N;
 
    // Calculate the upper
    // bound of K
    while (l < r) {
 
        // Stores mid element
        // of l and r
        var mid = parseInt((l + r) / 2);
 
        // If arr[mid] is less
        // than or equal to K
        if (arr[mid] <= K) {
 
            // Right subarray
            l = mid + 1;
        }
 
        else {
 
            // Left subarray
            r = mid;
        }
    }
    return l;
}
 
// Function to print all array elements
// whose frequency is greater than N / K
function NDivKWithFreq(arr, N, K)
{
 
    // Sort the array arr[]
    sortt(arr);
 
    // Stores index of
    // an array element
    var i = 0;
 
    // Traverse the array
    while (i < N) {
 
        // Stores upper bound of arr[i]
        var X = upperBound(arr, N, arr[i]);
 
        // If frequency of arr[i] is
        // greater than N / 4
        if ((X - i) > parseInt(N / 4)) {
 
            document.write( arr[i] + " ");
        }
 
        // Update i
        i = X;
    }
}
 
    // Given array arr[]
    var arr = [ 1, 2, 2, 6, 6, 6, 6, 7, 10 ];
 
    // Size of array
    var N = arr.length;
 
    var K = 4;
 
    // Function Call
    NDivKWithFreq(arr, N, K);
 
//This code in contributed by SoumikMondal
</script>

Output:

Time Complexity: O(N * log₂N)
Auxiliary Space: O(1)

Another Approach: Using Built-in Python functions:

Count the frequencies of every element using Counter() function.
Traverse the frequency array and print all the elements which occur at more than n/k times.

Below is the implementation:

Python3

# Python3 implementation
from collections import Counter
 
# Function to find the number of array
# elements with frequency more than n/k times
def printElements(arr, n, k):
 
    # Calculating n/k
    x = n//k
 
    # Counting frequency of every
    # element using Counter
    mp = Counter(arr)
     
    # Traverse the map and print all
    # the elements with occurrence atleast n/k times
    for it in mp:
        if mp[it] > x:
            print(it)
 
 
# Driver code
arr = [1, 2, 2, 6, 6, 6, 6, 7, 10]
 
# Size of array
n = len(arr)
k = 4
 
printElements(arr, n, k)
 
# This code is contributed by vikkycirus

Output:

Time Complexity: O(N)

Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the idea is to store the frequency of each distinct array element into a Map. Finally, traverse the map and check if its frequency is greater than (N / K) or not. If found to be true, then print the array element. Refer to this article for the discussion of this approach.

Time Complexity: O(N)
Auxiliary Space: O(N)

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