Given an array of size n and a number k, fin all elements that appear more than ” n/k ” times.

3 comments on “Given an array of size n and a number k, fin all elements that appear more than ” n/k ” times.”

• 
  

  ankithchavan822

  def appn(ar,n,k):
  t = n//k
  s = set(ar)
  f=[]
  for i in s:
  if ar.count(i) > t:
  f.append(i)
  return f

  n=int(input(‘Enter n:’))
  print(‘Enter the ar elements:’)
  ar=list(map(int,input().split()))
  k=int(input(‘Enter k:’))

  print(‘List of elements that appear more than n/k times are:’)
  appn(ar,n,k)

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• 
  

  Manisha

  # Python3 code to find elements whose
  # frequency is more than n/k

  def morethanNbyK(arr, n, k):
  x = n // k

  # unordered_map initialization
  freq = {}

  for i in range(n):
  if arr[i] in freq:
  freq[arr[i]] += 1
  else:
  freq[arr[i]] = 1

  # Traversing the map
  for i in freq:

  # Checking if value of a key-value pair
  # is greater than x (where x=n/k)
  if (freq[i] > x):

  # Print the key of whose value
  # is greater than x
  print(i)

  # Driver code
  if __name__ == ‘__main__’:
  arr = [1, 1, 2, 2, 3, 5, 4, 2, 2, 3, 1, 1, 1]
  n = len(arr)
  k = 4

  morethanNbyK(arr, n, k)

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• 
  

  amarnadhamarnadh206

  a=[1,2,3,3,2,1,3,2,7,7]
  k=5
  d=dict()
  for i in a:
  if i in d:
  d[i]+=1
  else:
  d[i]=1
  for key,value in d.items():
  if value>(len(a)/k):
  print(key)

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