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Zoek een index van het maximaal voorkomende element met gelijke waarschijnlijkheid

Gegeven een array van gehele getallen, zoek het meest voorkomende element van de array en retourneer een van de indexen willekeurig met gelijke waarschijnlijkheid.
Voorbeelden: 
 

  Input:    arr[] = [-1 4 9 7 7 2 7 3 0 9 6 5 7 8 9]   Output:    Element with maximum frequency present at index 6 OR Element with maximum frequency present at Index 3 OR Element with maximum frequency present at index 4 OR Element with maximum frequency present at index 12 All outputs above have equal probability.


 




Het idee is om de array één keer te doorlopen en het maximaal voorkomende element en de frequentie n ervan te achterhalen. Vervolgens genereren we een willekeurig getal r tussen 1 en n en retourneren uiteindelijk het r'de voorkomen van het maximaal voorkomende element in de array.
Hieronder vindt u de implementatie van het bovenstaande idee - 
 

C++
// C++ program to return index of most occurring element // of the array randomly with equal probability #include    #include  #include  using namespace std; // Function to return index of most occurring element // of the array randomly with equal probability void findRandomIndexOfMax(int arr[] int n) {  // freq store frequency of each element in the array  unordered_map<int int> freq;  for (int i = 0; i < n; i++)  freq[arr[i]] += 1;  int max_element; // stores max occurring element  // stores count of max occurring element  int max_so_far = INT_MIN;  // traverse each pair in map and find maximum  // occurring element and its frequency  for (pair<int int> p : freq)  {  if (p.second > max_so_far)  {  max_so_far = p.second;  max_element = p.first;  }  }  // generate a random number between [1 max_so_far]  int r = (rand() % max_so_far) + 1;  // traverse array again and return index of rth  // occurrence of max element  for (int i = 0 count = 0; i < n; i++)  {  if (arr[i] == max_element)  count++;  // print index of rth occurrence of max element  if (count == r)  {  cout << 'Element with maximum frequency present '  'at index ' << i << endl;  break;  }  } } // Driver code int main() {  // input array  int arr[] = { -1 4 9 7 7 2 7 3 0 9 6 5  7 8 9 };  int n = sizeof(arr) / sizeof(arr[0]);  // randomize seed  srand(time(NULL));  findRandomIndexOfMax(arr n);  return 0; } 
Java
// Java program to return index of most occurring element // of the array randomly with equal probability import java.util.*; class GFG { // Function to return index of most occurring element // of the array randomly with equal probability static void findRandomIndexOfMax(int arr[] int n) {  // freq store frequency of each element in the array  HashMap<Integer Integer> mp = new HashMap<Integer Integer>();  for (int i = 0; i < n; i++)  if(mp.containsKey(arr[i]))  {  mp.put(arr[i] mp.get(arr[i]) + 1);  }  else  {  mp.put(arr[i] 1);  }  int max_element = Integer.MIN_VALUE; // stores max occurring element  // stores count of max occurring element  int max_so_far = Integer.MIN_VALUE;  // traverse each pair in map and find maximum  // occurring element and its frequency  for (Map.Entry<Integer Integer> p : mp.entrySet())   {  if (p.getValue() > max_so_far)  {  max_so_far = p.getValue();  max_element = p.getKey();  }  }    // generate a random number between [1 max_so_far]  int r = (int) ((new Random().nextInt(max_so_far) % max_so_far) + 1);  // traverse array again and return index of rth  // occurrence of max element  for (int i = 0 count = 0; i < n; i++)  {  if (arr[i] == max_element)  count++;  // print index of rth occurrence of max element  if (count == r)  {  System.out.print('Element with maximum frequency present '  +'at index ' + i +'n');  break;  }  } } // Driver code public static void main(String[] args) {  // input array  int arr[] = { -1 4 9 7 7 2 7 3 0 9 6 5  7 8 9 };  int n = arr.length;  findRandomIndexOfMax(arr n); } } // This code is contributed by Rajput-Ji 
Python3
# Python3 program to return index of most occurring element # of the array randomly with equal probability import random # Function to return index of most occurring element # of the array randomly with equal probability def findRandomIndexOfMax(arr n): # freq store frequency of each element in the array mp = dict() for i in range(n) : if(arr[i] in mp): mp[arr[i]] = mp[arr[i]] + 1 else: mp[arr[i]] = 1 max_element = -323567 # stores max occurring element # stores count of max occurring element max_so_far = -323567 # traverse each pair in map and find maximum # occurring element and its frequency for p in mp : if (mp[p] > max_so_far): max_so_far = mp[p] max_element = p # generate a random number between [1 max_so_far] r = int( ((random.randrange(1 max_so_far 2) % max_so_far) + 1)) i = 0 count = 0 # traverse array again and return index of rth # occurrence of max element while ( i < n ): if (arr[i] == max_element): count = count + 1 # Print index of rth occurrence of max element if (count == r): print('Element with maximum frequency present at index '  i ) break i = i + 1 # Driver code # input array arr = [-1 4 9 7 7 2 7 3 0 9 6 5 7 8 9] n = len(arr) findRandomIndexOfMax(arr n) # This code is contributed by Arnab Kundu 
C#
using System; using System.Collections.Generic; class GFG { // Function to return index of most occurring element // of the array randomly with equal probability static void findRandomIndexOfMax(int[] arr int n) {  // freq store frequency of each element in the array  Dictionary<int int> mp = new Dictionary<int int>();  for (int i = 0; i < n; i++)  {  if (mp.ContainsKey(arr[i]))  {  mp[arr[i]]++;  }  else  {  mp[arr[i]] = 1;  }  }  int max_element = int.MinValue; // stores max occurring element  // stores count of max occurring element  int max_so_far = int.MinValue;  // traverse each pair in map and find maximum  // occurring element and its frequency  foreach (KeyValuePair<int int> p in mp)  {  if (p.Value > max_so_far)  {  max_so_far = p.Value;  max_element = p.Key;  }  }  // generate a random number between [1 max_so_far]  Random rand = new Random();  int r = rand.Next(max_so_far) + 1;  // traverse array again and return index of rth  // occurrence of max element  for (int i = 0 count = 0; i < n; i++)  {  if (arr[i] == max_element)  count++;  // print index of rth occurrence of max element  if (count == r)  {  Console.WriteLine('Element with maximum frequency present '  + 'at index ' + i + 'n');  break;  }  } } // Driver code public static void Main() {  // input array  int[] arr = { -1 4 9 7 7 2 7 3 0 9 6 5  7 8 9 };  int n = arr.Length;  findRandomIndexOfMax(arr n); } } 
JavaScript
<script> // Javascript program to return index of most occurring element // of the array randomly with equal probability // Function to return index of most occurring element // of the array randomly with equal probability function findRandomIndexOfMax(arrn) {  // freq store frequency of each element in the array  let mp = new Map();  for (let i = 0; i < n; i++)  if(mp.has(arr[i]))  {  mp.set(arr[i] mp.get(arr[i]) + 1);  }  else  {  mp.set(arr[i] 1);  }    let max_element = Number.MIN_VALUE; // stores max occurring element    // stores count of max occurring element  let max_so_far = Number.MIN_VALUE;    // traverse each pair in map and find maximum  // occurring element and its frequency  for (let [key value] of mp.entries())   {  if (value > max_so_far)  {  max_so_far = value;  max_element = key;  }  }    // generate a random number between [1 max_so_far]    let r = Math.floor(((Math.random() * max_so_far)% max_so_far)+ 1)    // traverse array again and return index of rth  // occurrence of max element  for (let i = 0 count = 0; i < n; i++)  {  if (arr[i] == max_element)  count++;    // print index of rth occurrence of max element  if (count == r)  {  document.write('Element with maximum frequency present '  +'at index ' + i +'  
'
); break; } } } // Driver code let arr=[-1 4 9 7 7 2 7 3 0 9 6 5 7 8 9 ]; let n = arr.length; findRandomIndexOfMax(arr n); // This code is contributed by avanitrachhadiya2155 </script>

Uitgang:  
 

Element with maximum frequency present at index 4


Tijdcomplexiteit van bovenstaande oplossing is O(n). 
Hulpruimte gebruikt door het programma is O(n).