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Знайдіть індекс максимального елемента, що зустрічається з рівною ймовірністю

Дано масив цілих чисел, знайдіть елемент масиву, який найчастіше зустрічається, і поверніть будь-який з його індексів випадковим чином з рівною ймовірністю.
приклади: 
 

  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.


 




Ідея полягає в тому, щоб один раз переглянути масив і знайти максимальний елемент, що зустрічається, і його частоту n. Потім ми генеруємо випадкове число r між 1 і n і, нарешті, повертаємо r-е входження максимального елемента в масиві.
Нижче наведено реалізацію вищезазначеної ідеї – 
 

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>

Вихід:  
 

Element with maximum frequency present at index 4


Часова складність вищезгаданого рішення дорівнює O(n). 
Допоміжне приміщення використовується програмою – це O(n).