Atsižvelgiant į sveikųjų skaičių masyvą, suraskite dažniausiai pasitaikantį masyvo elementą ir atsitiktinai su tokia pačia tikimybe grąžinkite bet kurį iš jo indeksų.
Pavyzdžiai:
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.
Idėja yra vieną kartą perskaityti masyvą ir sužinoti didžiausią pasitaikantį elementą ir jo dažnį n. Tada sugeneruojame atsitiktinį skaičių r tarp 1 ir n ir galiausiai grąžiname r'-ąjį maksimalaus elemento pasireiškimą masyve.
Žemiau yra aukščiau pateiktos idėjos įgyvendinimas -
// 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>
Išvestis:
Element with maximum frequency present at index 4
Laiko sudėtingumas aukščiau pateikto tirpalo yra O(n).
Pagalbinė erdvė programos naudojamas yra O(n).