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Find an index of maximum occurring element with equal probability
• Difficulty Level : Easy
• Last Updated : 28 Feb, 2020

Given an array of integers, find the most occurring element of the array and return any one of its indexes randomly with equal probability.

Examples:

```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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to iterate through the array once and find out the maximum occurring element and its frequency n. Then we generate a random number r between 1 and n and finally return the r’th occurrence of maximum occurring element in the array.

Below are implementation of above idea –

## 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);`` ` `    ``// 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 mp = ``new` `HashMap();``    ``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 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`

Output:

```Element with maximum frequency present at index 4
```

Time complexity of above solution is O(n).
Auxiliary space used by the program is O(n).

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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