# Find an index of maximum occurring element with equal probability

• Difficulty Level : Easy
• Last Updated : 06 Jul, 2021

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.```

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`

## Javascript

 ``

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).
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