# Largest number in the Array having frequency same as value

Given an array arr containing N integers, the task is to find the largest number in the array whose frequency is equal to its value. If no such number exists, then print -1.

Examples:

Input: arr = [3, 2, 5, 2, 4, 5]
Output: 2
Explanation:
In this given array frequency of 2 is 2, whereas the frequency of the remaining numbers doesnt match with itself. So the answer is 2.

Input: arr = [3, 3, 3, 4, 4, 4, 4]
Output: 4
Explanation:
In this given array frequency of 3 is 3 and 4 is 4 but largest number is 4. So the answer is 4.

Input: arr = [1, 1, 1, 2, 3, 3]
Output: -1
Explanation:
There is no such number in the given array whose frequency is equal to itself. Thus the output is -1.

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

Simple Approach:

1. Create a new array to keep the count of the occurrences in the given array.
2. Traverse the new array in reverse order.
3. Return the first number whose count is equal to itself.

Below is the implementation of the above approach:

## C++

 `// C++ solution to the above problem ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the largest number ` `// whose frequency is equal to itself. ` `int` `findLargestNumber(vector<``int``>& arr) ` `{ ` ` `  `    ``// Find the maximum element in the array ` `    ``int` `k = *max_element(arr.begin(), ` `                         ``arr.end()); ` `    ``int` `m[k] = {}; ` ` `  `    ``for` `(``auto` `n : arr) ` `        ``++m[n]; ` ` `  `    ``for` `(``auto` `n = arr.size(); n > 0; --n) { ` `        ``if` `(n == m[n]) ` `            ``return` `n; ` `    ``} ` `    ``return` `-1; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``vector<``int``> arr = { 3, 2, 5, 2, 4, 5 }; ` ` `  `    ``cout << findLargestNumber(arr) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java solution to the above problem ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Function to find the largest number ` `// whose frequency is equal to itself. ` `static` `int` `findLargestNumber(``int``[] arr) ` `{ ` `     `  `    ``// Find the maximum element in the array ` `    ``int` `k = Arrays.stream(arr).max().getAsInt(); ` `    ``int` `[]m = ``new` `int``[k + ``1``]; ` ` `  `    ``for``(``int` `n : arr) ` `       ``++m[n]; ` ` `  `    ``for``(``int` `n = arr.length - ``1``; n > ``0``; --n)  ` `    ``{ ` `       ``if` `(n == m[n]) ` `           ``return` `n; ` `    ``} ` `    ``return` `-``1``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int``[] arr = { ``3``, ``2``, ``5``, ``2``, ``4``, ``5` `}; ` ` `  `    ``System.out.print(findLargestNumber(arr) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by amal kumar choubey `

## C#

 `// C# solution to the above problem ` `using` `System; ` `using` `System.Linq; ` ` `  `class` `GFG{ ` ` `  `// Function to find the largest number ` `// whose frequency is equal to itself. ` `static` `int` `findLargestNumber(``int``[] arr) ` `{ ` `     `  `    ``// Find the maximum element in the array ` `    ``int` `k = arr.Max(); ` `    ``int` `[]m = ``new` `int``[k + 1]; ` ` `  `    ``foreach``(``int` `n ``in` `arr) ` `        ``++m[n]; ` ` `  `    ``for``(``int` `n = arr.Length - 1; n > 0; --n)  ` `    ``{ ` `       ``if` `(n == m[n]) ` `           ``return` `n; ` `    ``} ` `    ``return` `-1; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int``[] arr = { 3, 2, 5, 2, 4, 5 }; ` ` `  `    ``Console.Write(findLargestNumber(arr) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by amal kumar choubey `

Output:

```2
```

Time Complexity: O(N)
Auxiliary Space Complexity: O(N)

Another Approach:
Note: This approach is valid only when the numbers in the given array are less than 65536 i.e. 216.

1. Here, use the input array to store the count.
2. Since the values are limited, simply use the first half(first 16 bits) of the integer for keeping the count by adding 65536.
3. Use the right shift operator(right shift by 16 bits) while traversing the array in reverse order and return the first number whose count is equal to itself.

Below is the implementation of the above approach:

 `// C++ code for the above problem ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the largest number ` `// whose frequency is equal to itself. ` `int` `findLargestNumber(vector<``int``>& arr) ` `{ ` `    ``for` `(``auto` `n : arr) { ` `        ``n &= 0xFFFF; ` `        ``if` `(n <= arr.size()) { ` `            ``// Adding 65536 to keep the ` `            ``// count of the current number ` `            ``arr[n - 1] += 0x10000; ` `        ``} ` `    ``} ` ` `  `    ``for` `(``auto` `i = arr.size(); i > 0; --i) { ` `        ``// right shifting by 16 bits ` `        ``// to find the count of the ` `        ``// number i ` `        ``if` `((arr[i - 1] >> 16) == i) ` `            ``return` `i; ` `    ``} ` `    ``return` `-1; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``vector<``int``> arr ` `        ``= { 3, 2, 5, 5, 2, 4, 5 }; ` ` `  `    ``cout << findLargestNumber(arr) ` `         ``<< endl; ` `    ``return` `0; ` `} `

Output:

```2
```

Time Complexity: O(N)
Auxiliary Space Complexity: O(1)

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