# Program to find the number of persons wearing white hat

There are N persons in a room, each of them wearing a hat which is either black or white. Every person counts the number of other persons wearing the white hat. Given the number of counts of each person, the task is to find the number of persons wearing white hats, or print -1 if the given counts don’t correspond to a valid situation.

Examples:

```Input : arr[] = {2, 1, 1}.
Output : 2
First person sees two white hats. Second
and third persons see one white hat. The
first person must be wearing a black hat
and other two must be wearing a white hat.

Input : arr[] = {2, 2, 2}
Output : 3
All are wearing white hats.

Input : arr[] = {10, 10}
Output : -1
There are only two persons, count can't be 10.
```

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

There are only two kinds of persons. If each person counts correctly (valid situation), then the count value of each person wearing white hat is same. And also, the count value of each person wearing black hat is same. So there will be only one or two types of value in the array.
Let the number of white hats be i, 0 <= i <= N-1.
Now observe for each person wearing the white hat, the count value will be i – 1. So there will be i persons whose count will be i-1.
Also the number of persons wearing the black hats will be, N – i and their given count value will be i.
An interesting case is with zero white hats. If all values are 0, then everybody is wearing a black hat. Otherwise there can be at most one zero for the case when there is single person wearing a white hat. In case of one zero, all other entries must be 1.

Algorithm for solving this problem:

```1. Count the frequency of each element of the array.
2. Since there are one or two types, say x and y.
a) If the number of x's equal to x + 1 and number
of y's equal to n - y. The Number of hats equal
to y or x + 1.
b) Otherwise print -1. ```

Explained example:

```Suppose, N = 5, the number of white hats can be range
from 0 to 4.
For white hats = 1, array will be {0, 1, 1, 1, 1}.
Number of 0's = 0 + 1 = 1.
Number of 1's = 5 - 1 = 4.

For white hats = 2, array will be {1, 1, 2, 2, 2}.
Number of 1's = 1 + 1 = 2.
Number of 2's = 5 - 3 = 2.

For white hats = 3, array will be {2, 2, 2, 3, 3}.
Number of 2's = 2 + 1 = 3.
Number of 3's = 5 - 3 = 2.

For white hats = 5, array will be {4, 4, 4, 4, 4}.
Number of 4's = 4 + 1 = 5.
Number of 5's = 5 - 5 = 0. ```

Below is the implementation of this approach:

## CPP

 `// C++ program to count number of white hats ` `#include ` `using` `namespace` `std; ` ` `  `// Given counts of White hats seen by n people, ` `// return count of white hats. ` `int` `numOfWhiteHats(``int` `arr[], ``int` `n) ` `{ ` `    ``// Counting frequencies of all values in given ` `    ``// array ` `    ``int` `freq[n+1]; ` `    ``memset``(freq, 0, ``sizeof``(freq)); ` `    ``for` `(``int` `i=0; i= n) ` `            ``return` `-1; ` `        ``freq[arr[i]]++; ` `    ``} ` ` `  `    ``// Counting number of different frequencies ` `    ``int` `diffFreq = 0; ` `    ``for` `(``int` `i = n-1; i >= 0; i--) ` `        ``if` `(freq[i]) ` `            ``diffFreq++; ` ` `  `    ``// Cases where all the persons wearing white hat. ` `    ``if` `(diffFreq == 1 && freq[n-1] == n) ` `        ``return` `n; ` ` `  `    ``// Case where no one wearing white hat. ` `    ``if` `(diffFreq == 1 && freq[0] == n) ` `        ``return` `0; ` ` `  `    ``// Else : number of distinct frequency must be 2. ` `    ``if` `(diffFreq != 2) ` `        ``return` `-1; ` ` `  `    ``// Finding the last frequency with non zero value. ` `    ``// Note that we traverse from right side. ` `    ``int` `k; ` `    ``for` `(k = n-1; k >= 1; k--) ` `        ``if` `(freq[k]) ` `            ``break``; ` ` `  `    ``// Checking number of k's must be n - k. ` `    ``// And number of (k-1)'s must be k. ` `    ``if` `(freq[k-1] == k && freq[k] + k == n) ` `        ``return` `freq[k-1]; ` `    ``else` `        ``return` `-1; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = {2, 2, 2, 3, 3}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]); ` `    ``cout << numOfWhiteHats(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to count number of white hats ` `import` `java.util.Arrays; ` ` `  `class` `GFG { ` `     `  `    ``// Given counts of White hats seen by n  ` `    ``// people, return count of white hats. ` `    ``static` `int` `numOfWhiteHats(``int` `arr[], ``int` `n) ` `    ``{ ` `         `  `        ``// Counting frequencies of all values  ` `        ``// in given array ` `        ``int` `freq[] = ``new` `int``[n + ``1``]; ` `        ``Arrays.fill(freq, ``0``); ` `         `  `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `             `  `            ``// Count of White hats cannot be  ` `            ``// more than n for n persons. ` `            ``if` `(arr[i] >= n) ` `                ``return` `-``1``; ` `                 `  `            ``freq[arr[i]]++; ` `        ``} ` ` `  `        ``// Counting number of different  ` `        ``// frequencies ` `        ``int` `diffFreq = ``0``; ` `         `  `        ``for` `(``int` `i = n - ``1``; i >= ``0``; i--) ` `            ``if` `(freq[i] > ``0``) ` `                ``diffFreq++; ` ` `  `        ``// Cases where all the persons wearing  ` `        ``// white hat. ` `        ``if` `(diffFreq == ``1` `&& freq[n - ``1``] == n) ` `            ``return` `n; ` ` `  `        ``// Case where no one wearing white hat. ` `        ``if` `(diffFreq == ``1` `&& freq[``0``] == n) ` `            ``return` `0``; ` ` `  `        ``// Else : number of distinct frequency  ` `        ``// must be 2. ` `        ``if` `(diffFreq != ``2``) ` `            ``return` `-``1``; ` ` `  `        ``// Finding the last frequency with non  ` `        ``// zero value. ` `        ``// Note that we traverse from right side. ` `        ``int` `k; ` `         `  `        ``for` `(k = n - ``1``; k >= ``1``; k--) ` `            ``if` `(freq[k] > ``0``) ` `                ``break``; ` ` `  `        ``// Checking number of k's must be n - k. ` `        ``// And number of (k-1)'s must be k. ` `        ``if` `(freq[k - ``1``] == k && freq[k] + k == n) ` `            ``return` `freq[k - ``1``]; ` `        ``else` `            ``return` `-``1``; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``2``, ``2``, ``2``, ``3``, ``3` `}; ` `        ``int` `n = arr.length; ` `        ``System.out.print(numOfWhiteHats(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# python program to count ` `# number of white hats ` ` `  `def` `numOfWhiteHats(arr, n): ` ` `  `    ``# Counting frequencies of ` `    ``# all values in given ` `    ``# array ` `    ``freq``=``[``0` `for` `i ``in` `range``(n ``+` `1` `+` `1``)] ` `    ``for` `i ``in` `range``(n): ` `     `  `        ``# Count of White hats ` `        ``# cannot be more than ` `        ``# n for n persons. ` `        ``if` `(arr[i] >``=` `n): ` `            ``return` `-``1` `        ``freq[arr[i]]``+``=``1` `     `  `  `  `    ``# Counting number of ` `    ``# different frequencies ` `    ``diffFreq ``=` `0` `    ``for` `i ``in` `range``(n``-``1``,``-``1``,``-``1``): ` `        ``if` `(freq[i]): ` `            ``diffFreq``+``=``1` `  `  `    ``# Cases where all the ` `    ``# persons wearing white hat. ` `    ``if` `(diffFreq ``=``=` `1` `and` `freq[n``-``1``] ``=``=` `n): ` `        ``return` `n ` `  `  `    ``# Case where no one ` `    ``# wearing white hat. ` `    ``if` `(diffFreq ``=``=` `1` `and` `freq[``0``] ``=``=` `n): ` `        ``return` `0` `  `  `    ``# Else : number of distinct ` `    ``# frequency must be 2. ` `    ``if` `(diffFreq !``=` `2``): ` `        ``return` `-``1` `  `  `    ``# Finding the last frequency ` `    ``# with non zero value. ` `    ``# Note that we traverse ` `    ``# from right side. ` `    ``for` `k ``in` `range``(n ``-` `1``, ``0``, ``-``1``): ` `        ``if` `(freq[k]): ` `            ``break` `  `  `    ``# Checking number of k's ` `    ``# must be n - k. ` `    ``# And number of (k-1)'s ` `    ``# must be k. ` `    ``if` `(freq[k``-``1``] ``=``=` `k ``and` `freq[k] ``+` `k ``=``=` `n): ` `        ``return` `freq[k``-``1``] ` `    ``else``: ` `        ``return` `-``1` `  `  `# Driver code ` ` `  `arr``=` `[``2``, ``2``, ``2``, ``3``, ``3``] ` `n``=` `len``(arr) ` `print``(numOfWhiteHats(arr, n)) ` ` `  `# This code is contributed ` `# by Anant Agarwal. `

## C#

 `// C# program to count number of white hats ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Given counts of White hats seen by n  ` `    ``// people, return count of white hats. ` `    ``static` `int` `numOfWhiteHats(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``// Counting frequencies of all values  ` `        ``// in given array ` `        ``int` `[]freq = ``new` `int``[n + 1]; ` `        ``//Arrays.fill(freq, 0); ` `         `  `        ``for` `(``int` `i = 0; i < n; i++) { ` `             `  `            ``// Count of White hats cannot be  ` `            ``// more than n for n persons. ` `            ``if` `(arr[i] >= n) ` `                ``return` `-1; ` `                 `  `            ``freq[arr[i]]++; ` `        ``} ` ` `  `        ``// Counting number of different  ` `        ``// frequencies ` `        ``int` `diffFreq = 0; ` `         `  `        ``for` `(``int` `i = n - 1; i >= 0; i--) ` `            ``if` `(freq[i] > 0) ` `                ``diffFreq++; ` ` `  `        ``// Cases where all the persons wearing  ` `        ``// white hat. ` `        ``if` `(diffFreq == 1 && freq[n - 1] == n) ` `            ``return` `n; ` ` `  `        ``// Case where no one wearing white hat. ` `        ``if` `(diffFreq == 1 && freq[0] == n) ` `            ``return` `0; ` ` `  `        ``// Else : number of distinct frequency  ` `        ``// must be 2. ` `        ``if` `(diffFreq != 2) ` `            ``return` `-1; ` ` `  `        ``// Finding the last frequency with non  ` `        ``// zero value. ` `        ``// Note that we traverse from right side. ` `        ``int` `k; ` `         `  `        ``for` `(k = n - 1; k >= 1; k--) ` `            ``if` `(freq[k] > 0) ` `                ``break``; ` ` `  `        ``// Checking number of k's must be n - k. ` `        ``// And number of (k-1)'s must be k. ` `        ``if` `(freq[k - 1] == k && freq[k] + k == n) ` `            ``return` `freq[k - 1]; ` `        ``else` `            ``return` `-1; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]arr = {2, 2, 2, 3, 3}; ` `        ``int` `n = arr.Length; ` `        ``Console.WriteLine(numOfWhiteHats(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

Output:

```3
```

This article is contributed by Anuj Chauhan(anuj0503). 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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Improved By : vt_m, shubham_singh

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