In the problem, “Special Array With X Elements Greater Than or Equal X”, we are given an array of non-negative integers. We need to find some integer X such that there are exactly X elements greater than or equal to it in the array. If there is no possible X that satisfies this condition, we need to output -1.

Table of Contents

## Example

1 2 3 4

-1

1 3 4 5

3

## Approach(Brute Force)

It is certain that if there exists a solution X, then **X will always be unique.**

**Proof:**

Consider there are two solutions X and Y such that X != Y. Lets assume X < Y. Now, number of elements greater than or equal to X should be X as we consider it as a solution. But, since Y > X, there are Y elements greater than or equal to X such that X != Y. Hence our assumption of X being a solution is wrong unless X and Y are equal.

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So, there is always a unique solution, if a solution exists. Now, it can also be concluded that if X is a solution, then the number of elements greater than/equal to it = X, which should be less than N, where N = size of the array(as the maximum number of elements possible = N).

So, if a solution X exists, then it **must lie in the range [1, N].**

We can check for every integer in the range [1, N] if the number of elements in the array that are greater than or equal to any integer in the range is equal to that integer itself. For example, consider Array = {1 , 3 , 4 , 5}. Now, 1 and 2 have 4 and 3 elements greater than or equal to them in the array respectively. Number of elements greater than/equal to 3 = 3. So, 3 is the only solution. Since we traverse the whole tree to find the number of elements greater than for every integer in the range: [1, N], this approach is slow.

### Algorithm

- Run a loop from i = 1 to i = N to check for solution:
- Count the number of elements greater than or equal to ‘
*i*‘ in the array- If the count is equal to ‘
*i*‘, return i

- If the count is equal to ‘

- Count the number of elements greater than or equal to ‘
- Return
**-1**;

### Implementation of Special Array With X Elements Greater Than or Equal X Leetcode Solution

#### C++ Program

#include <bits/stdc++.h> using namespace std; int specialArray(vector <int> &a) { int n = a.size() , counter = 0; for(int i = 1 ; i <= n ; i++) { counter = 0; for(int j = 0 ; j < n ; j++) if(a[j] >= i) counter++; if(counter == i) return i; } return -1; } int main() { vector <int> a = {1 , 3 , 4 , 5}; cout << specialArray(a) << '\n'; }

#### Java Program

class special_array { public static void main(String args[]) { int[] a = {1 , 3 , 4 , 5}; System.out.println(specialArray(a)); } static int specialArray(int[] a) { int n = a.length , counter = 0; for(int i = 1 ; i <= n ; i++) { counter = 0; for(int j = 0 ; j < n ; j++) if(a[j] >= i) counter++; if(counter == i) return i; } return -1; } }

3

### Complexity Analysis of Special Array With X Elements Greater Than or Equal X Leetcode Solution

#### Time Complexity

**O(N * N)**, N = Size of the array. In the worst case, when X = N is the solution, we make O(N * N) comparisons.

#### Space complexity

**O(1)**. Only constant memory space is used.

## Approach(Optimal)

We can store the number of occurrences of all elements in a table, using an array. Note that for any element with a value greater than N, we can count it as an occurrence of an element with value N because the maximum value of the solution can be N.

Now, the frequency of elements greater than or equal to X = frequency of X + frequency of all elements greater than X. In order to find this for every element in the range [1, N], we need to have a suffix frequency array.

So, for every integer in the array, we need to set **frequency[i] = frequency[i] + frequency[i + 1]**, for every ‘*i*‘ in range [N – 1 , 1], which will create a suffix frequency array, storing number of elements greater than or equal to **‘ i‘** as

**frequency[i].**

Now, because of the lookup table, we can check a solution for any integer in the range [1, N] in **O(1)** time. This is a case where we trade off memory to improve on time. Since the table is of size N, we have no worries of memory limits as well.

### Algorithm

- Initialize a
**frequency**array of size N, having each value as**zero** - For every element
*i*in the array:- If
*i*> N, increment frequency[N] - Increment frequency[i]

- If
- Create a suffix sum array from
**frequency**as:- For every element ranging from
*i*=**N – 1**to*i*=**set**frequency[*i*] = frequency[*i*] + frequency[*i*+ 1]

- For every element ranging from
- Run a loop from
*i*= 1 to*i*= N- If
*i*== frequency[*i*], return*i*

- If
- return -1

### Implementation of Special Array With X Elements Greater Than or Equal X Leetcode Solution

#### C++ Program

#include <bits/stdc++.h> using namespace std; int specialArray(vector<int>& a) { int n = a.size(); vector <int> frequency(n + 1 , 0); for(int i = 0 ; i < n ; i++) if(a[i] > n) frequency[n]++; else frequency[a[i]]++; for(int i = n - 1 ; i >= 0 ; i--) frequency[i] += frequency[i + 1]; for(int i = 0 ; i <= n ; i++) if(frequency[i] == i) return i; return -1; } int main() { vector <int> a = {1 , 3 , 4 , 5}; cout << specialArray(a) << '\n'; }

#### Java Program

class special_array { public static void main(String args[]) { int[] a = {1 , 3 , 4 , 5}; System.out.println(specialArray(a)); } static int specialArray(int[] a) { int n = a.length; int[] frequency = new int[n + 1]; for(int i = 0 ; i < n ; i++) frequency[i] = 0; for(int i = 0 ; i < n ; i++) if(a[i] > n) frequency[n]++; else frequency[a[i]]++; for(int i = n - 1 ; i >= 0 ; i--) frequency[i] += frequency[i + 1]; for(int i = 0 ; i <= n ; i++) if(frequency[i] == i) return i; return -1; } }

3

### Complexity Analysis of Special Array With X Elements Greater Than or Equal X Leetcode Solution

#### Time Complexity

**O(N)**, N = Size of the array. We can check for any integer in O(1) time, so in the worst case, we use O(N) time.

#### Space complexity

**O(N)**. Linear memory space is used for storing frequencies.