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Problem 1608 Special Array With X Elements Greater Than or Equal X

Problem Statement

Link - Problem 1608

Question

You are given an array nums of non-negative integers. nums is considered special if there exists a number x such that there are exactly x numbers in nums that are greater than or equal to x.

Notice that x does not have to be an element in nums.

Return x if the array is special, otherwise, return -1. It can be proven that if nums is special, the value for x is unique.

Example 1:

Input: nums = [3,5]
Output: 2
Explanation: There are 2 values (3 and 5) that are greater than or equal to 2.

Example 2:

Input: nums = [0,0]
Output: -1
Explanation: No numbers fit the criteria for x.
If x = 0, there should be 0 numbers >= x, but there are 2.
If x = 1, there should be 1 number >= x, but there are 0.
If x = 2, there should be 2 numbers >= x, but there are 0.
x cannot be greater since there are only 2 numbers in nums.

Example 3:

Input: nums = [0,4,3,0,4]
Output: 3
Explanation: There are 3 values that are greater than or equal to 3.

Constraints:

• 1 <= nums.length <= 100
• 0 <= nums[i] <= 1000

Solution

class Solution {
public:
    int specialArray(vector<int>& nums) {
        int bucket[1001] = {0};
        for (int num : nums) {
            bucket[num]++;
        }
        int total = nums.size();
        for (int i = 0; i < 1001; i++) {
            if (i == total) {
                return i;
            }
            total -= bucket[i];
        }
        return -1;
    }
};

Complexity Analysis

| Algorithm       | Time Complexity | Space Complexity |
| --------------- | --------------- | ---------------- |
| Bucket Counting | O(n)            | O(1)             |

Explanation

Intial Thoughts

To approach this problem, we should first understand what a special array is. It’s an array where there exists a number x such that there are exactly x numbers in the array that are greater than or equal to x. We can think of it as finding a balance point where the number of elements greater than or equal to x is equal to x itself. We should consider all possible values of x and check if the condition is met. We can use a systematic approach to iterate through all possible values of x and count the numbers greater than or equal to x.

Intuitive Analysis

Intuitively, we can solve this problem by iterating through all possible values of x and checking if the condition is met. We can imagine a bucket system where each bucket represents a value of x, and we count the numbers greater than or equal to x. We can start from the smallest possible value of x, which is 0, and increment it until we find a match or reach the maximum possible value. We can also consider the fact that x does not have to be an element in the array, so we need to check all possible values, not just the ones present in the array. By systematically checking all possible values of x, we can find the special number if it exists.

1. Intuition

• The problem requires finding a special number x such that there are exactly x numbers in the array nums that are greater than or equal to x.
• To approach this problem, we need to think about the possible range of x and how to efficiently check each possible value.
• The key insight is that x must be within the range of the array length, as there cannot be more numbers greater than or equal to x than the total length of the array.
• We can iterate through the possible values of x and check if the condition is met by counting the numbers greater than or equal to x.
• This solution works because it systematically checks all possible values of x and returns the first one that meets the condition.
• The algorithmic choice of using a bucket array to store the frequency of each number in nums allows for efficient counting of numbers greater than or equal to x.
• The time complexity of this solution is O(n + m), where n is the length of nums and m is the maximum possible value of x.
• The space complexity is O(m), which is used to store the bucket array.

2. Implementation

• We start by initializing a bucket array bucket of size 1001 with all elements set to 0, where bucket[i] represents the frequency of i in nums.
• We then iterate through nums and update the frequency of each number in the bucket array using bucket[num]++.
• Next, we initialize a variable total to the length of nums and iterate through the possible values of x from 0 to 1000.
• For each x, we check if the condition is met by comparing x with total, and if it is, we return x.
• If the condition is not met, we update total by subtracting the frequency of x from it using total -= bucket[i].
• If no x is found that meets the condition, we return -1.
• The use of bucket array allows us to efficiently count the numbers greater than or equal to x in O(1) time.
• The loop iterates through the possible values of x and checks the condition, resulting in a time complexity of O(n + m).

Complexity Analysis

Time Complexity:

• The time complexity is driven by two main loops: the first loop for (int num : nums) iterates over each element in the input array, resulting in a time complexity of O(n), where n is the number of elements in the array.
• The second loop for (int i = 0; i < 1001; i++) runs a constant number of times (1001 times), resulting in a time complexity of O(1), which is dominated by the O(n) complexity of the first loop.
• Using the rule of sum for Big O notation, the overall time complexity is O(n) + O(1) = O(n), justifying the Big O classification as linear time complexity.

Space Complexity:

• The space complexity is analyzed by considering the memory usage of the algorithm, which includes the input array nums and the bucket array of size 1001.
• The bucket array has a fixed size of 1001, regardless of the size of the input array, resulting in a space complexity of O(1), as the memory usage does not grow with the size of the input array.
• The input array nums is not included in the space complexity analysis because it is part of the input, not the auxiliary space used by the algorithm, thus justifying the space complexity as constant.

Footnote

This question is rated as Easy difficulty.

Hints

Count the number of elements greater than or equal to x for each x in the range [0, nums.length].

If for any x, the condition satisfies, return that x. Otherwise, there is no answer.