Special Array With X Elements Greater Than or Equal X
EasyPrompt
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 <= 1000 <= nums[i] <= 1000
Approaches
3 approaches with complexity analysis and trade-offs.
The most straightforward approach is to test every possible value for x and see if it meets the problem's criteria. The problem states that x is the number of elements greater than or equal to x. Since there are n elements in the array, the count of elements satisfying the condition can be at most n. Therefore, the value of x cannot be greater than n. This means we only need to check integers for x in the range [0, n]. For each candidate x, we can simply iterate through the entire nums array and count how many numbers are greater than or equal to x. If this count matches x, we've found our answer.
Algorithm
- Let
nbe the length of thenumsarray. - Iterate through all possible values of
xfrom0ton. - For each
x, we need to check if it satisfies the condition. To do this, we count the number of elements innumsthat are greater than or equal tox. - Initialize a counter,
count, to 0. - Iterate through the
numsarray. For each elementnum, ifnum >= x, incrementcount. - After checking all elements in
nums, ifcountis equal tox, we have found the special number. Returnx. - If the loop finishes without finding any such
x, it means the array is not special. Return-1.
Walkthrough
This method involves a nested loop structure. The outer loop iterates through each possible candidate for the special number x, from 0 to the length of the array, n. The inner loop iterates through the nums array to count how many elements are greater than or equal to the current candidate x. If at any point the count equals x, we have found the unique special number and can return it immediately. If the outer loop completes without finding a match, it implies no such x exists, and we return -1.
class Solution { public int specialArray(int[] nums) { int n = nums.length; for (int x = 0; x <= n; x++) { int count = 0; for (int num : nums) { if (num >= x) { count++; } } if (count == x) { return x; } } return -1; }}Complexity
Time
O(n^2), where `n` is the number of elements in `nums`. The outer loop runs `n+1` times, and for each iteration, the inner loop runs `n` times.
Space
O(1), as we only use a few variables to store the count and loop indices.
Trade-offs
Pros
Simple to understand and implement.
Requires no extra space, O(1) space complexity.
Cons
This approach is inefficient for larger arrays as it has a quadratic time complexity.
It repeatedly scans the entire array for each potential value of
x.
We can improve the efficiency by sorting the array first. Once the array nums is sorted, we can more quickly determine how many elements are greater than or equal to a certain value. If a number x is the special number, it means there are exactly x elements in nums that are >= x. In a sorted array, these would be the last x elements. This observation allows us to check the condition in a single pass after the initial sort.
Algorithm
- First, sort the input array
numsin non-decreasing order. - Let
nbe the length of the array. - Iterate from
i = 0ton-1. In each iteration, we consider a potential special numberx = n - i. Thisxrepresents the count of elements from indexito the end of the sorted array. - For
xto be the special number, two conditions must be met:- All these
xelements must be greater than or equal tox. Since the array is sorted, we only need to check the smallest of them, which isnums[i]. So,nums[i] >= x. - There must be exactly
xsuch elements. This means the element just before this block,nums[i-1], must be less thanx. Ifiis 0, there is no preceding element, so this condition is implicitly met.
- All these
- If both
nums[i] >= xand(i == 0 || nums[i-1] < x)are true, thenxis the special number. Returnx. - If the loop completes without finding a solution, return -1.
Walkthrough
After sorting nums in ascending order, we can iterate through the array and check potential values for x. Let's say we are checking if x is the special number. In the sorted array, if the condition holds, the last x elements must be >= x, and the element at index n-x-1 must be < x. We can formulate this into a single loop. We iterate i from 0 to n-1, and for each i, we test the candidate x = n - i. The x elements from nums[i] to nums[n-1] are the x largest elements. The condition nums[i] >= x ensures all of them are large enough. The condition (i == 0 || nums[i-1] < x) ensures that we have exactly x such elements. If we find such an x, we return it.
import java.util.Arrays; class Solution { public int specialArray(int[] nums) { Arrays.sort(nums); int n = nums.length; // Check possible values for x from n down to 1. // x = n - i for (int i = 0; i < n; i++) { int x = n - i; if (nums[i] >= x && (i == 0 || nums[i - 1] < x)) { return x; } } return -1; }}Complexity
Time
O(n log n), dominated by the sorting step. The subsequent linear scan takes O(n) time.
Space
O(log n) to O(n), depending on the implementation of the sorting algorithm. For instance, Java's `Arrays.sort` for primitives uses a dual-pivot quicksort, which has an average space complexity of O(log n).
Trade-offs
Pros
Significantly more efficient than the brute-force approach with O(n log n) time complexity.
The logic is sound and directly checks the necessary conditions on the sorted array.
Cons
The original array is modified due to sorting. If the original order must be preserved, a copy of the array is needed, which uses O(n) space.
Slightly more complex to reason about than the brute-force approach.
Given the constraints that nums.length is at most 100 and the values in nums are between 0 and 1000, we can use a counting-based approach for a highly efficient solution. This method avoids comparisons and sorting by directly counting the occurrences of each number. We can then use these counts to find the special number x in a single pass.
Algorithm
- Let
Mbe the maximum possible value innumsplus one (e.g., 1001 based on constraints). - Create a frequency array,
freq, of sizeM, initialized to all zeros. - Iterate through the input array
numsand populate the frequency array: for eachnuminnums, incrementfreq[num]. - Initialize a variable
count_ge = 0, which will keep track of the number of elements greater than or equal to the currentx. - Iterate
xdownwards fromM-1to0. - In each step, add the frequency of the current number
xto our running total:count_ge += freq[x]. - After updating
count_ge, it now holds the total count of numbers in the original array that are greater than or equal tox. - Check if
count_ge == x. If they are equal, we have found the special number, so returnx. - If the loop finishes and no such
xis found, return -1.
Walkthrough
The core idea is to build a frequency map (or an array since the values are bounded) of the numbers in nums. We create an array freq of size 1001. freq[i] will store how many times the number i appears in nums. After populating this frequency array, we can determine the number of elements >= x for any x. A clever way to do this is to iterate x downwards from 1000 to 0. We maintain a running sum count_ge of elements greater than or equal to the current x. When we are at x, we add freq[x] to count_ge. Then, count_ge represents the total number of elements in nums that are >= x. We check if this count_ge is equal to x. If it is, we've found our answer. This approach processes the array in a way that gives us the counts we need with minimal repeated work.
class Solution { public int specialArray(int[] nums) { // The maximum value for x is n, and the max value in nums is 1000. // We use a frequency array up to 1001. int[] freq = new int[1001]; for (int num : nums) { freq[num]++; } int count_ge = 0; // count of numbers greater than or equal to x // Iterate x from the maximum possible value down to 0. for (int x = 1000; x >= 0; x--) { count_ge += freq[x]; if (count_ge == x) { return x; } } return -1; }}Complexity
Time
O(N + M), where `N` is the length of `nums` and `M` is the range of values. We iterate through `nums` once (O(N)) and then through the frequency array once (O(M)).
Space
O(M), where `M` is the maximum possible value in `nums` (1001 in this case). This space is used for the frequency array.
Trade-offs
Pros
This is the most efficient approach with a linear time complexity.
It's simple to implement and does not require complex data structures or algorithms like sorting.
Cons
Requires extra space proportional to the maximum possible value in the input array, which could be large if the constraints were different.
Solutions
Solution
class Solution {public int specialArray(int[] nums) { for (int x = 1; x <= nums.length; ++x) { int cnt = 0; for (int v : nums) { if (v >= x) { ++cnt; } } if (cnt == x) { return x; } } return -1; }}class Solution {public: int specialArray(vector<int> &nums) { for (int x = 1; x <= nums.size(); ++x) { int cnt = 0; for (int v : nums) cnt += v >= x; if (cnt == x) return x; } return -1; }};class Solution: def specialArray(self, nums: List[int]) -> int: for x in range(1, len(nums) + 1): cnt = sum(v >= x for v in nums) if cnt == x: return x return - 1Video walkthrough
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