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.
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; }}Video walkthrough
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