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Leetcode 1608. Special Array With X Elements Greater Than or Equal X

Problem Statement

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, then there is a unique x that satisfies the condition.

Clarifying Questions

  1. Q: What range of values can the elements of nums take? A: The elements in nums are non-negative integers, i.e., they can range from 0 to a large positive value subject to the problem constraints.

  2. Q: What is the size range of the array nums? A: The size of nums can vary, but usually, in such problems, we can expect it to follow standard constraints like up to 10^4 elements unless specified otherwise.

  3. Q: Is there any specific edge case we need to consider? A: Edge cases could include an empty array or an array with all elements being the same.

Strategy

  1. Sort the Array: Start by sorting the array. This allows us to easily determine how many numbers are greater than or equal to any given number.

  2. Iterate and Check: Iterate through the array to check at each position if the count of elements greater than or equal to the current element matches the condition.

  3. Binary Search Optimization: Since the array is sorted, we can leverage binary search to optimize the check for the condition if required.

Code

#include <vector>
#include <algorithm>

class Solution {
public:
    int specialArray(std::vector<int>& nums) {
        std::sort(nums.begin(), nums.end());
        int n = nums.size();

        for (int i = 0; i <= n; ++i) {
            auto it = std::lower_bound(nums.begin(), nums.end(), i);
            int count = nums.end() - it;
            if (count == i) {
                return i;
            }
        }
        
        return -1;
    }
};

Explanation

  1. Sorting: We sort nums to make it easier to find the count of elements greater than or equal to a given value.
  2. Iteration: We iterate through possible values of x from 0 to n (length of nums).
  3. Count Check: Use std::lower_bound to find the first element not less than i. The number of elements greater than or equal to i is then calculated by nums.end() - it.
  4. Return Result: If count == i, then we return i as it satisfies the condition. If no such i is found, return -1.

Time Complexity

  1. Sorting: (O(n \log n))
  2. Iteration with Lower Bound: Each call to lower_bound is (O(\log n)). Hence, iterating from 0 to n results in (O(n \log n)).

Thus, the overall time complexity is (O(n \log n)). This is efficient enough for typical input sizes in competitive programming.

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