Special Array With X Elements Greater Than or Equal X

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

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.

Input & Output

Example 1 — Basic Special Array

$ Input: nums = [3,5]

› Output: 2

💡 Note: There are exactly 2 numbers greater than or equal to 2: both 3 and 5. So x = 2 makes this a special array.

Example 2 — Not Special Array

$ Input: nums = [0,0]

› Output: -1

💡 Note: For x=0: 2 elements ≥ 0 (2≠0). For x=1: 0 elements ≥ 1 (0≠1). For x=2: 0 elements ≥ 2 (0≠2). No valid x exists.

Example 3 — Edge Case with Zero

$ Input: nums = [0,4,3,0,4]

› Output: 3

💡 Note: There are exactly 3 numbers greater than or equal to 3: [4,3,4]. So x = 3 makes this a special array.

Constraints

1 ≤ nums.length ≤ 100
0 ≤ nums[i] ≤ 1000

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that x must be between 0 and array length, and we need exactly x elements ≥ x. The brute force approach checks every possible x value. The sort-first approach leverages sorting for efficient validation. Best approach depends on array size: brute force for small arrays, sorting for medium, binary search for very large. Time: O(n²) to O(n log n), Space: O(1).

Common Approaches

✓ Binary Search on Answer

⏱️ Time: O(n log n) Space: O(1)

Since x must be between 0 and array length, we can binary search on this range. For each candidate x, count elements >= x efficiently and check if count equals x.

Brute Force - Check All Possible Values

⏱️ Time: O(n²) Space: O(1)

For each potential value x from 0 to array length, count how many elements are greater than or equal to x. If the count equals x, we found our answer.

Sort First Approach

⏱️ Time: O(n log n) Space: O(1)

After sorting in descending order, the first i elements are the largest i elements. If nums[i-1] >= i, then exactly i elements are >= i, making i our special value.

Binary Search on Answer — Algorithm Steps

Set search range: left=0, right=n
For mid value, count elements >= mid
If count == mid, found answer
If count > mid, search right half, else left half

Visualization

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Step-by-Step Walkthrough

Set Range

Search x in [0, array_length]

Check Middle

Count elements >= mid

Narrow Range

Adjust search bounds based on count

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int left = 0, right = numsSize;
    
    while (left <= right) {
        int mid = (left + right) / 2;
        int count = 0;
        for (int i = 0; i < numsSize; i++) {
            if (nums[i] >= mid) {
                count++;
            }
        }
        
        if (count == mid) {
            return mid;
        } else if (count > mid) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    
    return -1;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0) {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Binary search runs O(log n) iterations, each taking O(n) time to count elements

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra space for binary search variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Binary Search on Answer — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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