LeetCode #1608 — EASY

Special Array With X Elements Greater Than or Equal X

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

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, 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[3,5]

Example 2

[0,0]

Example 3

[0,4,3,0,4]

Step 02

Core Insight

What unlocks the optimal approach

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.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1608: Special Array With X Elements Greater Than or Equal X
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;
    }
}

// Accepted solution for LeetCode #1608: Special Array With X Elements Greater Than or Equal X
func specialArray(nums []int) int {
	for x := 1; x <= len(nums); x++ {
		cnt := 0
		for _, v := range nums {
			if v >= x {
				cnt++
			}
		}
		if cnt == x {
			return x
		}
	}
	return -1
}

# Accepted solution for LeetCode #1608: Special Array With X Elements Greater Than or Equal X
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 -1

// Accepted solution for LeetCode #1608: Special Array With X Elements Greater Than or Equal X
impl Solution {
    pub fn special_array(nums: Vec<i32>) -> i32 {
        let n = nums.len() as i32;
        for i in 0..=n {
            let mut count = 0;
            for &num in nums.iter() {
                if num >= i {
                    count += 1;
                }
            }
            if count == i {
                return i;
            }
        }
        -1
    }
}

// Accepted solution for LeetCode #1608: Special Array With X Elements Greater Than or Equal X
function specialArray(nums: number[]): number {
    const n = nums.length;
    for (let i = 0; i <= n; i++) {
        if (i === nums.reduce((r, v) => r + (v >= i ? 1 : 0), 0)) {
            return i;
        }
    }
    return -1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.