Description
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
Solutions
Solution 1: Brute Force Enumeration
We enumerate x in the range of [1..n], and then count the number of elements in the array that are greater than or equal to x, denoted as cnt. If there exists cnt equal to x, return x directly.
The time complexity is O(n2), where n is the length of the array. The space complexity is O(1).
PythonJavaC++GoTypeScriptRust
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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
}
}(code-box)
Solution 2: Sorting + Binary Search
We can also sort nums first.
Next, we still enumerate x, and use binary search to find the first element in nums that is greater than or equal to x, quickly counting the number of elements in nums that are greater than or equal to x.
The time complexity is O(n × log n), and the space complexity is O(log n). Where n is the length of the array.
PythonJavaC++GoTypeScriptRust
class Solution:
def specialArray(self, nums: List[int]) -> int:
nums.sort()
n = len(nums)
for x in range(1, n + 1):
cnt = n - bisect_left(nums, x)
if cnt == x:
return x
return -1(code-box)
class Solution {
public int specialArray(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
for (int x = 1; x <= n; ++x) {
int left = 0, right = n;
while (left < right) {
int mid = (left + right) >> 1;
if (nums[mid] >= x) {
right = mid;
} else {
left = mid + 1;
}
}
int cnt = n - left;
if (cnt == x) {
return x;
}
}
return -1;
}
}(code-box)
class Solution {
public:
int specialArray(vector<int>& nums) {
int n = nums.size();
sort(nums.begin(), nums.end());
for (int x = 1; x <= n; ++x) {
int cnt = n - (lower_bound(nums.begin(), nums.end(), x) - nums.begin());
if (cnt == x) return x;
}
return -1;
}
};(code-box)
func specialArray(nums []int) int {
sort.Ints(nums)
n := len(nums)
for x := 1; x <= n; x++ {
left, right := 0, n
for left < right {
mid := (left + right) >> 1
if nums[mid] >= x {
right = mid
} else {
left = mid + 1
}
}
cnt := n - left
if cnt == x {
return x
}
}
return -1
}(code-box)
function specialArray(nums: number[]): number {
const n = nums.length;
let left = 0;
let right = n + 1;
while (left < right) {
const mid = (left + right) >> 1;
const count = nums.reduce((r, v) => r + (v >= mid ? 1 : 0), 0);
if (count === mid) {
return mid;
}
if (count > mid) {
left = mid + 1;
} else {
right = mid;
}
}
return -1;
}(code-box)
use std::cmp::Ordering;
impl Solution {
pub fn special_array(nums: Vec<i32>) -> i32 {
let n = nums.len() as i32;
let mut left = 0;
let mut right = n + 1;
while left < right {
let mid = left + (right - left) / 2;
let mut count = 0;
for &num in nums.iter() {
if num >= mid {
count += 1;
}
}
match count.cmp(&mid) {
Ordering::Equal => {
return mid;
}
Ordering::Less => {
right = mid;
}
Ordering::Greater => {
left = mid + 1;
}
}
}
-1
}
}(code-box)
