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.
Example 4:
Input: nums = [3,6,7,7,0] Output: -1
Constraints:
1 <= nums.length <= 100
0 <= nums[i] <= 1000
Related Topics:
Array
// OJ: https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int specialArray(vector<int>& A) {
sort(begin(A), end(A));
int N = A.size();
for (int i = 0; i <= N; ++i) {
int cnt = N - (lower_bound(begin(A), end(A), i) - begin(A));
if (cnt == i) return i;
}
return -1;
}
};
We store the count of numbers in a cnt
array. And we just need to iterate from max(A)
towards 0
to find the valid value, because for any k > max(A)
, there are 0 elements >= k
and thus must be invalid.
// OJ: https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/
// Author: github.com/lzl124631x
// Time: O(N + max(A))
// Space: O(max(A))
class Solution {
public:
int specialArray(vector<int>& A) {
int cnt[1001] = {};
for (int n : A) cnt[n]++;
for (int i = 999; i >= 0; --i) {
cnt[i] += cnt[i + 1];
if (cnt[i] == i) return i;
}
return -1;
}
};
The sorting part takes O(NlogN)
. But the rest just takes O(logN)
.
// OJ: https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int specialArray(vector<int>& A) {
sort(begin(A), end(A));
int N = A.size(), L = 0, R = N;
while (L <= R) {
int M = (L + R) / 2;
bool cur = M == 0 || A[N - M] >= M, prev = M == N || A[N - M - 1] < M;
if (cur && prev) return M;
if (!cur) R = M - 1;
else L = M + 1;
}
return -1;
}
};