Description
You are given a 0-indexed integer array nums of even length.
As long as nums is not empty, you must repetitively:
- Find the minimum number in
nums and remove it.
- Find the maximum number in
nums and remove it.
- Calculate the average of the two removed numbers.
The average of two numbers a and b is (a + b) / 2.
- For example, the average of
2 and 3 is (2 + 3) / 2 = 2.5.
Return the number of distinct averages calculated using the above process.
Note that when there is a tie for a minimum or maximum number, any can be removed.
Example 1:
Input: nums = [4,1,4,0,3,5]
Output: 2
Explanation:
1. Remove 0 and 5, and the average is (0 + 5) / 2 = 2.5. Now, nums = [4,1,4,3].
2. Remove 1 and 4. The average is (1 + 4) / 2 = 2.5, and nums = [4,3].
3. Remove 3 and 4, and the average is (3 + 4) / 2 = 3.5.
Since there are 2 distinct numbers among 2.5, 2.5, and 3.5, we return 2.
Example 2:
Input: nums = [1,100]
Output: 1
Explanation:
There is only one average to be calculated after removing 1 and 100, so we return 1.
Constraints:
2 <= nums.length <= 100nums.length is even.0 <= nums[i] <= 100
Solutions
Solution 1: Sorting
The problem requires us to find the minimum and maximum values in the array nums each time, delete them, and then calculate the average of the two deleted numbers. Therefore, we can first sort the array nums, then take the first and last elements of the array each time, calculate their sum, use a hash table or array cnt to record the number of times each sum appears, and finally count the number of different sums.
The time complexity is O(n × log n), and the space complexity is O(n). Here, n is the length of the array nums.
PythonJavaC++GoTypeScriptRust
class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
return len(set(nums[i] + nums[-i - 1] for i in range(len(nums) >> 1)))(code-box)
class Solution {
public int distinctAverages(int[] nums) {
Arrays.sort(nums);
Set<Integer> s = new HashSet<>();
int n = nums.length;
for (int i = 0; i < n >> 1; ++i) {
s.add(nums[i] + nums[n - i - 1]);
}
return s.size();
}
}(code-box)
class Solution {
public:
int distinctAverages(vector<int>& nums) {
sort(nums.begin(), nums.end());
unordered_set<int> s;
int n = nums.size();
for (int i = 0; i < n >> 1; ++i) {
s.insert(nums[i] + nums[n - i - 1]);
}
return s.size();
}
};(code-box)
func distinctAverages(nums []int) (ans int) {
sort.Ints(nums)
n := len(nums)
s := map[int]struct{}{}
for i := 0; i < n>>1; i++ {
s[nums[i]+nums[n-i-1]] = struct{}{}
}
return len(s)
}(code-box)
function distinctAverages(nums: number[]): number {
nums.sort((a, b) => a - b);
const s: Set<number> = new Set();
const n = nums.length;
for (let i = 0; i < n >> 1; ++i) {
s.add(nums[i] + nums[n - i - 1]);
}
return s.size;
}(code-box)
impl Solution {
pub fn distinct_averages(nums: Vec<i32>) -> i32 {
let mut nums = nums;
nums.sort();
let n = nums.len();
let mut cnt = vec![0; 201];
let mut ans = 0;
for i in 0..n >> 1 {
let x = (nums[i] + nums[n - i - 1]) as usize;
cnt[x] += 1;
if cnt[x] == 1 {
ans += 1;
}
}
ans
}
}(code-box)
Solution 2
PythonJavaC++GoTypeScriptRust
class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
ans = 0
cnt = Counter()
for i in range(len(nums) >> 1):
x = nums[i] + nums[-i - 1]
cnt[x] += 1
if cnt[x] == 1:
ans += 1
return ans(code-box)
class Solution {
public int distinctAverages(int[] nums) {
Arrays.sort(nums);
int[] cnt = new int[201];
int n = nums.length;
int ans = 0;
for (int i = 0; i < n >> 1; ++i) {
if (++cnt[nums[i] + nums[n - i - 1]] == 1) {
++ans;
}
}
return ans;
}
}(code-box)
class Solution {
public:
int distinctAverages(vector<int>& nums) {
sort(nums.begin(), nums.end());
int cnt[201]{};
int n = nums.size();
int ans = 0;
for (int i = 0; i < n >> 1; ++i) {
if (++cnt[nums[i] + nums[n - i - 1]] == 1) {
++ans;
}
}
return ans;
}
};(code-box)
func distinctAverages(nums []int) (ans int) {
sort.Ints(nums)
n := len(nums)
cnt := [201]int{}
for i := 0; i < n>>1; i++ {
x := nums[i] + nums[n-i-1]
cnt[x]++
if cnt[x] == 1 {
ans++
}
}
return
}(code-box)
function distinctAverages(nums: number[]): number {
nums.sort((a, b) => a - b);
const cnt: number[] = Array(201).fill(0);
let ans = 0;
const n = nums.length;
for (let i = 0; i < n >> 1; ++i) {
if (++cnt[nums[i] + nums[n - i - 1]] === 1) {
++ans;
}
}
return ans;
}(code-box)
use std::collections::HashMap;
impl Solution {
pub fn distinct_averages(nums: Vec<i32>) -> i32 {
let mut h = HashMap::new();
let mut nums = nums;
let mut ans = 0;
let n = nums.len();
nums.sort();
for i in 0..n >> 1 {
let x = nums[i] + nums[n - i - 1];
*h.entry(x).or_insert(0) += 1;
if *h.get(&x).unwrap() == 1 {
ans += 1;
}
}
ans
}
}(code-box)
Solution 3
Rust
use std::collections::HashSet;
impl Solution {
pub fn distinct_averages(nums: Vec<i32>) -> i32 {
let mut set = HashSet::new();
let mut ans = 0;
let n = nums.len();
let mut nums = nums;
nums.sort();
for i in 0..n >> 1 {
let x = nums[i] + nums[n - i - 1];
if set.contains(&x) {
continue;
}
set.insert(x);
ans += 1;
}
ans
}
}(code-box)
