LeetCode 2028. Find Missing Observations Solution in Java, C++, Python & More | Explanation + Code

Description

You have observations of n + m 6-sided dice rolls with each face numbered from 1 to 6. n of the observations went missing, and you only have the observations of m rolls. Fortunately, you have also calculated the average value of the n + m rolls.

You are given an integer array rolls of length m where rolls[i] is the value of the ith observation. You are also given the two integers mean and n.

Return an array of length n containing the missing observations such that the average value of the n + m rolls is exactly mean. If there are multiple valid answers, return any of them. If no such array exists, return an empty array.

The average value of a set of k numbers is the sum of the numbers divided by k.

Note that mean is an integer, so the sum of the n + m rolls should be divisible by n + m.

 

Example 1:

Input: rolls = [3,2,4,3], mean = 4, n = 2
Output: [6,6]
Explanation: The mean of all n + m rolls is (3 + 2 + 4 + 3 + 6 + 6) / 6 = 4.

Example 2:

Input: rolls = [1,5,6], mean = 3, n = 4
Output: [2,3,2,2]
Explanation: The mean of all n + m rolls is (1 + 5 + 6 + 2 + 3 + 2 + 2) / 7 = 3.

Example 3:

Input: rolls = [1,2,3,4], mean = 6, n = 4
Output: []
Explanation: It is impossible for the mean to be 6 no matter what the 4 missing rolls are.

 

Constraints:

• m == rolls.length
• 1 <= n, m <= 105
• 1 <= rolls[i], mean <= 6

Solutions

Solution 1: Construction

According to the problem description, the sum of all numbers is (n + m) × mean, and the sum of known numbers is ∑_{i=0}^m-1 rolls[i]. Therefore, the sum of the missing numbers is s = (n + m) × mean - ∑_{i=0}^m-1 rolls[i].

If s \gt n × 6 or s \lt n, it means there is no answer that satisfies the conditions, so we return an empty array.

Otherwise, we can evenly distribute s to n numbers, that is, the value of each number is s / n, and the value of s \bmod n numbers is increased by 1.

The time complexity is O(n + m), where n and m are the number of missing numbers and known numbers, respectively. Ignoring the space consumption of the answer, the space complexity is O(1).

PythonJavaC++GoTypeScriptRust

class Solution: def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]: m = len(rolls) s = (n + m) * mean - sum(rolls) if s > n * 6 or s < n: return [] ans = [s // n] * n for i in range(s % n): ans[i] += 1 return ans(code-box)

class Solution { public int[] missingRolls(int[] rolls, int mean, int n) { int m = rolls.length; int s = (n + m) * mean; for (int v : rolls) { s -= v; } if (s > n * 6 || s < n) { return new int[0]; } int[] ans = new int[n]; Arrays.fill(ans, s / n); for (int i = 0; i < s % n; ++i) { ++ans[i]; } return ans; } }(code-box)

class Solution { public: vector<int> missingRolls(vector<int>& rolls, int mean, int n) { int m = rolls.size(); int s = (n + m) * mean - accumulate(rolls.begin(), rolls.end(), 0); if (s > n * 6 || s < n) { return {}; } vector<int> ans(n, s / n); for (int i = 0; i < s % n; ++i) { ++ans[i]; } return ans; } };(code-box)

func missingRolls(rolls []int, mean int, n int) []int { m := len(rolls) s := (n + m) * mean for _, v := range rolls { s -= v } if s > n*6 || s < n { return []int{} } ans := make([]int, n) for i, j := 0, 0; i < n; i, j = i+1, j+1 { ans[i] = s / n if j < s%n { ans[i]++ } } return ans }(code-box)

function missingRolls(rolls: number[], mean: number, n: number): number[] { const m = rolls.length; const s = (n + m) * mean - rolls.reduce((a, b) => a + b, 0); if (s > n * 6 || s < n) { return []; } const ans: number[] = Array(n).fill((s / n) | 0); for (let i = 0; i < s % n; ++i) { ans[i]++; } return ans; }(code-box)

impl Solution { pub fn missing_rolls(rolls: Vec<i32>, mean: i32, n: i32) -> Vec<i32> { let m = rolls.len() as i32; let s = (n + m) * mean - rolls.iter().sum::<i32>(); if s > n * 6 || s < n { return vec![]; } let mut ans = vec![s / n; n as usize]; for i in 0..(s % n) as usize { ans[i] += 1; } ans } }(code-box)