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2028. Find Missing Observations

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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 i^th 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) \times \textit{mean}, and the sum of known numbers is \sum_{i=0}^{m-1} \textit{rolls}[i]. Therefore, the sum of the missing numbers is s = (n + m) \times \textit{mean} - \sum_{i=0}^{m-1} \textit{rolls}[i].

If s \gt n \times 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).

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}