Formatted question description: https://leetcode.ca/all/2028.html

2028. Find Missing Observations (Medium)

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.

Example 4:

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

Constraints:

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

Similar Questions:

Solution 1.

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;
    }
}

class Solution {
public:
    vector<int> missingRolls(vector<int>& rolls, int mean, int n) {
        int m = rolls.size();
        int s = (n + m) * mean;
        for (int& v : rolls) s -= v;
        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;
    }
};

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

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
}

function missingRolls(rolls: number[], mean: number, n: number): number[] {
    const len = rolls.length + n;
    const sum = rolls.reduce((p, v) => p + v);
    const max = n * 6;
    const min = n;
    if ((sum + max) / len < mean || (sum + min) / len > mean) {
        return [];
    }

    const res = new Array(n);
    for (let i = min; i <= max; i++) {
        if ((sum + i) / len === mean) {
            const num = Math.floor(i / n);
            res.fill(num);
            let count = i - n * num;
            let j = 0;
            while (count != 0) {
                if (res[j] === 6) {
                    j++;
                } else {
                    res[j]++;
                    count--;
                }
            }
            break;
        }
    }
    return res;
}

impl Solution {
    pub fn missing_rolls(rolls: Vec<i32>, mean: i32, n: i32) -> Vec<i32> {
        let n = n as usize;
        let mean = mean as usize;
        let len = rolls.len() + n;
        let sum: i32 = rolls.iter().sum();
        let sum = sum as usize;
        let max = n * 6;
        let min = n;
        if (sum + max) < mean * len || (sum + min) > mean * len {
            return vec![];
        }

        let mut res = vec![0; n];
        for i in min..=max {
            if (sum + i) / len == mean {
                let num = i / n;
                res.fill(num as i32);
                let mut count = i - n * num;
                let mut j = 0;
                while count != 0 {
                    if res[j] == 6 {
                        j += 1;
                    } else {
                        res[j] += 1;
                        count -= 1;
                    }
                }
                break;
            }
        }
        res
    }
}

2028 - Find Missing Observations

2028. Find Missing Observations (Medium)

Solution 1.

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2028. Find Missing Observations (Medium)

Solution 1.

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