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2028. Find Missing Observations

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 $(n + m) \times mean$, and the sum of known numbers is sum(rolls). Therefore, the sum of the missing numbers is $s = (n + m) \times mean - sum(rolls)$.

If $s > n \times 6$ or $s < n$, it means there is no answer that satisfies the conditions, so 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 $O(n + m)$, and the space complexity is $O(1)$. Here, $n$ and $m$ are the number of missing numbers and known numbers, respectively.

  • 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
        }
    }
    
    

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