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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.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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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