# [2028. 找出缺失的观测数据](https://leetcode.cn/problems/find-missing-observations)

[English Version](/solution/2000-2099/2028.Find%20Missing%20Observations/README_EN.md)

## 题目描述

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<p>现有一份 <code>n + m</code>&nbsp;次投掷单个<strong> 六面</strong> 骰子的观测数据，骰子的每个面从 <code>1</code> 到 <code>6</code> 编号。观测数据中缺失了 <code>n</code> 份，你手上只拿到剩余&nbsp;<code>m</code> 次投掷的数据。幸好你有之前计算过的这 <code>n + m</code> 次投掷数据的 <strong>平均值</strong> 。</p>

<p>给你一个长度为 <code>m</code> 的整数数组 <code>rolls</code> ，其中&nbsp;<code>rolls[i]</code> 是第 <code>i</code> 次观测的值。同时给你两个整数 <code>mean</code> 和 <code>n</code> 。</p>

<p>返回一个长度为<em> </em><code>n</code><em> </em>的数组，包含所有缺失的观测数据，且满足这<em> </em><code>n + m</code><em> </em>次投掷的 <strong>平均值</strong> 是<em> </em><code>mean</code> 。如果存在多组符合要求的答案，只需要返回其中任意一组即可。如果不存在答案，返回一个空数组。</p>

<p><code>k</code>&nbsp;个数字的 <strong>平均值</strong> 为这些数字求和后再除以&nbsp;<code>k</code> 。</p>

<p>注意 <code>mean</code> 是一个整数，所以 <code>n + m</code> 次投掷的总和需要被&nbsp;<code>n + m</code>&nbsp;整除。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<pre>
<strong>输入：</strong>rolls = [3,2,4,3], mean = 4, n = 2
<strong>输出：</strong>[6,6]
<strong>解释：</strong>所有 n + m 次投掷的平均值是 (3 + 2 + 4 + 3 + 6 + 6) / 6 = 4 。
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>rolls = [1,5,6], mean = 3, n = 4
<strong>输出：</strong>[2,3,2,2]
<strong>解释：</strong>所有 n + m 次投掷的平均值是 (1 + 5 + 6 + 2 + 3 + 2 + 2) / 7 = 3 。
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>rolls = [1,2,3,4], mean = 6, n = 4
<strong>输出：</strong>[]
<strong>解释：</strong>无论丢失的 4 次数据是什么，平均值都不可能是 6 。
</pre>

<p><strong>示例 4：</strong></p>

<pre>
<strong>输入：</strong>rolls = [1], mean = 3, n = 1
<strong>输出：</strong>[5]
<strong>解释：</strong>所有 n + m 次投掷的平均值是 (1 + 5) / 2 = 3 。
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>m == rolls.length</code></li>
	<li><code>1 &lt;= n, m &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= rolls[i], mean &lt;= 6</code></li>
</ul>

## 解法

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### **Python3**

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```python
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**

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```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;
    }
}
```

### **TypeScript**

```ts
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;
}
```

### **Rust**

```rust
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
    }
}
```

### **C++**

```cpp
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;
    }
};
```

### **Go**

```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
}
```

### **...**

```

```

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