# [2028. Find Missing Observations](https://leetcode.com/problems/find-missing-observations)

[中文文档](/solution/2000-2099/2028.Find%20Missing%20Observations/README.md)

## Description

<p>You have observations of <code>n + m</code> <strong>6-sided</strong> dice rolls with each face numbered from <code>1</code> to <code>6</code>. <code>n</code> of the observations went missing, and you only have the observations of <code>m</code> rolls. Fortunately, you have also calculated the <strong>average value</strong> of the <code>n + m</code> rolls.</p>

<p>You are given an integer array <code>rolls</code> of length <code>m</code> where <code>rolls[i]</code> is the value of the <code>i<sup>th</sup></code> observation. You are also given the two integers <code>mean</code> and <code>n</code>.</p>

<p>Return <em>an array of length </em><code>n</code><em> containing the missing observations such that the <strong>average value </strong>of the </em><code>n + m</code><em> rolls is <strong>exactly</strong> </em><code>mean</code>. If there are multiple valid answers, return <em>any of them</em>. If no such array exists, return <em>an empty array</em>.</p>

<p>The <strong>average value</strong> of a set of <code>k</code> numbers is the sum of the numbers divided by <code>k</code>.</p>

<p>Note that <code>mean</code> is an integer, so the sum of the <code>n + m</code> rolls should be divisible by <code>n + m</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> rolls = [3,2,4,3], mean = 4, n = 2
<strong>Output:</strong> [6,6]
<strong>Explanation:</strong> The mean of all n + m rolls is (3 + 2 + 4 + 3 + 6 + 6) / 6 = 4.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> rolls = [1,5,6], mean = 3, n = 4
<strong>Output:</strong> [2,3,2,2]
<strong>Explanation:</strong> The mean of all n + m rolls is (1 + 5 + 6 + 2 + 3 + 2 + 2) / 7 = 3.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> rolls = [1,2,3,4], mean = 6, n = 4
<strong>Output:</strong> []
<strong>Explanation:</strong> It is impossible for the mean to be 6 no matter what the 4 missing rolls are.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</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>

## Solutions

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### **Python3**

```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**

```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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