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feat: update solutions to lc problem: No.2028 #2917

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81 changes: 25 additions & 56 deletions solution/2000-2099/2028.Find Missing Observations/README.md
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Expand Up @@ -82,13 +82,13 @@ tags:

### 方法一：构造

根据题目描述，所有数字之和为 $(n + m) \times mean$，已知的数字之和为 `sum(rolls)`，那么缺失的数字之和为 $s = (n + m) \times mean - sum(rolls)$。
根据题目描述，所有数字之和为 $(n + m) \times \text{mean}$，已知的数字之和为 $\sum_{i=0}^{m-1} \text{rolls}[i]$，那么缺失的数字之和为 $s = (n + m) \times \text{mean} - \sum_{i=0}^{m-1} \text{rolls}[i]$。

如果 $s \gt n \times 6$ 或者 $s \lt n$，说明不存在满足条件的答案，返回空数组。

否则，我们可以将 $s$ 平均分配到 $n$ 个数字上，即每个数字的值为 $s / n$，其中 $s \bmod n$ 个数字的值再加上 $1$。

时间复杂度 $O(n + m)$，空间复杂度 $O(1)$。其中 $n$ 和 $m$ 分别为缺失的数字个数和已知的数字个数。
时间复杂度 $O(n + m)$，其中 $n$ 和 $m$ 分别为缺失的数字个数和已知的数字个数。忽略答案的空间消耗，空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -137,11 +137,14 @@ 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 {};
int s = (n + m) * mean - accumulate(rolls.begin(), rolls.end(), 0);
if (s > n * 6 || s < n) {
return {};
}
vector<int> ans(n, s / n);
for (int i = 0; i < s % n; ++i) ++ans[i];
for (int i = 0; i < s % n; ++i) {
++ans[i];
}
return ans;
}
};
Expand Down Expand Up @@ -174,33 +177,16 @@ func missingRolls(rolls []int, mean int, n int) []int {

```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) {
const m = rolls.length;
const s = (n + m) * mean - rolls.reduce((a, b) => a + b, 0);
if (s > n * 6 || s < n) {
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;
}
const ans: number[] = Array(n).fill((s / n) | 0);
for (let i = 0; i < s % n; ++i) {
ans[i]++;
}
return res;
return ans;
}
```

Expand All @@ -209,36 +195,19 @@ function missingRolls(rolls: number[], mean: number, n: number): number[] {
```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 {
let m = rolls.len() as i32;
let s = (n + m) * mean - rolls.iter().sum::<i32>();

if s > n * 6 || s < n {
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;
}
let mut ans = vec![s / n; n as usize];
for i in 0..(s % n) as usize {
ans[i] += 1;
}
res

ans
}
}
```
Expand Down
83 changes: 26 additions & 57 deletions solution/2000-2099/2028.Find Missing Observations/README_EN.md
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Expand Up @@ -72,13 +72,13 @@ tags:

### 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 $s = (n + m) \times mean - sum(rolls)$.
According to the problem description, the sum of all numbers is $(n + m) \times \text{mean}$, and the sum of known numbers is $\sum_{i=0}^{m-1} \text{rolls}[i]$. Therefore, the sum of the missing numbers is $s = (n + m) \times \text{mean} - \sum_{i=0}^{m-1} \text{rolls}[i]$.

If $s > n \times 6$ or $s < n$, it means there is no answer that satisfies the conditions, so return an empty array.
If $s \gt n \times 6$ or $s \lt n$, it means there is no answer that satisfies the conditions, so we 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.
The time complexity is $O(n + m)$, where $n$ and $m$ are the number of missing numbers and known numbers, respectively. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down Expand Up @@ -127,11 +127,14 @@ 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 {};
int s = (n + m) * mean - accumulate(rolls.begin(), rolls.end(), 0);
if (s > n * 6 || s < n) {
return {};
}
vector<int> ans(n, s / n);
for (int i = 0; i < s % n; ++i) ++ans[i];
for (int i = 0; i < s % n; ++i) {
++ans[i];
}
return ans;
}
};
Expand Down Expand Up @@ -164,33 +167,16 @@ func missingRolls(rolls []int, mean int, n int) []int {

```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) {
const m = rolls.length;
const s = (n + m) * mean - rolls.reduce((a, b) => a + b, 0);
if (s > n * 6 || s < n) {
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;
}
const ans: number[] = Array(n).fill((s / n) | 0);
for (let i = 0; i < s % n; ++i) {
ans[i]++;
}
return res;
return ans;
}
```

Expand All @@ -199,36 +185,19 @@ function missingRolls(rolls: number[], mean: number, n: number): number[] {
```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 {
let m = rolls.len() as i32;
let s = (n + m) * mean - rolls.iter().sum::<i32>();

if s > n * 6 || s < n {
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;
}
let mut ans = vec![s / n; n as usize];
for i in 0..(s % n) as usize {
ans[i] += 1;
}
res

ans
}
}
```
Expand Down
11 changes: 7 additions & 4 deletions solution/2000-2099/2028.Find Missing Observations/Solution.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -2,11 +2,14 @@ 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 {};
int s = (n + m) * mean - accumulate(rolls.begin(), rolls.end(), 0);
if (s > n * 6 || s < n) {
return {};
}
vector<int> ans(n, s / n);
for (int i = 0; i < s % n; ++i) ++ans[i];
for (int i = 0; i < s % n; ++i) {
++ans[i];
}
return ans;
}
};
35 changes: 9 additions & 26 deletions solution/2000-2099/2028.Find Missing Observations/Solution.rs
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Original file line number Diff line number Diff line change
@@ -1,34 +1,17 @@
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 {
let m = rolls.len() as i32;
let s = (n + m) * mean - rolls.iter().sum::<i32>();

if s > n * 6 || s < n {
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;
}
let mut ans = vec![s / n; n as usize];
for i in 0..(s % n) as usize {
ans[i] += 1;
}
res

ans
}
}
31 changes: 7 additions & 24 deletions solution/2000-2099/2028.Find Missing Observations/Solution.ts
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Original file line number Diff line number Diff line change
@@ -1,29 +1,12 @@
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) {
const m = rolls.length;
const s = (n + m) * mean - rolls.reduce((a, b) => a + b, 0);
if (s > n * 6 || s < n) {
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;
}
const ans: number[] = Array(n).fill((s / n) | 0);
for (let i = 0; i < s % n; ++i) {
ans[i]++;
}
return res;
return ans;
}
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