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feat: add solutions to lc problem: No.3528 #4380

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dcf0373
feat: add solutions to lc problem: No.3528
yanglbme Apr 30, 2025
a25670e
fix: update go code
yanglbme Apr 30, 2025
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147 changes: 129 additions & 18 deletions solution/3500-3599/3528.Unit Conversion I/README.md
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Expand Up @@ -34,8 +34,8 @@ tags:
<p><strong>解释：</strong></p>

<ul>
<li>使用 <code>conversions[0]</code>：将一个 0 类型单位转换为 2 个 1 类型单位。</li>
<li>使用&nbsp;<code>conversions[0]</code>&nbsp;和&nbsp;<code>conversions[1]</code>&nbsp;将一个 0 类型单位转换为 6 个 2 类型单位。</li>
<li>使用 <code>conversions[0]</code>：将一个 0 类型单位转换为 2 个 1 类型单位。</li>
<li>使用&nbsp;<code>conversions[0]</code>&nbsp;和&nbsp;<code>conversions[1]</code>&nbsp;将一个 0 类型单位转换为 6 个 2 类型单位。</li>
</ul>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3500-3599/3528.Unit%20Conversion%20I/images/1745660099-FZhVTM-example1.png" style="width: 545px; height: 119px;" /></div>

Expand All @@ -49,13 +49,13 @@ tags:
<p><strong>解释：</strong></p>

<ul>
<li>使用 <code>conversions[0]</code>&nbsp;将一个 0 类型单位转换为 2 个 1 类型单位。</li>
<li>使用 <code>conversions[1]</code>&nbsp;将一个 0 类型单位转换为 3 个 2 类型单位。</li>
<li>使用 <code>conversions[0]</code> 和 <code>conversions[2]</code>&nbsp;将一个 0 类型单位转换为 8 个 3 类型单位。</li>
<li>使用 <code>conversions[0]</code> 和 <code>conversions[3]</code>&nbsp;将一个 0 类型单位转换为 10 个 4 类型单位。</li>
<li>使用 <code>conversions[1]</code> 和 <code>conversions[4]</code>&nbsp;将一个 0 类型单位转换为 6 个 5 类型单位。</li>
<li>使用 <code>conversions[0]</code>、<code>conversions[3]</code> 和 <code>conversions[5]</code>&nbsp;将一个 0 类型单位转换为 30 个 6 类型单位。</li>
<li>使用 <code>conversions[1]</code>、<code>conversions[4]</code> 和 <code>conversions[6]</code>&nbsp;将一个 0 类型单位转换为 24 个 7 类型单位。</li>
<li>使用 <code>conversions[0]</code>&nbsp;将一个 0 类型单位转换为 2 个 1 类型单位。</li>
<li>使用 <code>conversions[1]</code>&nbsp;将一个 0 类型单位转换为 3 个 2 类型单位。</li>
<li>使用 <code>conversions[0]</code> 和 <code>conversions[2]</code>&nbsp;将一个 0 类型单位转换为 8 个 3 类型单位。</li>
<li>使用 <code>conversions[0]</code> 和 <code>conversions[3]</code>&nbsp;将一个 0 类型单位转换为 10 个 4 类型单位。</li>
<li>使用 <code>conversions[1]</code> 和 <code>conversions[4]</code>&nbsp;将一个 0 类型单位转换为 6 个 5 类型单位。</li>
<li>使用 <code>conversions[0]</code>、<code>conversions[3]</code> 和 <code>conversions[5]</code>&nbsp;将一个 0 类型单位转换为 30 个 6 类型单位。</li>
<li>使用 <code>conversions[1]</code>、<code>conversions[4]</code> 和 <code>conversions[6]</code>&nbsp;将一个 0 类型单位转换为 24 个 7 类型单位。</li>
</ul>
</div>

Expand All @@ -64,11 +64,11 @@ tags:
<p><strong>提示：</strong></p>

<ul>
<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
<li><code>conversions.length == n - 1</code></li>
<li><code>0 &lt;= sourceUnit<sub>i</sub>, targetUnit<sub>i</sub> &lt; n</code></li>
<li><code>1 &lt;= conversionFactor<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
<li>保证单位&nbsp;0 可以通过&nbsp;<strong>唯一&nbsp;</strong>的转换路径（不需要反向转换）转换为任何其他单位。</li>
<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
<li><code>conversions.length == n - 1</code></li>
<li><code>0 &lt;= sourceUnit<sub>i</sub>, targetUnit<sub>i</sub> &lt; n</code></li>
<li><code>1 &lt;= conversionFactor<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
<li>保证单位&nbsp;0 可以通过&nbsp;<strong>唯一&nbsp;</strong>的转换路径（不需要反向转换）转换为任何其他单位。</li>
</ul>

<!-- description:end -->
Expand All @@ -77,32 +77,143 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：DFS

由于题目保证了单位 0 可以通过唯一的转换路径转换为其他单位，因此我们可以使用深度优先搜索（DFS）来遍历所有单位的转换关系。另外，由于 $\textit{conversions}$ 数组的长度为 $n - 1$，表示有 $n - 1$ 条转换关系，因此我们可以将单位转换关系看作一棵树，根节点为单位 0，其他节点为其他单位。

我们可以用一个邻接表 $g$ 来表示单位转换关系，其中 $g[i]$ 表示单位 $i$ 可以转换到的单位和对应的转换因子。

然后，我们从根节点 $0$ 开始进行深度优先搜索，即调函数 $\textit{dfs}(s, \textit{mul})$，其中 $s$ 表示当前单位，$\textit{mul}$ 表示从单位 $0$ 转换到单位 $s$ 的转换因子。初始时 $s = 0$, $\textit{mul} = 1$。在每次递归中，我们将当前单位 $s$ 的转换因子 $\textit{mul}$ 存储到答案数组中，然后遍历当前单位 $s$ 的所有邻接单位 $t$，递归调用 $\textit{dfs}(t, \textit{mul} \times w \mod (10^9 + 7))$，其中 $w$ 为单位 $s$ 转换到单位 $t$ 的转换因子。

最后，我们返回答案数组即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为单位的数量。

<!-- tabs:start -->

#### Python3

```python

class Solution:
def baseUnitConversions(self, conversions: List[List[int]]) -> List[int]:
def dfs(s: int, mul: int) -> None:
ans[s] = mul
for t, w in g[s]:
dfs(t, mul * w % mod)

mod = 10**9 + 7
n = len(conversions) + 1
g = [[] for _ in range(n)]
for s, t, w in conversions:
g[s].append((t, w))
ans = [0] * n
dfs(0, 1)
return ans
```

#### Java

```java

class Solution {
private final int mod = (int) 1e9 + 7;
private List<int[]>[] g;
private int[] ans;
private int n;

public int[] baseUnitConversions(int[][] conversions) {
n = conversions.length + 1;
g = new List[n];
Arrays.setAll(g, k -> new ArrayList<>());
ans = new int[n];
for (var e : conversions) {
g[e[0]].add(new int[] {e[1], e[2]});
}
dfs(0, 1);
return ans;
}

private void dfs(int s, long mul) {
ans[s] = (int) mul;
for (var e : g[s]) {
dfs(e[0], mul * e[1] % mod);
}
}
}
```

#### C++

```cpp

class Solution {
public:
vector<int> baseUnitConversions(vector<vector<int>>& conversions) {
const int mod = 1e9 + 7;
int n = conversions.size() + 1;
vector<vector<pair<int, int>>> g(n);
vector<int> ans(n);
for (const auto& e : conversions) {
g[e[0]].push_back({e[1], e[2]});
}
auto dfs = [&](this auto&& dfs, int s, long long mul) -> void {
ans[s] = mul;
for (auto [t, w] : g[s]) {
dfs(t, mul * w % mod);
}
};
dfs(0, 1);
return ans;
}
};
```

#### Go

```go
func baseUnitConversions(conversions [][]int) []int {
const mod = int(1e9 + 7)
n := len(conversions) + 1

g := make([][]struct{ t, w int }, n)
for _, e := range conversions {
s, t, w := e[0], e[1], e[2]
g[s] = append(g[s], struct{ t, w int }{t, w})
}

ans := make([]int, n)

var dfs func(s int, mul int)
dfs = func(s int, mul int) {
ans[s] = mul
for _, e := range g[s] {
dfs(e.t, mul*e.w%mod)
}
}

dfs(0, 1)
return ans
}
```

#### TypeScript

```ts
function baseUnitConversions(conversions: number[][]): number[] {
const mod = BigInt(1e9 + 7);
const n = conversions.length + 1;
const g: { t: number; w: number }[][] = Array.from({ length: n }, () => []);
for (const [s, t, w] of conversions) {
g[s].push({ t, w });
}
const ans: number[] = Array(n).fill(0);
const dfs = (s: number, mul: number) => {
ans[s] = mul;
for (const { t, w } of g[s]) {
dfs(t, Number((BigInt(mul) * BigInt(w)) % mod));
}
};
dfs(0, 1);
return ans;
}
```

<!-- tabs:end -->
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147 changes: 129 additions & 18 deletions solution/3500-3599/3528.Unit Conversion I/README_EN.md
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Expand Up @@ -33,8 +33,8 @@ tags:
<p><strong>Explanation:</strong></p>

<ul>
<li>Convert a single unit of type 0 into 2 units of type 1 using <code>conversions[0]</code>.</li>
<li>Convert a single unit of type 0 into 6 units of type 2 using <code>conversions[0]</code>, then <code>conversions[1]</code>.</li>
<li>Convert a single unit of type 0 into 2 units of type 1 using <code>conversions[0]</code>.</li>
<li>Convert a single unit of type 0 into 6 units of type 2 using <code>conversions[0]</code>, then <code>conversions[1]</code>.</li>
</ul>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3500-3599/3528.Unit%20Conversion%20I/images/example1.png" style="width: 545px; height: 118px;" /></div>

Expand All @@ -48,25 +48,25 @@ tags:
<p><strong>Explanation:</strong></p>

<ul>
<li>Convert a single unit of type 0 into 2 units of type 1 using <code>conversions[0]</code>.</li>
<li>Convert a single unit of type 0 into 3 units of type 2 using <code>conversions[1]</code>.</li>
<li>Convert a single unit of type 0 into 8 units of type 3 using <code>conversions[0]</code>, then <code>conversions[2]</code>.</li>
<li>Convert a single unit of type 0 into 10 units of type 4 using <code>conversions[0]</code>, then <code>conversions[3]</code>.</li>
<li>Convert a single unit of type 0 into 6 units of type 5 using <code>conversions[1]</code>, then <code>conversions[4]</code>.</li>
<li>Convert a single unit of type 0 into 30 units of type 6 using <code>conversions[0]</code>, <code>conversions[3]</code>, then <code>conversions[5]</code>.</li>
<li>Convert a single unit of type 0 into 24 units of type 7 using <code>conversions[1]</code>, <code>conversions[4]</code>, then <code>conversions[6]</code>.</li>
<li>Convert a single unit of type 0 into 2 units of type 1 using <code>conversions[0]</code>.</li>
<li>Convert a single unit of type 0 into 3 units of type 2 using <code>conversions[1]</code>.</li>
<li>Convert a single unit of type 0 into 8 units of type 3 using <code>conversions[0]</code>, then <code>conversions[2]</code>.</li>
<li>Convert a single unit of type 0 into 10 units of type 4 using <code>conversions[0]</code>, then <code>conversions[3]</code>.</li>
<li>Convert a single unit of type 0 into 6 units of type 5 using <code>conversions[1]</code>, then <code>conversions[4]</code>.</li>
<li>Convert a single unit of type 0 into 30 units of type 6 using <code>conversions[0]</code>, <code>conversions[3]</code>, then <code>conversions[5]</code>.</li>
<li>Convert a single unit of type 0 into 24 units of type 7 using <code>conversions[1]</code>, <code>conversions[4]</code>, then <code>conversions[6]</code>.</li>
</ul>
</div>

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

<ul>
<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
<li><code>conversions.length == n - 1</code></li>
<li><code>0 &lt;= sourceUnit<sub>i</sub>, targetUnit<sub>i</sub> &lt; n</code></li>
<li><code>1 &lt;= conversionFactor<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
<li>It is guaranteed that unit 0 can be converted into any other unit through a <strong>unique</strong> combination of conversions without using any conversions in the opposite direction.</li>
<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
<li><code>conversions.length == n - 1</code></li>
<li><code>0 &lt;= sourceUnit<sub>i</sub>, targetUnit<sub>i</sub> &lt; n</code></li>
<li><code>1 &lt;= conversionFactor<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
<li>It is guaranteed that unit 0 can be converted into any other unit through a <strong>unique</strong> combination of conversions without using any conversions in the opposite direction.</li>
</ul>

<!-- description:end -->
Expand All @@ -75,32 +75,143 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: DFS

Since the problem guarantees that unit 0 can be converted to any other unit through a unique conversion path, we can use Depth-First Search (DFS) to traverse all unit conversion relationships. Additionally, since the length of the $\textit{conversions}$ array is $n - 1$, representing $n - 1$ conversion relationships, we can treat the unit conversion relationships as a tree, where the root node is unit 0, and the other nodes are the other units.

We can use an adjacency list $g$ to represent the unit conversion relationships, where $g[i]$ represents the units that unit $i$ can convert to and the corresponding conversion factors.

Then, we start the DFS from the root node $0$, i.e., call the function $\textit{dfs}(s, \textit{mul})$, where $s$ represents the current unit, and $\textit{mul}$ represents the conversion factor from unit $0$ to unit $s$. Initially, $s = 0$, $\textit{mul} = 1$. In each recursion, we store the conversion factor $\textit{mul}$ of the current unit $s$ into the answer array, then traverse all adjacent units $t$ of the current unit $s$, and recursively call $\textit{dfs}(t, \textit{mul} \times w \mod (10^9 + 7))$, where $w$ is the conversion factor from unit $s$ to unit $t$.

Finally, we return the answer array.

The complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of units.

<!-- tabs:start -->

#### Python3

```python

class Solution:
def baseUnitConversions(self, conversions: List[List[int]]) -> List[int]:
def dfs(s: int, mul: int) -> None:
ans[s] = mul
for t, w in g[s]:
dfs(t, mul * w % mod)

mod = 10**9 + 7
n = len(conversions) + 1
g = [[] for _ in range(n)]
for s, t, w in conversions:
g[s].append((t, w))
ans = [0] * n
dfs(0, 1)
return ans
```

#### Java

```java

class Solution {
private final int mod = (int) 1e9 + 7;
private List<int[]>[] g;
private int[] ans;
private int n;

public int[] baseUnitConversions(int[][] conversions) {
n = conversions.length + 1;
g = new List[n];
Arrays.setAll(g, k -> new ArrayList<>());
ans = new int[n];
for (var e : conversions) {
g[e[0]].add(new int[] {e[1], e[2]});
}
dfs(0, 1);
return ans;
}

private void dfs(int s, long mul) {
ans[s] = (int) mul;
for (var e : g[s]) {
dfs(e[0], mul * e[1] % mod);
}
}
}
```

#### C++

```cpp

class Solution {
public:
vector<int> baseUnitConversions(vector<vector<int>>& conversions) {
const int mod = 1e9 + 7;
int n = conversions.size() + 1;
vector<vector<pair<int, int>>> g(n);
vector<int> ans(n);
for (const auto& e : conversions) {
g[e[0]].push_back({e[1], e[2]});
}
auto dfs = [&](this auto&& dfs, int s, long long mul) -> void {
ans[s] = mul;
for (auto [t, w] : g[s]) {
dfs(t, mul * w % mod);
}
};
dfs(0, 1);
return ans;
}
};
```

#### Go

```go
func baseUnitConversions(conversions [][]int) []int {
const mod = int(1e9 + 7)
n := len(conversions) + 1

g := make([][]struct{ t, w int }, n)
for _, e := range conversions {
s, t, w := e[0], e[1], e[2]
g[s] = append(g[s], struct{ t, w int }{t, w})
}

ans := make([]int, n)

var dfs func(s int, mul int)
dfs = func(s int, mul int) {
ans[s] = mul
for _, e := range g[s] {
dfs(e.t, mul*e.w%mod)
}
}

dfs(0, 1)
return ans
}
```

#### TypeScript

```ts
function baseUnitConversions(conversions: number[][]): number[] {
const mod = BigInt(1e9 + 7);
const n = conversions.length + 1;
const g: { t: number; w: number }[][] = Array.from({ length: n }, () => []);
for (const [s, t, w] of conversions) {
g[s].push({ t, w });
}
const ans: number[] = Array(n).fill(0);
const dfs = (s: number, mul: number) => {
ans[s] = mul;
for (const { t, w } of g[s]) {
dfs(t, Number((BigInt(mul) * BigInt(w)) % mod));
}
};
dfs(0, 1);
return ans;
}
```

<!-- tabs:end -->
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