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feat: add solutions to lc problem: No.2145 (doocs#3511)
No.2145.Count the Hidden Sequences
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‎solution/2100-2199/2145.Count the Hidden Sequences/README.md

Lines changed: 41 additions & 21 deletions
Original file line numberDiff line numberDiff line change
@@ -85,7 +85,13 @@ tags:
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<!-- solution:start -->
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### 方法一
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### 方法一：前缀和
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由于数组 $\textit{differences}$ 已经确定，那么数组 $\textit{hidden}$ 的元素最大值与最小值之差也是固定的，我们只要确保差值不超过 $\textit{upper} - \textit{lower}$ 即可。
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我们不妨假设数组 $\textit{hidden}$ 的第一个元素为 $0$，那么 $\textit{hidden}[i] = \textit{hidden}[i - 1] + \textit{differences}[i - 1]$，其中 $1 \leq i \leq n$。记数组 $\textit{hidden}$ 的最大值为 $mx$，最小值为 $mi$，如果 $mx - mi \leq \textit{upper} - \textit{lower}$，那么我们就可以构造出一个合法的 $\textit{hidden}$ 数组，可以构造的个数为 $\textit{upper} - \textit{lower} - (mx - mi) + 1$。否则，无法构造出合法的 $\textit{hidden}$ 数组，返回 $0$。
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时间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{differences}$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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@@ -94,26 +100,26 @@ tags:
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```python
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class Solution:
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def numberOfArrays(self, differences: List[int], lower: int, upper: int) -> int:
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num = mi = mx = 0
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x = mi = mx = 0
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for d in differences:
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num += d
100-
mi = min(mi, num)
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mx = max(mx, num)
102-
return max(0, upper - lower - (mx - mi) + 1)
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x += d
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mi = min(mi, x)
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mx = max(mx, x)
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return max(upper - lower - (mx - mi) + 1, 0)
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```
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#### Java
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```java
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class Solution {
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public int numberOfArrays(int[] differences, int lower, int upper) {
110-
long num = 0, mi = 0, mx = 0;
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long x = 0, mi = 0, mx = 0;
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for (int d : differences) {
112-
num += d;
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mi = Math.min(mi, num);
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mx = Math.max(mx, num);
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x += d;
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mi = Math.min(mi, x);
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mx = Math.max(mx, x);
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}
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return Math.max(0, (int) (upper - lower - (mx - mi) + 1));
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return (int) Math.max(upper - lower - (mx - mi) + 1, 0);
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}
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}
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```
@@ -124,13 +130,13 @@ class Solution {
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class Solution {
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public:
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int numberOfArrays(vector<int>& differences, int lower, int upper) {
127-
long long num = 0, mi = 0, mx = 0;
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for (int& d : differences) {
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num += d;
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mi = min(mi, num);
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mx = max(mx, num);
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long long x = 0, mi = 0, mx = 0;
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for (int d : differences) {
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x += d;
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mi = min(mi, x);
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mx = max(mx, x);
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}
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return max(0, (int) (upper - lower - (mx - mi) + 1));
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return max(upper - lower - (mx - mi) + 1, 0LL);
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}
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};
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```
@@ -139,16 +145,30 @@ public:
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```go
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func numberOfArrays(differences []int, lower int, upper int) int {
142-
num, mi, mx := 0, 0, 0
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x, mi, mx := 0, 0, 0
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for _, d := range differences {
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num += d
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mi = min(mi, num)
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mx = max(mx, num)
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x += d
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mi = min(mi, x)
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mx = max(mx, x)
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}
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return max(0, upper-lower-(mx-mi)+1)
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}
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```
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#### TypeScript
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```ts
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function numberOfArrays(differences: number[], lower: number, upper: number): number {
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let [x, mi, mx] = [0, 0, 0];
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for (const d of differences) {
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x += d;
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mi = Math.min(mi, x);
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mx = Math.max(mx, x);
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}
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return Math.max(0, upper - lower - (mx - mi) + 1);
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/2100-2199/2145.Count the Hidden Sequences/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -86,7 +86,13 @@ Thus, we return 4.
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Prefix Sum
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Since the array $\textit{differences}$ is already determined, the difference between the maximum and minimum values of the elements in the array $\textit{hidden}$ is also fixed. We just need to ensure that this difference does not exceed $\textit{upper} - \textit{lower}$.
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Let's assume the first element of the array $\textit{hidden}$ is $0$. Then, $\textit{hidden}[i] = \textit{hidden}[i - 1] + \textit{differences}[i - 1]$, where $1 \leq i \leq n$. Let the maximum value of the array $\textit{hidden}$ be $mx$ and the minimum value be $mi$. If $mx - mi \leq \textit{upper} - \textit{lower}$, then we can construct a valid $\textit{hidden}$ array. The number of possible constructions is $\textit{upper} - \textit{lower} - (mx - mi) + 1$. Otherwise, it is impossible to construct a valid $\textit{hidden}$ array, and we return $0$.
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The time complexity is $O(n)$, where $n$ is the length of the array $\textit{differences}$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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@@ -95,26 +101,26 @@ Thus, we return 4.
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```python
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class Solution:
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def numberOfArrays(self, differences: List[int], lower: int, upper: int) -> int:
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num = mi = mx = 0
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x = mi = mx = 0
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for d in differences:
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num += d
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mi = min(mi, num)
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mx = max(mx, num)
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return max(0, upper - lower - (mx - mi) + 1)
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x += d
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mi = min(mi, x)
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mx = max(mx, x)
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return max(upper - lower - (mx - mi) + 1, 0)
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```
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#### Java
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```java
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class Solution {
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public int numberOfArrays(int[] differences, int lower, int upper) {
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long num = 0, mi = 0, mx = 0;
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long x = 0, mi = 0, mx = 0;
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for (int d : differences) {
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num += d;
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mi = Math.min(mi, num);
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mx = Math.max(mx, num);
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x += d;
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mi = Math.min(mi, x);
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mx = Math.max(mx, x);
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}
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return Math.max(0, (int) (upper - lower - (mx - mi) + 1));
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return (int) Math.max(upper - lower - (mx - mi) + 1, 0);
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}
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}
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```
@@ -125,13 +131,13 @@ class Solution {
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class Solution {
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public:
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int numberOfArrays(vector<int>& differences, int lower, int upper) {
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long long num = 0, mi = 0, mx = 0;
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for (int& d : differences) {
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num += d;
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mi = min(mi, num);
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mx = max(mx, num);
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long long x = 0, mi = 0, mx = 0;
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for (int d : differences) {
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x += d;
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mi = min(mi, x);
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mx = max(mx, x);
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}
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return max(0, (int) (upper - lower - (mx - mi) + 1));
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return max(upper - lower - (mx - mi) + 1, 0LL);
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}
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};
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```
@@ -140,16 +146,30 @@ public:
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```go
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func numberOfArrays(differences []int, lower int, upper int) int {
143-
num, mi, mx := 0, 0, 0
149+
x, mi, mx := 0, 0, 0
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for _, d := range differences {
145-
num += d
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mi = min(mi, num)
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mx = max(mx, num)
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x += d
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mi = min(mi, x)
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mx = max(mx, x)
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}
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return max(0, upper-lower-(mx-mi)+1)
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}
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```
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#### TypeScript
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```ts
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function numberOfArrays(differences: number[], lower: number, upper: number): number {
163+
let [x, mi, mx] = [0, 0, 0];
164+
for (const d of differences) {
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x += d;
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mi = Math.min(mi, x);
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mx = Math.max(mx, x);
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}
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return Math.max(0, upper - lower - (mx - mi) + 1);
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/2100-2199/2145.Count the Hidden Sequences/Solution.cpp

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class Solution {
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public:
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int numberOfArrays(vector<int>& differences, int lower, int upper) {
4-
long long num = 0, mi = 0, mx = 0;
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for (int& d : differences) {
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num += d;
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mi = min(mi, num);
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mx = max(mx, num);
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long long x = 0, mi = 0, mx = 0;
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for (int d : differences) {
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x += d;
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mi = min(mi, x);
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mx = max(mx, x);
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}
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return max(0, (int) (upper - lower - (mx - mi) + 1));
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return max(upper - lower - (mx - mi) + 1, 0LL);
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}
12-
};
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};

‎solution/2100-2199/2145.Count the Hidden Sequences/Solution.go

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Original file line numberDiff line numberDiff line change
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func numberOfArrays(differences []int, lower int, upper int) int {
2-
num, mi, mx := 0, 0, 0
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x, mi, mx := 0, 0, 0
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for _, d := range differences {
4-
num += d
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mi = min(mi, num)
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mx = max(mx, num)
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x += d
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mi = min(mi, x)
6+
mx = max(mx, x)
77
}
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return max(0, upper-lower-(mx-mi)+1)
9-
}
9+
}

‎solution/2100-2199/2145.Count the Hidden Sequences/Solution.java

Lines changed: 6 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -1,11 +1,11 @@
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class Solution {
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public int numberOfArrays(int[] differences, int lower, int upper) {
3-
long num = 0, mi = 0, mx = 0;
3+
long x = 0, mi = 0, mx = 0;
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for (int d : differences) {
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num += d;
6-
mi = Math.min(mi, num);
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mx = Math.max(mx, num);
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x += d;
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mi = Math.min(mi, x);
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mx = Math.max(mx, x);
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}
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return Math.max(0, (int) (upper - lower - (mx - mi) + 1));
9+
return (int) Math.max(upper - lower - (mx - mi) + 1, 0);
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}
11-
}
11+
}

‎solution/2100-2199/2145.Count the Hidden Sequences/Solution.py

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Original file line numberDiff line numberDiff line change
@@ -1,8 +1,8 @@
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class Solution:
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def numberOfArrays(self, differences: List[int], lower: int, upper: int) -> int:
3-
num = mi = mx = 0
3+
x = mi = mx = 0
44
for d in differences:
5-
num += d
6-
mi = min(mi, num)
7-
mx = max(mx, num)
8-
return max(0, upper - lower - (mx - mi) + 1)
5+
x += d
6+
mi = min(mi, x)
7+
mx = max(mx, x)
8+
return max(upper - lower - (mx - mi) + 1, 0)

‎solution/2100-2199/2145.Count the Hidden Sequences/Solution.ts

Lines changed: 9 additions & 0 deletions
Original file line numberDiff line numberDiff line change
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1+
function numberOfArrays(differences: number[], lower: number, upper: number): number {
2+
let [x, mi, mx] = [0, 0, 0];
3+
for (const d of differences) {
4+
x += d;
5+
mi = Math.min(mi, x);
6+
mx = Math.max(mx, x);
7+
}
8+
return Math.max(0, upper - lower - (mx - mi) + 1);
9+
}

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