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Aug 23, 2023
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feat: add solutions to lc problem: No.2143 #1493

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157 changes: 153 additions & 4 deletions solution/2100-2199/2143.Choose Numbers From Two Arrays in Range/README.md
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Expand Up @@ -71,30 +71,179 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一：动态规划**

我们定义 $f[i][j]$ 表示以第 $i$ 个元素结尾，且从 $nums1$ 中选取的数字和与从 $nums2$ 中选取的数字和之差为 $j$ 的平衡区间的个数。由于差值可能为负数，因此，我们统一将 $j$ 加上 $s_2 = \sum_{k=0}^{n-1}nums2[k]$，这样就可以保证 $j$ 为非负整数。

考虑 $f[i][j]$，我们可以单独将第 $i$ 个元素视为一个区间，那么 $f[i][nums1[i] + s_2]$ 和 $f[i][-nums2[i] + s_2]$ 都会增加 $1$。此外，如果 $i \gt 0$，我们也可以将第 $i$ 个元素添加到前面的某个区间中，我们在 $[0, s_1 + s_2]$ 范围内枚举 $j$，如果 $j \geq a$，那么 $f[i][j]$ 会增加 $f[i - 1][j - a]$，如果 $j + b \leq s_1 + s_2$，那么 $f[i][j]$ 会增加 $f[i - 1][j + b]$。

答案为 $\sum_{i=0}^{n-1}f[i][s_2]$。

时间复杂度 $O(n \times M)$，空间复杂度 $O(n \times M)$。其中 $n$ 和 $M$ 分别为数组 $nums1$ 的长度以及数字和的最大值。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def countSubranges(self, nums1: List[int], nums2: List[int]) -> int:
n = len(nums1)
s1, s2 = sum(nums1), sum(nums2)
f = [[0] * (s1 + s2 + 1) for _ in range(n)]
ans = 0
mod = 10**9 + 7
for i, (a, b) in enumerate(zip(nums1, nums2)):
f[i][a + s2] += 1
f[i][-b + s2] += 1
if i:
for j in range(s1 + s2 + 1):
if j >= a:
f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod
if j + b < s1 + s2 + 1:
f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod
ans = (ans + f[i][s2]) % mod
return ans
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
class Solution {
public int countSubranges(int[] nums1, int[] nums2) {
int n = nums1.length;
int s1 = Arrays.stream(nums1).sum();
int s2 = Arrays.stream(nums2).sum();
int[][] f = new int[n][s1 + s2 + 1];
int ans = 0;
final int mod = (int) 1e9 + 7;
for (int i = 0; i < n; ++i) {
int a = nums1[i], b = nums2[i];
f[i][a + s2]++;
f[i][-b + s2]++;
if (i > 0) {
for (int j = 0; j <= s1 + s2; ++j) {
if (j >= a) {
f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
}
if (j + b <= s1 + s2) {
f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
}
}
}
ans = (ans + f[i][s2]) % mod;
}
return ans;
}
}
```

### **C++**

```cpp
class Solution {
public:
int countSubranges(vector<int>& nums1, vector<int>& nums2) {
int n = nums1.size();
int s1 = accumulate(nums1.begin(), nums1.end(), 0);
int s2 = accumulate(nums2.begin(), nums2.end(), 0);
int f[n][s1 + s2 + 1];
memset(f, 0, sizeof(f));
int ans = 0;
const int mod = 1e9 + 7;
for (int i = 0; i < n; ++i) {
int a = nums1[i], b = nums2[i];
f[i][a + s2]++;
f[i][-b + s2]++;
if (i) {
for (int j = 0; j <= s1 + s2; ++j) {
if (j >= a) {
f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
}
if (j + b <= s1 + s2) {
f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
}
}
}
ans = (ans + f[i][s2]) % mod;
}
return ans;
}
};
```

### **TypeScript**
### **Go**

```go
func countSubranges(nums1 []int, nums2 []int) (ans int) {
n := len(nums1)
s1, s2 := sum(nums1), sum(nums2)
f := make([][]int, n)
for i := range f {
f[i] = make([]int, s1+s2+1)
}
const mod int = 1e9 + 7
for i, a := range nums1 {
b := nums2[i]
f[i][a+s2]++
f[i][-b+s2]++
if i > 0 {
for j := 0; j <= s1+s2; j++ {
if j >= a {
f[i][j] = (f[i][j] + f[i-1][j-a]) % mod
}
if j+b <= s1+s2 {
f[i][j] = (f[i][j] + f[i-1][j+b]) % mod
}
}
}
ans = (ans + f[i][s2]) % mod
}
return
}

func sum(nums []int) (ans int) {
for _, x := range nums {
ans += x
}
return
}
```

<!-- 这里可写当前语言的特殊实现逻辑 -->
### **TypeScript**

```ts

function countSubranges(nums1: number[], nums2: number[]): number {
const n = nums1.length;
const s1 = nums1.reduce((a, b) => a + b, 0);
const s2 = nums2.reduce((a, b) => a + b, 0);
const f: number[][] = Array(n)
.fill(0)
.map(() => Array(s1 + s2 + 1).fill(0));
const mod = 1e9 + 7;
let ans = 0;
for (let i = 0; i < n; ++i) {
const [a, b] = [nums1[i], nums2[i]];
f[i][a + s2]++;
f[i][-b + s2]++;
if (i) {
for (let j = 0; j <= s1 + s2; ++j) {
if (j >= a) {
f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
}
if (j + b <= s1 + s2) {
f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
}
}
}
ans = (ans + f[i][s2]) % mod;
}
return ans;
}
```

### **...**
Expand Down
145 changes: 143 additions & 2 deletions solution/2100-2199/2143.Choose Numbers From Two Arrays in Range/README_EN.md
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Expand Up @@ -72,19 +72,160 @@ In the second balanced range, we choose nums2[1] and in the third balanced range
### **Python3**

```python

class Solution:
def countSubranges(self, nums1: List[int], nums2: List[int]) -> int:
n = len(nums1)
s1, s2 = sum(nums1), sum(nums2)
f = [[0] * (s1 + s2 + 1) for _ in range(n)]
ans = 0
mod = 10**9 + 7
for i, (a, b) in enumerate(zip(nums1, nums2)):
f[i][a + s2] += 1
f[i][-b + s2] += 1
if i:
for j in range(s1 + s2 + 1):
if j >= a:
f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod
if j + b < s1 + s2 + 1:
f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod
ans = (ans + f[i][s2]) % mod
return ans
```

### **Java**

```java
class Solution {
public int countSubranges(int[] nums1, int[] nums2) {
int n = nums1.length;
int s1 = Arrays.stream(nums1).sum();
int s2 = Arrays.stream(nums2).sum();
int[][] f = new int[n][s1 + s2 + 1];
int ans = 0;
final int mod = (int) 1e9 + 7;
for (int i = 0; i < n; ++i) {
int a = nums1[i], b = nums2[i];
f[i][a + s2]++;
f[i][-b + s2]++;
if (i > 0) {
for (int j = 0; j <= s1 + s2; ++j) {
if (j >= a) {
f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
}
if (j + b <= s1 + s2) {
f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
}
}
}
ans = (ans + f[i][s2]) % mod;
}
return ans;
}
}
```

### **C++**

```cpp
class Solution {
public:
int countSubranges(vector<int>& nums1, vector<int>& nums2) {
int n = nums1.size();
int s1 = accumulate(nums1.begin(), nums1.end(), 0);
int s2 = accumulate(nums2.begin(), nums2.end(), 0);
int f[n][s1 + s2 + 1];
memset(f, 0, sizeof(f));
int ans = 0;
const int mod = 1e9 + 7;
for (int i = 0; i < n; ++i) {
int a = nums1[i], b = nums2[i];
f[i][a + s2]++;
f[i][-b + s2]++;
if (i) {
for (int j = 0; j <= s1 + s2; ++j) {
if (j >= a) {
f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
}
if (j + b <= s1 + s2) {
f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
}
}
}
ans = (ans + f[i][s2]) % mod;
}
return ans;
}
};
```

### **Go**

```go
func countSubranges(nums1 []int, nums2 []int) (ans int) {
n := len(nums1)
s1, s2 := sum(nums1), sum(nums2)
f := make([][]int, n)
for i := range f {
f[i] = make([]int, s1+s2+1)
}
const mod int = 1e9 + 7
for i, a := range nums1 {
b := nums2[i]
f[i][a+s2]++
f[i][-b+s2]++
if i > 0 {
for j := 0; j <= s1+s2; j++ {
if j >= a {
f[i][j] = (f[i][j] + f[i-1][j-a]) % mod
}
if j+b <= s1+s2 {
f[i][j] = (f[i][j] + f[i-1][j+b]) % mod
}
}
}
ans = (ans + f[i][s2]) % mod
}
return
}

func sum(nums []int) (ans int) {
for _, x := range nums {
ans += x
}
return
}
```

### **TypeScript**

```ts

function countSubranges(nums1: number[], nums2: number[]): number {
const n = nums1.length;
const s1 = nums1.reduce((a, b) => a + b, 0);
const s2 = nums2.reduce((a, b) => a + b, 0);
const f: number[][] = Array(n)
.fill(0)
.map(() => Array(s1 + s2 + 1).fill(0));
const mod = 1e9 + 7;
let ans = 0;
for (let i = 0; i < n; ++i) {
const [a, b] = [nums1[i], nums2[i]];
f[i][a + s2]++;
f[i][-b + s2]++;
if (i) {
for (let j = 0; j <= s1 + s2; ++j) {
if (j >= a) {
f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
}
if (j + b <= s1 + s2) {
f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
}
}
}
ans = (ans + f[i][s2]) % mod;
}
return ans;
}
```

### **...**
Expand Down
29 changes: 29 additions & 0 deletions solution/2100-2199/2143.Choose Numbers From Two Arrays in Range/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -0,0 +1,29 @@
class Solution {
public:
int countSubranges(vector<int>& nums1, vector<int>& nums2) {
int n = nums1.size();
int s1 = accumulate(nums1.begin(), nums1.end(), 0);
int s2 = accumulate(nums2.begin(), nums2.end(), 0);
int f[n][s1 + s2 + 1];
memset(f, 0, sizeof(f));
int ans = 0;
const int mod = 1e9 + 7;
for (int i = 0; i < n; ++i) {
int a = nums1[i], b = nums2[i];
f[i][a + s2]++;
f[i][-b + s2]++;
if (i) {
for (int j = 0; j <= s1 + s2; ++j) {
if (j >= a) {
f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
}
if (j + b <= s1 + s2) {
f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
}
}
}
ans = (ans + f[i][s2]) % mod;
}
return ans;
}
};
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