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feat: add solutions to lc problem: No.2143 (doocs#1493)
No.2143.Choose Numbers From Two Arrays in Range
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‎solution/2100-2199/2143.Choose Numbers From Two Arrays in Range/README.md

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Original file line numberDiff line numberDiff line change
@@ -71,30 +71,179 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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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。
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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 $f[i - 1][j + b]$。
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答案为 $\sum_{i=0}^{n-1}f[i][s_2]$。
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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。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def countSubranges(self, nums1: List[int], nums2: List[int]) -> int:
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n = len(nums1)
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s1, s2 = sum(nums1), sum(nums2)
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f = [[0] * (s1 + s2 + 1) for _ in range(n)]
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ans = 0
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mod = 10**9 + 7
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for i, (a, b) in enumerate(zip(nums1, nums2)):
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f[i][a + s2] += 1
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f[i][-b + s2] += 1
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if i:
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for j in range(s1 + s2 + 1):
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if j >= a:
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f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod
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if j + b < s1 + s2 + 1:
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f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod
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ans = (ans + f[i][s2]) % mod
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int countSubranges(int[] nums1, int[] nums2) {
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int n = nums1.length;
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int s1 = Arrays.stream(nums1).sum();
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int s2 = Arrays.stream(nums2).sum();
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int[][] f = new int[n][s1 + s2 + 1];
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int ans = 0;
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final int mod = (int) 1e9 + 7;
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for (int i = 0; i < n; ++i) {
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int a = nums1[i], b = nums2[i];
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f[i][a + s2]++;
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f[i][-b + s2]++;
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if (i > 0) {
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for (int j = 0; j <= s1 + s2; ++j) {
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if (j >= a) {
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f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
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}
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if (j + b <= s1 + s2) {
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f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
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}
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}
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}
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ans = (ans + f[i][s2]) % mod;
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countSubranges(vector<int>& nums1, vector<int>& nums2) {
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int n = nums1.size();
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int s1 = accumulate(nums1.begin(), nums1.end(), 0);
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int s2 = accumulate(nums2.begin(), nums2.end(), 0);
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int f[n][s1 + s2 + 1];
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memset(f, 0, sizeof(f));
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int ans = 0;
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const int mod = 1e9 + 7;
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for (int i = 0; i < n; ++i) {
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int a = nums1[i], b = nums2[i];
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f[i][a + s2]++;
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f[i][-b + s2]++;
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if (i) {
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for (int j = 0; j <= s1 + s2; ++j) {
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if (j >= a) {
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f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
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}
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if (j + b <= s1 + s2) {
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f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
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}
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}
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}
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ans = (ans + f[i][s2]) % mod;
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}
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return ans;
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}
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};
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```
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### **TypeScript**
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### **Go**
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```go
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func countSubranges(nums1 []int, nums2 []int) (ans int) {
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n := len(nums1)
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s1, s2 := sum(nums1), sum(nums2)
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f := make([][]int, n)
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for i := range f {
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f[i] = make([]int, s1+s2+1)
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}
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const mod int = 1e9 + 7
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for i, a := range nums1 {
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b := nums2[i]
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f[i][a+s2]++
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f[i][-b+s2]++
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if i > 0 {
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for j := 0; j <= s1+s2; j++ {
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if j >= a {
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f[i][j] = (f[i][j] + f[i-1][j-a]) % mod
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}
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if j+b <= s1+s2 {
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f[i][j] = (f[i][j] + f[i-1][j+b]) % mod
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}
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}
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}
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ans = (ans + f[i][s2]) % mod
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}
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return
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}
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func sum(nums []int) (ans int) {
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for _, x := range nums {
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ans += x
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}
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return
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}
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```
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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### **TypeScript**
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```ts
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function countSubranges(nums1: number[], nums2: number[]): number {
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const n = nums1.length;
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const s1 = nums1.reduce((a, b) => a + b, 0);
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const s2 = nums2.reduce((a, b) => a + b, 0);
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const f: number[][] = Array(n)
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.fill(0)
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.map(() => Array(s1 + s2 + 1).fill(0));
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const mod = 1e9 + 7;
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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const [a, b] = [nums1[i], nums2[i]];
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f[i][a + s2]++;
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f[i][-b + s2]++;
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if (i) {
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for (let j = 0; j <= s1 + s2; ++j) {
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if (j >= a) {
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f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
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}
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if (j + b <= s1 + s2) {
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f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
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}
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}
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}
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ans = (ans + f[i][s2]) % mod;
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}
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return ans;
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}
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```
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### **...**

‎solution/2100-2199/2143.Choose Numbers From Two Arrays in Range/README_EN.md

Lines changed: 143 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -72,19 +72,160 @@ In the second balanced range, we choose nums2[1] and in the third balanced range
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### **Python3**
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```python
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class Solution:
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def countSubranges(self, nums1: List[int], nums2: List[int]) -> int:
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n = len(nums1)
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s1, s2 = sum(nums1), sum(nums2)
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f = [[0] * (s1 + s2 + 1) for _ in range(n)]
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ans = 0
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mod = 10**9 + 7
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for i, (a, b) in enumerate(zip(nums1, nums2)):
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f[i][a + s2] += 1
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f[i][-b + s2] += 1
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if i:
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for j in range(s1 + s2 + 1):
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if j >= a:
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f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod
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if j + b < s1 + s2 + 1:
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f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod
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ans = (ans + f[i][s2]) % mod
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return ans
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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 countSubranges(int[] nums1, int[] nums2) {
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int n = nums1.length;
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int s1 = Arrays.stream(nums1).sum();
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int s2 = Arrays.stream(nums2).sum();
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int[][] f = new int[n][s1 + s2 + 1];
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int ans = 0;
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final int mod = (int) 1e9 + 7;
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for (int i = 0; i < n; ++i) {
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int a = nums1[i], b = nums2[i];
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f[i][a + s2]++;
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f[i][-b + s2]++;
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if (i > 0) {
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for (int j = 0; j <= s1 + s2; ++j) {
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if (j >= a) {
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f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
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}
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if (j + b <= s1 + s2) {
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f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
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}
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}
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}
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ans = (ans + f[i][s2]) % mod;
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countSubranges(vector<int>& nums1, vector<int>& nums2) {
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int n = nums1.size();
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int s1 = accumulate(nums1.begin(), nums1.end(), 0);
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int s2 = accumulate(nums2.begin(), nums2.end(), 0);
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int f[n][s1 + s2 + 1];
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memset(f, 0, sizeof(f));
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int ans = 0;
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const int mod = 1e9 + 7;
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for (int i = 0; i < n; ++i) {
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int a = nums1[i], b = nums2[i];
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f[i][a + s2]++;
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f[i][-b + s2]++;
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if (i) {
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for (int j = 0; j <= s1 + s2; ++j) {
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if (j >= a) {
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f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
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}
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if (j + b <= s1 + s2) {
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f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
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}
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}
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}
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ans = (ans + f[i][s2]) % mod;
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func countSubranges(nums1 []int, nums2 []int) (ans int) {
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n := len(nums1)
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s1, s2 := sum(nums1), sum(nums2)
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f := make([][]int, n)
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for i := range f {
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f[i] = make([]int, s1+s2+1)
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}
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const mod int = 1e9 + 7
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for i, a := range nums1 {
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b := nums2[i]
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f[i][a+s2]++
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f[i][-b+s2]++
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if i > 0 {
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for j := 0; j <= s1+s2; j++ {
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if j >= a {
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f[i][j] = (f[i][j] + f[i-1][j-a]) % mod
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}
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if j+b <= s1+s2 {
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f[i][j] = (f[i][j] + f[i-1][j+b]) % mod
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}
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}
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}
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ans = (ans + f[i][s2]) % mod
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}
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return
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}
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func sum(nums []int) (ans int) {
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for _, x := range nums {
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ans += x
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function countSubranges(nums1: number[], nums2: number[]): number {
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const n = nums1.length;
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const s1 = nums1.reduce((a, b) => a + b, 0);
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const s2 = nums2.reduce((a, b) => a + b, 0);
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const f: number[][] = Array(n)
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.fill(0)
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.map(() => Array(s1 + s2 + 1).fill(0));
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const mod = 1e9 + 7;
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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const [a, b] = [nums1[i], nums2[i]];
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f[i][a + s2]++;
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f[i][-b + s2]++;
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if (i) {
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for (let j = 0; j <= s1 + s2; ++j) {
217+
if (j >= a) {
218+
f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
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}
220+
if (j + b <= s1 + s2) {
221+
f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
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}
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}
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}
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ans = (ans + f[i][s2]) % mod;
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}
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return ans;
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}
88229
```
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### **...**

‎solution/2100-2199/2143.Choose Numbers From Two Arrays in Range/Solution.cpp

Lines changed: 29 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,29 @@
1+
class Solution {
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public:
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int countSubranges(vector<int>& nums1, vector<int>& nums2) {
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int n = nums1.size();
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int s1 = accumulate(nums1.begin(), nums1.end(), 0);
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int s2 = accumulate(nums2.begin(), nums2.end(), 0);
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int f[n][s1 + s2 + 1];
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memset(f, 0, sizeof(f));
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int ans = 0;
10+
const int mod = 1e9 + 7;
11+
for (int i = 0; i < n; ++i) {
12+
int a = nums1[i], b = nums2[i];
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f[i][a + s2]++;
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f[i][-b + s2]++;
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if (i) {
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for (int j = 0; j <= s1 + s2; ++j) {
17+
if (j >= a) {
18+
f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
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}
20+
if (j + b <= s1 + s2) {
21+
f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
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}
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}
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}
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ans = (ans + f[i][s2]) % mod;
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}
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return ans;
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}
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};

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