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feat: add solutions to lc problem: No.1498
No.1498.Number of Subsequences That Satisfy the Given Sum Condition
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‎solution/1400-1499/1498.Number of Subsequences That Satisfy the Given Sum Condition/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：排序 + 枚举贡献 + 二分查找**
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由于题目中描述的是子序列，并且涉及到最小元素与最大元素的和，因此我们可以先对数组 `nums` 进行排序。
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然后我们枚举最小元素 $nums[i]$，对于每个 $nums[i]$，我们可以在 $nums[i + 1]$ 到 $nums[n - 1]$ 中找到最大元素 $nums[j]$，使得 $nums[i] + nums[j] \leq target$，此时满足条件的子序列数目为 $2^{j - i}$，其中 $2^{j - i}$ 表示从 $nums[i + 1]$ 到 $nums[j]$ 的所有子序列的数目。我们将所有的子序列数目累加即可。
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时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 `nums` 的长度。
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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 numSubseq(self, nums: List[int], target: int) -> int:
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mod = 10**9 + 7
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nums.sort()
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n = len(nums)
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f = [1] + [0] * n
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for i in range(1, n + 1):
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f[i] = f[i - 1] * 2 % mod
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ans = 0
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for i, x in enumerate(nums):
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if x * 2 > target:
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break
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j = bisect_right(nums, target - x, i + 1) - 1
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ans = (ans + f[j - i]) % 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 numSubseq(int[] nums, int target) {
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Arrays.sort(nums);
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final int mod = (int) 1e9 + 7;
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int n = nums.length;
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int[] f = new int[n + 1];
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f[0] = 1;
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for (int i = 1; i <= n; ++i) {
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f[i] = (f[i - 1] * 2) % mod;
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}
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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if (nums[i] * 2L > target) {
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break;
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}
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int j = search(nums, target - nums[i], i + 1) - 1;
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ans = (ans + f[j - i]) % mod;
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}
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return ans;
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}
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private int search(int[] nums, int x, int left) {
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int right = nums.length;
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while (left < right) {
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int mid = (left + right) >> 1;
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if (nums[mid] > x) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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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 numSubseq(vector<int>& nums, int target) {
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sort(nums.begin(), nums.end());
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const int mod = 1e9 + 7;
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int n = nums.size();
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int f[n + 1];
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f[0] = 1;
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for (int i = 1; i <= n; ++i) {
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f[i] = (f[i - 1] * 2) % mod;
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}
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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if (nums[i] * 2L > target) {
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break;
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}
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int j = upper_bound(nums.begin() + i + 1, nums.end(), target - nums[i]) - nums.begin() - 1;
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ans = (ans + f[j - i]) % 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 numSubseq(nums []int, target int) (ans int) {
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sort.Ints(nums)
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n := len(nums)
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f := make([]int, n+1)
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f[0] = 1
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const mod int = 1e9 + 7
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for i := 1; i <= n; i++ {
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f[i] = f[i-1] * 2 % mod
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}
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for i, x := range nums {
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if x*2 > target {
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break
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}
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j := sort.SearchInts(nums[i+1:], target-x+1) + i
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ans = (ans + f[j-i]) % mod
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}
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return
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}
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```
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### **...**

‎solution/1400-1499/1498.Number of Subsequences That Satisfy the Given Sum Condition/README_EN.md

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@@ -55,13 +55,110 @@ Number of valid subsequences (63 - 2 = 61).
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### **Python3**
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```python
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class Solution:
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def numSubseq(self, nums: List[int], target: int) -> int:
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mod = 10**9 + 7
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nums.sort()
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n = len(nums)
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f = [1] + [0] * n
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for i in range(1, n + 1):
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f[i] = f[i - 1] * 2 % mod
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ans = 0
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for i, x in enumerate(nums):
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if x * 2 > target:
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break
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j = bisect_right(nums, target - x, i + 1) - 1
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ans = (ans + f[j - i]) % 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 numSubseq(int[] nums, int target) {
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Arrays.sort(nums);
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final int mod = (int) 1e9 + 7;
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int n = nums.length;
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int[] f = new int[n + 1];
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f[0] = 1;
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for (int i = 1; i <= n; ++i) {
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f[i] = (f[i - 1] * 2) % mod;
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}
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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if (nums[i] * 2L > target) {
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break;
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}
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int j = search(nums, target - nums[i], i + 1) - 1;
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ans = (ans + f[j - i]) % mod;
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}
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return ans;
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}
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private int search(int[] nums, int x, int left) {
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int right = nums.length;
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while (left < right) {
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int mid = (left + right) >> 1;
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if (nums[mid] > x) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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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 numSubseq(vector<int>& nums, int target) {
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sort(nums.begin(), nums.end());
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const int mod = 1e9 + 7;
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int n = nums.size();
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int f[n + 1];
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f[0] = 1;
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for (int i = 1; i <= n; ++i) {
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f[i] = (f[i - 1] * 2) % mod;
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}
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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if (nums[i] * 2L > target) {
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break;
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}
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int j = upper_bound(nums.begin() + i + 1, nums.end(), target - nums[i]) - nums.begin() - 1;
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ans = (ans + f[j - i]) % 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 numSubseq(nums []int, target int) (ans int) {
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sort.Ints(nums)
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n := len(nums)
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f := make([]int, n+1)
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f[0] = 1
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const mod int = 1e9 + 7
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for i := 1; i <= n; i++ {
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f[i] = f[i-1] * 2 % mod
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}
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for i, x := range nums {
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if x*2 > target {
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break
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}
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j := sort.SearchInts(nums[i+1:], target-x+1) + i
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ans = (ans + f[j-i]) % mod
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}
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return
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}
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```
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### **...**

‎solution/1400-1499/1498.Number of Subsequences That Satisfy the Given Sum Condition/Solution.cpp

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class Solution {
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public:
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int numSubseq(vector<int>& nums, int target) {
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sort(nums.begin(), nums.end());
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const int mod = 1e9 + 7;
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int n = nums.size();
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int f[n + 1];
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f[0] = 1;
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for (int i = 1; i <= n; ++i) {
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f[i] = (f[i - 1] * 2) % mod;
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}
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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if (nums[i] * 2L > target) {
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break;
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}
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int j = upper_bound(nums.begin() + i + 1, nums.end(), target - nums[i]) - nums.begin() - 1;
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ans = (ans + f[j - i]) % mod;
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}
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return ans;
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}
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};

‎solution/1400-1499/1498.Number of Subsequences That Satisfy the Given Sum Condition/Solution.go

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func numSubseq(nums []int, target int) (ans int) {
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sort.Ints(nums)
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n := len(nums)
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f := make([]int, n+1)
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f[0] = 1
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const mod int = 1e9 + 7
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for i := 1; i <= n; i++ {
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f[i] = f[i-1] * 2 % mod
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}
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for i, x := range nums {
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if x*2 > target {
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break
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}
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j := sort.SearchInts(nums[i+1:], target-x+1) + i
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ans = (ans + f[j-i]) % mod
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}
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return
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}

‎solution/1400-1499/1498.Number of Subsequences That Satisfy the Given Sum Condition/Solution.java

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class Solution {
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public int numSubseq(int[] nums, int target) {
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Arrays.sort(nums);
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final int mod = (int) 1e9 + 7;
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int n = nums.length;
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int[] f = new int[n + 1];
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f[0] = 1;
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for (int i = 1; i <= n; ++i) {
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f[i] = (f[i - 1] * 2) % mod;
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}
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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if (nums[i] * 2L > target) {
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break;
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}
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int j = search(nums, target - nums[i], i + 1) - 1;
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ans = (ans + f[j - i]) % mod;
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}
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return ans;
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}
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private int search(int[] nums, int x, int left) {
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int right = nums.length;
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while (left < right) {
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int mid = (left + right) >> 1;
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if (nums[mid] > x) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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}
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}

‎solution/1400-1499/1498.Number of Subsequences That Satisfy the Given Sum Condition/Solution.py

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class Solution:
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def numSubseq(self, nums: List[int], target: int) -> int:
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mod = 10**9 + 7
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nums.sort()
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n = len(nums)
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f = [1] + [0] * n
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for i in range(1, n + 1):
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f[i] = f[i - 1] * 2 % mod
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ans = 0
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for i, x in enumerate(nums):
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if x * 2 > target:
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break
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j = bisect_right(nums, target - x, i + 1) - 1
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ans = (ans + f[j - i]) % mod
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return ans

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