Number of Subsequences That Satisfy the Given Sum Condition
<p>You are given an array of integers <code>nums</code> and an integer <code>target</code>.</p>

<p>Return <em>the number of <strong>non-empty</strong> subsequences of </em><code>nums</code><em> such that the sum of the minimum and maximum element on it is less or equal to </em><code>target</code>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,5,6,7], target = 9
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are 4 subsequences that satisfy the condition.
[3] -&gt; Min value + max value &lt;= target (3 + 3 &lt;= 9)
[3,5] -&gt; (3 + 5 &lt;= 9)
[3,5,6] -&gt; (3 + 6 &lt;= 9)
[3,6] -&gt; (3 + 6 &lt;= 9)
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,3,6,8], target = 10
<strong>Output:</strong> 6
<strong>Explanation:</strong> There are 6 subsequences that satisfy the condition. (nums can have repeated numbers).
[3] , [3] , [3,3], [3,6] , [3,6] , [3,3,6]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3,3,4,6,7], target = 12
<strong>Output:</strong> 61
<strong>Explanation:</strong> There are 63 non-empty subsequences, two of them do not satisfy the condition ([6,7], [7]).
Number of valid subsequences (63 - 2 = 61).
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
	<li><code>1 &lt;= target &lt;= 10<sup>6</sup></code></li>
</ul>
Array; Two Pointers; Binary Search; Sorting
class Solution {
public:
    int numSubseq(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        const int mod = 1e9 + 7;
        int n = nums.size();
        int f[n + 1];
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            f[i] = (f[i - 1] * 2) % mod;
        }
        int ans = 0;
        for (int i = 0; i < n && nums[i] * 2 <= target; ++i) {
            int j = upper_bound(nums.begin() + i + 1, nums.end(), target - nums[i]) - nums.begin() - 1;
            ans = (ans + f[j - i]) % mod;
        }
        return ans;
    }
};