Number of Subsequences That Satisfy the Given Sum Condition

Given an array of integers nums and an integer target.

Return the number of non-empty subsequences of nums such that the sum of the minimum and maximum element on it is less or equal than target.

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

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

Example 2:

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

Example 3:

Input: nums = [2,3,3,4,6,7], target = 12
Output: 61
Explanation: There are 63 non-empty subsequences, two of them don't satisfy the condition ([6,7], [7]).
Number of valid subsequences (63 - 2 = 61).

Example 4:

Input: nums = [5,2,4,1,7,6,8], target = 16
Output: 127
Explanation: All non-empty subset satisfy the condition (2^7 - 1) = 127

Constraints:

• 1 <= nums.length <= 10^5

• 1 <= nums[i] <= 10^6

• 1 <= target <= 10^6

class Solution {
    // In this question we only care about min, max of
    // a subsequence, thats why we can sort the array in starting.
    public int numSubseq(int[] nums, int target) {
        int n = nums.length;
        int mod = 1_000_000_007;
        // Record all 2 ^ n to save time
        int[] powerOfTwo = new int[n + 1];
        powerOfTwo[0] = 1;
        for (int i = 1; i <= n; ++i)
            powerOfTwo[i] = (powerOfTwo[i - 1] * 2) % mod;
        // Sorting part
        Arrays.sort(nums);
        // Try to use each number as the min number, fixing min number also makes sure
        // that min number alone will be a valid subsequence
        int ans = 0;
        for (int i = 0, j = n - 1; i <= j;) {
            // find the largest number that meets the sum condition
            if (nums[i] + nums[j] > target)
                j--;
            else {
                // add all possible combinations in [i + 1, j], and then in theory we will add
                // nums[i] to all those combinations to get the correct min & max
                ans = (ans + powerOfTwo[j - i]) % mod;
                i++;
            }
        }
        return ans;
    }
}