1498. Number of Subsequences That Satisfy the Given Sum Condition (Two Pointers) 20.6.28 Medium

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


Solution: O(nlogn) T and O(1) S
class Solution(object):
    def numSubseq(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        nums.sort()                                         # Order does not matter here.
        left, right = 0, len(nums) - 1
        mod = 10 ** 9 + 7
        res = 0
        while left <= right:
            if nums[left] + nums[right] > target:           
                right -= 1
            else:
                res += pow(2, right - left, mod)            # For each elements in the subarray A[i+1] ~ A[j],
                                                            # we can pick or not pick,
                                                            # so there are 2 ^ (j - i) subsequences in total.
                left += 1
        return res % mod