Solution Approach

The implementation follows the sorting and binary search strategy outlined in the reference approach:

Step 1: Preprocessing

• Sort the array nums in ascending order
• Pre-compute powers of 2 up to n and store them in array f, where f[i] = 2^i % mod
  • This avoids redundant power calculations during the main loop
  • Each f[i] = f[i-1] * 2 % mod

Step 2: Main Algorithm For each index i in the sorted array:

1. Set minimum element: Consider nums[i] as the minimum element of the subsequence

2. Early termination check: If nums[i] * 2 > target, break the loop

   • This means even a subsequence with just nums[i] (where min = max) exceeds the target
   • Since the array is sorted, all subsequent elements will also fail this check

3. Find maximum valid index using template-based binary search:

   • Define the feasible condition: nums[mid] > maxAllowed where maxAllowed = target - nums[i]
   • This creates a pattern: false, false, ..., true, true, true in the sorted array
   • Use the standard template with firstTrueIndex = -1, while left <= right, and right = mid - 1
   • The rightmost valid index j is firstTrueIndex - 1 (or n - 1 if no element exceeds maxAllowed)

4. Count valid subsequences: For the range [i, j], add 2^(j-i) to the answer

   • We must include nums[i] (minimum) and nums[j] (maximum)
   • For the j - i - 1 elements between them, each can be either included or excluded
   • This gives us 2^(j-i) total subsequences
   • Access the pre-computed value using f[j - i]

5. Apply modulo: Keep the answer within bounds using mod = 10^9 + 7

Time Complexity: O(n log n) for sorting + O(n log n) for n binary searches = O(n log n)

Space Complexity: O(n) for storing the powers of 2

Example Walkthrough

Let's walk through a concrete example with nums = [3, 5, 6, 7] and target = 9.

Step 1: Sort and Precompute Powers of 2

• Array is already sorted: [3, 5, 6, 7]
• Precompute powers of 2: f = [1, 2, 4, 8, 16] where f[k] = 2^k

Step 2: Process Each Element as Minimum

Iteration 1: i = 0, minimum = 3

• Check early termination: 3 * 2 = 6 ≤ 9, so continue
• Find maximum valid index using binary search:
  • Looking for largest j where nums[j] ≤ 9 - 3 = 6
  • Binary search finds j = 2 (nums[2] = 6)
• Count subsequences from index 0 to 2:
  • We have elements [3, 5, 6] where 3 is min, 6 is max
  • Number of subsequences = 2^(2-0) = 2^2 = 4
  • These subsequences are:
    1. [3, 6] - must include min and max
    2. [3, 5, 6] - include the middle element
    3. Actually, let me recalculate: for range [0,2], we fix nums[0]=3 as min and nums[2]=6 as max
    4. Between them we have 1 element (nums[1]=5) which can be included or excluded
    5. So we have 2^1 = 2 subsequences: [3, 6] and [3, 5, 6]
• Wait, I need to correct this. The formula is 2^(j-i-1) for elements between, but the code uses f[j-i]. Let me reconsider.
• Actually, looking at the code, it adds f[j - i] which is 2^(j-i). For j=2, i=0: f[2] = 4
• This counts all subsequences where 3 is minimum and any element up to index 2 can be maximum:
  • [3] (min=max=3, sum=6)
  • [3, 5] (min=3, max=5, sum=8)
  • [3, 6] (min=3, max=6, sum=9)
  • [3, 5, 6] (min=3, max=6, sum=9)
• Total so far: 4

Iteration 2: i = 1, minimum = 5

• Check early termination: 5 * 2 = 10 > 9, so break
• Since even the smallest possible sum (5+5=10) exceeds target, we stop

Final Answer: 4

The valid subsequences are:

1. [3] with min=3, max=3, sum=6 ≤ 9 ✓
2. [3, 5] with min=3, max=5, sum=8 ≤ 9 ✓
3. [3, 6] with min=3, max=6, sum=9 ≤ 9 ✓
4. [3, 5, 6] with min=3, max=6, sum=9 ≤ 9 ✓

Note how the algorithm efficiently counts these without explicitly generating each subsequence. By fixing the minimum element and finding the valid range for the maximum, it uses the power of 2 formula to count all possibilities at once.

Time and Space Complexity

Time Complexity: O(n × log n)

The time complexity breaks down as follows:

• Sorting the array takes O(n × log n) time
• Pre-computing powers of 2 in array f takes O(n) time
• The main loop iterates through each element once, taking O(n) iterations
• Inside each iteration, bisect_right performs binary search, which takes O(log n) time
• Overall: O(n × log n) + O(n) + O(n × log n) = O(n × log n)

Space Complexity: O(n)

The space complexity consists of:

• Array f stores n + 1 precomputed powers of 2, requiring O(n) space
• The sorting operation may use O(log n) to O(n) auxiliary space depending on the implementation (Python's Timsort uses up to O(n) in worst case)
• Other variables (mod, n, ans, i, x, j) use O(1) space
• Overall: O(n) space is required

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Using Wrong Binary Search Template Variant

A common mistake is using while left < right with right = mid instead of the standard template. This variant is harder to reason about and can lead to off-by-one errors.

Incorrect Implementation:

# Wrong template variant - harder to reason about
left, right = 0, n
while left < right:
    mid = (left + right) // 2
    if nums[mid] > max_allowed:
        right = mid
    else:
        left = mid + 1
return left

Correct Implementation:

# Standard template - tracks first true index explicitly
left, right = 0, n - 1
first_true_index = -1
while left <= right:
    mid = (left + right) // 2
    if nums[mid] > max_allowed:
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1
# Use first_true_index - 1 for rightmost valid index

2. Incorrect Power of 2 Calculation Without Modulo

Computing powers of 2 without applying modulo during the precomputation phase leads to integer overflow for large values of n.

Incorrect Implementation:

# This will cause overflow for large n
power_of_two = [1] * n
for i in range(1, n):
    power_of_two[i] = power_of_two[i - 1] * 2  # Missing modulo!

Correct Implementation:

power_of_two = [1] * n
for i in range(1, n):
    power_of_two[i] = (power_of_two[i - 1] * 2) % MOD

3. Not Handling first_true_index == -1 Edge Case

When all elements satisfy the condition (none exceed maxAllowed), first_true_index remains -1. Failing to handle this case leads to incorrect index calculation.

Incorrect Implementation:

# Crashes or gives wrong result when first_true_index is -1
right_idx = first_true_index - 1

Correct Implementation:

# Handle the case where no element exceeds maxAllowed
right_idx = first_true_index - 1 if first_true_index != -1 else n - 1

4. Forgetting to Sort the Array

The algorithm relies on the sorted property to correctly identify min/max elements and perform binary search. Without sorting, the logic completely breaks down.

Incorrect Implementation:

def numSubseq(self, nums: List[int], target: int) -> int:
    # Missing nums.sort()!
    n = len(nums)
    # Rest of the code...

Correct Implementation:

def numSubseq(self, nums: List[int], target: int) -> int:
    nums.sort()  # Essential for the algorithm to work
    n = len(nums)
    # Rest of the code...

5. Incorrect Subsequence Count Formula

Using 2^(j-i-1) instead of 2^(j-i) undercounts valid subsequences by not accounting for all possible combinations of intermediate elements.

Incorrect Implementation:

# This formula is wrong - it doesn't count all valid subsequences
subsequence_count = power_of_two[right_idx - left_idx - 1]

Correct Implementation:

# Correct formula: 2^(number of elements between and including boundaries)
subsequence_count = power_of_two[right_idx - left_idx]