Solution Approach

The implementation follows a sorting and binary search strategy.

Step 1: Sort the potions array

potions.sort()

We sort the potions array in ascending order. This preprocessing step enables us to use binary search efficiently.

Step 2: Binary Search Using first_true_index Template

For each spell with strength v, we need to find the first potion that forms a successful pair (where potion >= success / v).

Using the first_true_index template:

def bisect_left(nums, target):
    # Find first index where nums[idx] >= target
    left, right = 0, len(nums) - 1
    first_true_index = len(nums)  # Default: no element >= target

    while left <= right:
        mid = (left + right) // 2
        if nums[mid] >= target:
            first_true_index = mid
            right = mid - 1
        else:
            left = mid + 1

    return first_true_index

Counting Logic:

• first_true_index = bisect_left(potions, success / v) finds the first potion >= threshold
• Number of valid potions = len(potions) - first_true_index

Example walkthrough: If potions = [1, 2, 3, 4, 5], success = 7, and current spell strength is 2:

• Minimum potion needed: 7 / 2 = 3.5
• bisect_left([1, 2, 3, 4, 5], 3.5) returns 3 (first index where value >= 3.5)
• Valid potions are at indices 3 and 4 (values 4 and 5)
• Count of valid potions: 5 - 3 = 2

Time Complexity: O((m + n) × log m) - sort potions + binary search per spell.

Example Walkthrough

Let's walk through a concrete example to understand the solution approach:

Given:

• spells = [3, 1, 2]
• potions = [8, 5, 8]
• success = 16

Step 1: Sort the potions array

• potions = [5, 8, 8] (sorted in ascending order)
• m = 3 (length of potions)

Step 2: Process each spell using binary search

For spell[0] = 3:

• We need: 3 × potion ≥ 16
• Minimum potion strength needed: 16 / 3 = 5.33...
• Use bisect_left([5, 8, 8], 5.33) to find where 5.33 would be inserted
• Binary search finds index 1 (5.33 would go after the 5 at index 0)
• Valid potions are at indices 1 and 2 (values 8 and 8)
• Count: 3 - 1 = 2 successful pairs

For spell[1] = 1:

• We need: 1 × potion ≥ 16
• Minimum potion strength needed: 16 / 1 = 16
• Use bisect_left([5, 8, 8], 16) to find where 16 would be inserted
• Binary search finds index 3 (16 would go at the end)
• No potions meet this threshold
• Count: 3 - 3 = 0 successful pairs

For spell[2] = 2:

• We need: 2 × potion ≥ 16
• Minimum potion strength needed: 16 / 2 = 8
• Use bisect_left([5, 8, 8], 8) to find where 8 would be inserted
• Binary search finds index 1 (the leftmost position of value 8)
• Valid potions are at indices 1 and 2 (values 8 and 8)
• Count: 3 - 1 = 2 successful pairs

Final Result: [2, 0, 2]

This example demonstrates how sorting the potions array once allows us to efficiently find the count of valid pairs for each spell using binary search, avoiding the need to check every spell-potion combination individually.

Time and Space Complexity

Time Complexity: O(m log m + n log m) which simplifies to O((m + n) × log m)

• Sorting the potions array takes O(m log m) time, where m is the length of the potions array
• For each spell in the spells array, we perform a binary search using bisect_left on the sorted potions array, which takes O(log m) time per spell
• Since we have n spells, the total time for all binary searches is O(n log m)
• Overall time complexity: O(m log m + n log m) = O((m + n) × log m)

Space Complexity: O(log m)

• The sorting operation on the potions array uses O(log m) space for the recursion stack in Python's Timsort algorithm
• The list comprehension creates a new list of size n, but this is the output and typically not counted in space complexity analysis
• The binary search operation bisect_left uses O(1) additional space
• Overall auxiliary space complexity: O(log m)

Note: The reference answer states O(log n) for space complexity, but based on the code analysis, it should be O(log m) since the sorting is performed on the potions array of length m, not the spells array of length n.

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

Common Pitfalls

1. Using Wrong Binary Search Template

The problem uses bisect_left which finds the first index where element >= target.

Wrong approach:

while left < right:
    mid = (left + right) >> 1
    if spell * potions[mid] >= success:
        right = mid
    else:
        left = mid + 1
return left  # Returns left directly

Correct approach (first_true_index template):

first_true_index = len(potions)  # Default
while left <= right:
    mid = (left + right) // 2
    if spell * potions[mid] >= success:
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1
return first_true_index

2. Integer Division vs Float Division

The Pitfall: Using integer division (//) instead of float division (/) when calculating the minimum potion strength.

# WRONG - This will truncate the result
min_potion_strength = success // spell_strength

# CORRECT - Preserves decimal values
min_potion_strength = success / spell_strength

Why it matters: If success = 7 and spell_strength = 2, we need potions with strength ≥ 3.5. Using integer division would give us 3, incorrectly including potions with strength 3 (since 2 × 3 = 6 < 7).

3. Using bisect_right Instead of bisect_left

The Pitfall: Confusing bisect_right with bisect_left, which would give incorrect counts when exact matches exist.

# WRONG - May miss valid potions when exact threshold exists
insert_position = bisect_right(potions, min_potion_strength)

# CORRECT - Finds the leftmost valid position
insert_position = bisect_left(potions, min_potion_strength)

Example: If potions = [1, 2, 4, 4, 5] and we need min_potion_strength = 4:

• bisect_left returns index 2 (first occurrence of 4)
• bisect_right returns index 4 (after last occurrence of 4)
• Using bisect_right would incorrectly exclude the two potions with strength 4

4. Forgetting to Sort the Potions Array

The Pitfall: Applying binary search on an unsorted array, leading to incorrect results.

# WRONG - Binary search requires sorted array
result = [m - bisect_left(potions, success / v) for v in spells]

# CORRECT - Sort first, then apply binary search
potions.sort()
result = [m - bisect_left(potions, success / v) for v in spells]

5. Modifying Original Potions Array

The Pitfall: Sorting the original potions array in-place might be problematic if the original order needs to be preserved elsewhere.

Solution: Create a copy if needed:

sorted_potions = sorted(potions)  # Creates a new sorted list
# OR
sorted_potions = potions.copy()
sorted_potions.sort()

6. Edge Case: Very Small Spell Strength

The Pitfall: When spell strength approaches zero, success / spell_strength can become very large or cause floating-point precision issues.

Solution: While the problem states all values are positive integers, it's good practice to handle edge cases:

for spell_strength in spells:
    if spell_strength == 0:  # Though problem guarantees positive integers
        result.append(0)
        continue
    min_potion_strength = success / spell_strength
    # ... rest of the logic