Solution Approach

The solution implements a binary search to find the minimum possible maximum products per store.

Implementation Details:

1. Binary Search Setup: We use bisect_left from Python's bisect module to search in the range [1, 10^6]. The range starts at 1 (minimum possible products per store) and goes up to 10^6 (a reasonable upper bound for the maximum quantity).

2. Feasibility Check Function: The check(x) function determines if we can distribute all products with at most x items per store:

   def check(x):
       return sum((v + x - 1) // x for v in quantities) <= n

   • For each quantity v in quantities, we calculate (v + x - 1) // x
   • This formula computes ceil(v/x) using integer division, giving us the minimum number of stores needed for that product type
   • We sum the required stores for all product types
   • If the total is at most n, then x is feasible (returns True)

3. Binary Search Execution:

   return 1 + bisect_left(range(1, 10**6), True, key=check)

   • bisect_left finds the leftmost position where True can be inserted in the sorted sequence
   • The key=check parameter means we're searching based on the feasibility check
   • Since check(x) returns False for small values and True for large values, bisect_left finds the transition point
   • We add 1 to the result because bisect_left returns an index (0-based), but our range starts at 1

Why This Works:

• The check function creates a boolean sequence: [False, False, ..., False, True, True, ..., True]
• The transition from False to True happens at the minimum feasible x
• bisect_left efficiently finds this transition point in O(log(10^6)) time
• Each check takes O(m) time where m is the length of quantities
• Total time complexity: O(m * log(max_quantity))

Example Walkthrough

Let's walk through a concrete example with quantities = [15, 10, 10] and n = 7 stores.

Step 1: Set up binary search range

• Minimum possible x: 1 (each store gets at least 1 item)
• Maximum possible x: 15 (the largest quantity)
• We'll search in range [1, 15]

Step 2: Binary search iterations

Iteration 1: Try x = 8 (middle value)

• Product type 1 (15 items): needs ceil(15/8) = 2 stores (8 + 7)
• Product type 2 (10 items): needs ceil(10/8) = 2 stores (8 + 2)
• Product type 3 (10 items): needs ceil(10/8) = 2 stores (8 + 2)
• Total stores needed: 2 + 2 + 2 = 6 ≤ 7 ✓ (feasible)
• Since it works, try smaller values (search left half)

Iteration 2: Try x = 4 (middle of [1, 7])

• Product type 1 (15 items): needs ceil(15/4) = 4 stores (4 + 4 + 4 + 3)
• Product type 2 (10 items): needs ceil(10/4) = 3 stores (4 + 4 + 2)
• Product type 3 (10 items): needs ceil(10/4) = 3 stores (4 + 4 + 2)
• Total stores needed: 4 + 3 + 3 = 10 > 7 ✗ (not feasible)
• Since it doesn't work, try larger values (search right half)

Iteration 3: Try x = 6 (middle of [5, 7])

• Product type 1 (15 items): needs ceil(15/6) = 3 stores (6 + 6 + 3)
• Product type 2 (10 items): needs ceil(10/6) = 2 stores (6 + 4)
• Product type 3 (10 items): needs ceil(10/6) = 2 stores (6 + 4)
• Total stores needed: 3 + 2 + 2 = 7 ≤ 7 ✓ (feasible)
• Try smaller values

Iteration 4: Try x = 5

• Product type 1 (15 items): needs ceil(15/5) = 3 stores (5 + 5 + 5)
• Product type 2 (10 items): needs ceil(10/5) = 2 stores (5 + 5)
• Product type 3 (10 items): needs ceil(10/5) = 2 stores (5 + 5)
• Total stores needed: 3 + 2 + 2 = 7 ≤ 7 ✓ (feasible)
• Try smaller values one more time

Since x = 4 failed but x = 5 works, the answer is 5.

Final Distribution:

• Store 1: 5 items (type 1)
• Store 2: 5 items (type 1)
• Store 3: 5 items (type 1)
• Store 4: 5 items (type 2)
• Store 5: 5 items (type 2)
• Store 6: 5 items (type 3)
• Store 7: 5 items (type 3)

Maximum items in any store: 5 ✓

Time and Space Complexity

Time Complexity: O(m * log(max(quantities))) where m is the length of the quantities array.

• The binary search using bisect_left searches over a range from 1 to 10^6, which takes O(log(10^6)) iterations.
• For each iteration of the binary search, the check function is called.
• The check function iterates through all elements in quantities to calculate the sum, which takes O(m) time.
• Therefore, the overall time complexity is O(m * log(10^6)) which simplifies to O(m * log(max(quantities))) since 10^6 is used as an upper bound for the maximum possible value.

Space Complexity: O(1)

• The check function uses only a constant amount of extra space for the sum calculation and loop variable.
• The bisect_left function with a range object doesn't create the entire range in memory; it generates values on-the-fly, using O(1) space.
• No additional data structures are created that scale with the input size.
• Therefore, the space complexity is O(1) excluding the input array itself.

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

Common Pitfalls

1. Using Wrong Binary Search Template Variant

A common pitfall is using a different binary search variant that doesn't match the canonical template. The correct template uses while left <= right with first_true_index tracking.

Incorrect Implementation:

# WRONG: Using while left < right with right = mid
left, right = 1, max_value
while left < right:
    mid = (left + right) // 2
    if feasible(mid):
        right = mid
    else:
        left = mid + 1
return left

Correct Implementation:

# CORRECT: Using template with first_true_index
left, right = 1, max(quantities)
first_true_index = -1
while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1
return first_true_index

2. Using Wrong Search Bounds

Another pitfall is setting the upper bound too low or not considering edge cases.

The Problem:

# May fail if max(quantities) > 10000
right = 10000  # Too small

The Fix: Use the actual maximum quantity as the upper bound:

right = max(quantities)  # Exact bound

3. Inverted Boolean Logic in Feasible Function

A subtle pitfall is inverting the boolean logic, searching for False instead of True.

The Problem:

def feasible(max_products):
    # Inverted: returns True when NOT feasible
    return sum((q + max_products - 1) // max_products for q in quantities) > n

# This creates sequence: [True, True, ..., False, False]
# We want [False, False, ..., True, True]

The Fix: Keep the logic straightforward - return True when the distribution is feasible:

def feasible(max_products):
    return sum((q + max_products - 1) // max_products for q in quantities) <= n

4. Integer Overflow in Ceiling Division

While less common in Python, the ceiling division formula can be confusing.

The Problem:

# Incorrect ceiling division attempts
stores_needed = v // x + (1 if v % x > 0 else 0)  # Verbose
stores_needed = math.ceil(v / x)  # Requires import, potential float precision issues

The Fix: Use the integer-only formula that's both efficient and precise:

stores_needed = (v + x - 1) // x  # Equivalent to ceil(v/x) using only integers