Solution Approach

The solution uses the binary search template combined with a greedy validation function.

Key Insight: Inverting the Feasible Function Since we're searching for the maximum achievable value, we invert the feasible function. Instead of "can we achieve sweetness X?", we define feasible(x) as "can we NOT achieve sweetness X?":

def feasible(min_sweetness):
    current_sum = 0
    pieces_count = 0
    for chunk in sweetness:
        current_sum += chunk
        if current_sum >= min_sweetness:
            current_sum = 0
            pieces_count += 1
    # Cannot achieve if pieces_count <= k (need > k pieces)
    return pieces_count <= k

This creates a pattern like: [false, false, false, true, true] where false means "achievable" and true means "not achievable".

Binary Search Using the Template:

left, right = 1, total + 1
first_true_index = total + 1  # Default if all values achievable

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 - 1

The template finds the first infeasible sweetness. The answer is first_true_index - 1 (the last achievable sweetness).

Why the Greedy Check Works:

• We greedily form pieces from left to right
• Each piece is formed as soon as its total sweetness reaches mid
• If we can form more than k pieces (i.e., at least k+1 pieces), then mid is achievable
• The greedy approach is optimal: if there's any way to achieve sweetness mid, greedy will find it

Time Complexity: O(n × log(sum(sweetness))) where n is the length of the sweetness array.

Space Complexity: O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through a small example with sweetness = [1, 2, 2, 1, 2, 2, 1, 2, 2] and k = 2 (we need to make 2 cuts to create 3 pieces).

Step 1: Initialize Binary Search Range

• l = 0 (minimum possible sweetness)
• r = sum([1, 2, 2, 1, 2, 2, 1, 2, 2]) = 15 (maximum possible sweetness)

Step 2: Binary Search Process

Iteration 1:

• mid = (0 + 15 + 1) // 2 = 8
• Check if we can achieve minimum sweetness of 8:
  • Traverse array: [1, 2, 2, 1, 2, 2, 1, 2, 2]
  • Accumulate: 1 → 3 → 5 → 6 → 8 (≥8, form piece 1, reset)
  • Continue: 2 → 3 → 5 → 7 (can't reach 8, end of array)
  • Pieces formed: 1 (need 3)
  • Return false since 1 ≤ 2
• Result: r = 8 - 1 = 7

Iteration 2:

• mid = (0 + 7 + 1) // 2 = 4
• Check if we can achieve minimum sweetness of 4:
  • Traverse: [1, 2, 2, 1, 2, 2, 1, 2, 2]
  • Accumulate: 1 → 3 → 5 (≥4, form piece 1, reset)
  • Continue: 1 → 3 → 5 (≥4, form piece 2, reset)
  • Continue: 1 → 3 → 5 (≥4, form piece 3, reset)
  • Pieces formed: 3 (need 3)
  • Return true since 3 > 2
• Result: l = 4

Iteration 3:

• mid = (4 + 7 + 1) // 2 = 6
• Check if we can achieve minimum sweetness of 6:
  • Traverse: [1, 2, 2, 1, 2, 2, 1, 2, 2]
  • Accumulate: 1 → 3 → 5 → 6 (≥6, form piece 1, reset)
  • Continue: 2 → 4 → 5 → 7 (≥6, form piece 2, reset)
  • Continue: 2 (can't reach 6, end of array)
  • Pieces formed: 2 (need 3)
  • Return false since 2 ≤ 2
• Result: r = 6 - 1 = 5

Iteration 4:

• mid = (4 + 5 + 1) // 2 = 5
• Check if we can achieve minimum sweetness of 5:
  • Traverse: [1, 2, 2, 1, 2, 2, 1, 2, 2]
  • Accumulate: 1 → 3 → 5 (≥5, form piece 1, reset)
  • Continue: 1 → 3 → 5 (≥5, form piece 2, reset)
  • Continue: 1 → 3 → 5 (≥5, form piece 3, reset)
  • Pieces formed: 3 (need 3)
  • Return true since 3 > 2
• Result: l = 5

Step 3: Convergence

• Now l = 5 and r = 5, so the loop exits
• Return l = 5

Final Answer: 5

The optimal division would be: [1, 2, 2] (sweetness = 5), [1, 2, 2] (sweetness = 5), [1, 2, 2] (sweetness = 5). You eat the piece with minimum sweetness, which is 5.

Time and Space Complexity

Time Complexity: O(n × log(sum(sweetness)))

The algorithm uses binary search on the answer space. The search range is from 0 to sum(sweetness), which means the binary search will perform O(log(sum(sweetness))) iterations. In each iteration, the check function is called, which iterates through all n elements of the sweetness array exactly once, taking O(n) time. Therefore, the overall time complexity is O(n × log(sum(sweetness))), where n is the length of the sweetness array.

Space Complexity: O(1)

The algorithm only uses a constant amount of extra space. The variables l, r, mid in the main function and s, cnt, v in the check function are all scalar variables that don't depend on the input size. No additional data structures are created that scale with the input, so the space complexity is O(1).

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

Common Pitfalls

1. Using Wrong Binary Search Template Variant

Pitfall: For "find maximum" problems, using the non-standard while left < right with left = mid variant requires careful midpoint calculation.

Example of the Problem:

# Non-standard variant - prone to infinite loops
while left < right:
    mid = (left + right + 1) >> 1  # Must use +1 to avoid infinite loop
    if can_achieve(mid):
        left = mid
    else:
        right = mid - 1

Solution: Invert the feasible function and use the standard template:

def feasible(x):
    # Returns True if we CANNOT achieve x
    return pieces_count <= k

first_true_index = total + 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 - 1

2. Wrong Piece Count Comparison

Pitfall: Using pieces_count >= k instead of pieces_count > k in the validation.

Why it's problematic: We need k+1 pieces total. The inverted feasible returns true when we CANNOT achieve, so the condition should be pieces_count <= k.

Solution: Use return pieces_count <= k for the inverted feasible function.

3. Edge Case: k = 0

Pitfall: Not understanding that when k = 0, we need exactly 1 piece, so the answer is the sum of all sweetness.

Solution: The template handles this correctly - when k = 0, feasible returns false for all values up to the total sum, so first_true_index = total + 1 and the answer is total.

4. Integer Overflow in Large Inputs

Pitfall: In languages with fixed integer sizes, sum(sweetness) might overflow.

Solution: In Python this isn't an issue. In other languages, use long long in C++ or long in Java.

5. Misunderstanding the Greedy Strategy

Pitfall: Trying to optimize piece formation with DP or looking ahead.

Solution: Trust the greedy approach - cutting as soon as we reach the threshold is optimal for checking feasibility.