Minimum Speed to Arrive on Time

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Steps

setup

Initialize Search Range

Set the initial search range for speed from 1.0 to 10,000,000.0 and define the precision epsilon as 1e-6.

💡 Starting with a wide range ensures the minimum speed is within bounds; epsilon controls when to stop searching.

Line:

low, high = 1.0, 10**7
epsilon = 1e-6

💡 The algorithm begins with a broad search space to guarantee the minimum speed is included.

compare

Calculate Mid Speed

Calculate the mid speed as the average of low and high: (1.0 + 10,000,000.0) / 2 = 5,000,000.5.

💡 Mid speed is the candidate speed to test feasibility at this iteration.

Line:

mid = (low + high) / 2

💡 Mid speed divides the current search space in half, focusing the search.

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Check Feasibility at Mid Speed

Check if arriving on time is possible at speed 5,000,000 by calculating total travel time with ceil for all but last segment.

💡 Feasibility check determines if the candidate speed is sufficient to arrive on time.

Line:

if can_arrive(dist, hour, mid):

💡 At very high speeds, arrival is always feasible, so the algorithm will try to find a smaller speed.

shrink

Update High Bound to Mid

Since speed 5,000,000 is feasible, update high to mid to search for a smaller feasible speed.

💡 Reducing the upper bound narrows the search space towards the minimum speed.

Line:

high = mid

💡 The algorithm eliminates speeds greater than mid from consideration.

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Calculate New Mid Speed

Calculate new mid speed as (1 + 5,000,000) / 2 = 2,500,000.5.

💡 Mid speed recalculates the candidate speed in the reduced search space.

Line:

mid = (low + high) / 2

💡 The search space continues to halve each iteration.

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Check Feasibility at Mid Speed

Check if arriving on time is possible at speed 2,500,000 by calculating total travel time.

💡 Feasibility check guides whether to lower or raise the speed bounds.

Line:

if can_arrive(dist, hour, mid):

💡 Still feasible at this high speed, so the algorithm tries smaller speeds.

shrink

Update High Bound to Mid

Update high to mid = 2,500,000 to narrow the search space.

💡 Continuing to narrow the upper bound focuses the search on smaller speeds.

Line:

high = mid

💡 Speeds above mid are eliminated from consideration.

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Calculate Mid Speed Again

Calculate mid speed as (1 + 2,500,000) / 2 = 1,250,000.5.

💡 Mid speed recalculates candidate speed in the new range.

Line:

mid = (low + high) / 2

💡 The search space halves again, moving closer to the answer.

compare

Check Feasibility at Mid Speed

Check if arriving on time is possible at speed 1,250,000.

💡 Feasibility check continues to guide the search direction.

Line:

if can_arrive(dist, hour, mid):

💡 Still feasible, so the algorithm tries smaller speeds.

shrink

Update High Bound to Mid

Update high to mid = 1,250,000 to narrow the search space further.

💡 The upper bound moves closer to the minimum feasible speed.

Line:

high = mid

💡 Speeds above mid are eliminated.

compare

Calculate Mid Speed Closer to Answer

Calculate mid speed as (1 + 1250000) / 2 = 625000.5.

💡 Mid speed recalculates candidate speed in the narrowed range.

Line:

mid = (low + high) / 2

💡 The search space continues to halve, approaching the minimum speed.

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Check Feasibility at Mid Speed

Check if arriving on time is possible at speed 625,000.

💡 Feasibility check guides the binary search direction.

Line:

if can_arrive(dist, hour, mid):

💡 Still feasible, so the algorithm tries smaller speeds.

shrink

Update High Bound to Mid

Update high to mid = 625,000 to narrow the search space.

💡 Upper bound moves closer to the minimum feasible speed.

Line:

high = mid

💡 Speeds above mid are eliminated.

compare

Calculate Mid Speed Much Closer to Answer

Calculate mid speed as (1 + 625,000) / 2 = 312,500.5.

💡 Mid speed recalculates candidate speed in the narrowed range.

Line:

mid = (low + high) / 2

💡 The search space continues to halve, approaching the minimum speed.

compare

Check Feasibility at Mid Speed

Check if arriving on time is possible at speed 312,500.

💡 Feasibility check guides the binary search direction.

Line:

if can_arrive(dist, hour, mid):

💡 Still feasible, so the algorithm tries smaller speeds.

shrink

Update High Bound to Mid

Update high to mid = 312,500 to narrow the search space.

💡 Upper bound moves closer to the minimum feasible speed.

Line:

high = mid

💡 Speeds above mid are eliminated.

shrink

Converge to Minimum Speed

The binary search continues halving the search space until high - low <= epsilon, converging near 1.0.

💡 The algorithm stops when the search space is sufficiently small to determine the minimum speed.

Line:

while high - low > epsilon:

💡 The minimum feasible speed is found within the precision threshold.

reconstruct

Return Final Answer

Return the ceiling of high, which is 1, as the minimum integer speed to arrive on time.

💡 Ceiling ensures the speed is an integer that meets the time requirement.

Line:

return ceil(high)

💡 The algorithm outputs the smallest integer speed that allows on-time arrival.

from math import ceil

def can_arrive(dist, hour, speed):
    total_time = 0.0
    for i in range(len(dist) - 1):  # STEP 3,6,9,12,15
        total_time += ceil(dist[i] / speed)
    total_time += dist[-1] / speed
    return total_time <= hour

def min_speed_on_time(dist, hour):
    low, high = 1.0, 10**7  # STEP 1
    epsilon = 1e-6
    while high - low > epsilon:  # STEP 17
        mid = (low + high) / 2  # STEP 2,5,8,11,14
        if can_arrive(dist, hour, mid):  # STEP 3,6,9,12,15
            high = mid  # STEP 4,7,10,13,16
        else:
            low = mid
    return ceil(high)  # STEP 18

if __name__ == '__main__':
    print(min_speed_on_time([1,3,2], 6))  # Expected: 1

📊

Minimum Speed to Arrive on Time - Watch the Algorithm Execute, Step by Step

Watching each step shows how binary search on a continuous answer space works, revealing the logic behind choosing speeds and eliminating impossible ranges.

1

0

lo

3

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Next: lo=1 hi=10000000

Target: 6

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Next: lo=1 hi=10000000

Target: 6

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Next: lo=1 hi=5000000

Target: 6

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Next: lo=1 hi=5000000

Target: 6

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lo

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Next: lo=1 hi=5000000

Target: 6

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lo

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Next: lo=1 hi=2500000

Target: 6

1

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lo

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Next: lo=1 hi=2500000

Target: 6

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Next: lo=1 hi=2500000

Target: 6

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Next: lo=1 hi=1250000

Target: 6

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Next: lo=1 hi=1250000

Target: 6

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Next: lo=1 hi=1250000

Target: 6

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Next: lo=1 hi=625000

Target: 6

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Next: lo=1 hi=625000

Target: 6

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lo

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Next: lo=1 hi=625000

Target: 6

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Next: lo=1 hi=312500

Target: 6

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Next: lo=1 hi=312500

Target: 6

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Next: lo=1 hi=1.0000005

Target: 6

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lo

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Next: lo=1 hi=1.0000005

Key Takeaways

✓ Binary search on a continuous answer space efficiently narrows down the minimum feasible speed by halving the search range each iteration.

This insight is hard to see from code alone because the floating-point search space and feasibility checks are abstract concepts without visualization.

✓ Feasibility checks using ceil for all but the last segment ensure the algorithm respects the problem's rounding constraints.

Understanding why ceil is used only for intermediate segments clarifies how travel time is calculated and why the last segment differs.

✓ The decision to move the high or low bound depends entirely on feasibility, illustrating a classic binary search elimination pattern.

Seeing the search space shrink visually helps learners grasp how decisions prune impossible speeds.

Practice

1. Consider the following Python code implementing the min-heap approach to find the kth smallest element in a sorted matrix. Given the matrix [[1,3],[2,4]] and k = 3, what value does the function return?

easy

A. 2

B. 4

C. 3

D. 1

Solution

1. Step 1: Initialize min-heap with first elements

   Push (1,0,0) and (2,1,0) into heap.

2. Step 2: Extract elements until count == k

   Pop 1 (count=1), push 3 from row 0; pop 2 (count=2), push 4 from row 1; pop 3 (count=3), return 3.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Third smallest element is 3 [OK]

Hint: Trace heap pops and pushes carefully [OK]

Common Mistakes:

• Off-by-one counting of popped elements
• Returning wrong popped value
• Confusing row and column indices

2. Consider the following recursive binary search code for finding the insert position of target = 2 in nums = [1, 3, 5, 6]. What is the final returned value?

easy

A. 0

B. 1

C. 2

D. 3

Solution

1. Step 1: Trace binary_search(0, 3)

   mid = (0+3)//2 = 1, nums[1] = 3 > 2, recurse left: binary_search(0, 0)

2. Step 2: Trace binary_search(0, 0)

   mid = (0+0)//2 = 0, nums[0] = 1 < 2, recurse right: binary_search(1, 0)

3. Step 3: Base case binary_search(1, 0)

   left > right, return left = 1 as insertion index

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Target 2 fits between 1 and 3 at index 1 [OK]

Hint: Follow recursive calls and base case carefully [OK]

Common Mistakes:

• Returning right instead of left at base case
• Off-by-one in mid calculation
• Confusing left and right updates

3. Suppose the array is no longer sorted but you still need to find the first and last positions of a target value. Which approach is correct and efficient?

hard

A. Use a linear scan to find first and last occurrences directly

B. Sort the array first, then apply modified binary search to find positions

C. Use the same modified binary search approach since it works on any array

D. Use a hash map to store all indices of each value, then retrieve target positions

Solution

1. Step 1: Recognize that binary search requires sorted array

   Without sorting, binary search cannot guarantee correct positions.

2. Step 2: Evaluate alternatives

   Sorting changes indices, so linear scan or hash map are options; linear scan is simplest and O(n).

3. Step 3: Compare efficiency and correctness

   Linear scan finds first and last positions in one pass without modifying array.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Binary search requires sorted input; linear scan works otherwise [OK]

Hint: Binary search needs sorted array; linear scan works otherwise [OK]

Common Mistakes:

• Trying binary search on unsorted array
• Sorting and losing original indices
• Overcomplicating with hash maps

4. Suppose Koko can eat bananas from multiple piles simultaneously each hour (i.e., she can split her eating speed across piles). How does this change the approach to find the minimum eating speed to finish within h hours?

hard

A. Feasibility check simplifies to total bananas ≤ speed * h, so minimum speed = ceil(total / h)

B. We must use dynamic programming to consider all pile combinations per hour

C. The original binary search with feasibility check still applies without changes

D. Binary search is invalid; a greedy approach is required instead

Solution

1. Step 1: Understand the new eating model

   Koko can split speed across piles simultaneously, so total bananas eaten per hour equals speed.

2. Step 2: Simplify feasibility condition

   To finish in h hours, total bananas ≤ speed * h, so minimum speed = ceil(total bananas / h).

3. Step 3: Conclude algorithm change

   No binary search needed; direct calculation suffices.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Splitting speed removes per-pile constraints, simplifying problem [OK]

Hint: Splitting speed means total bananas divided by hours [OK]

Common Mistakes:

• Applying original binary search blindly
• Overcomplicating with DP

5. Suppose the problem is modified so that x can be negative, and you want to return the integer square root of its absolute value. Which modification to the Newton's method code below correctly handles this variant without breaking the algorithm?

hard

A. Remove the base case check and rely on the loop to handle all x.

B. Change the loop condition to 'while guess * guess >= x' to handle negatives.

C. Add 'x = abs(x)' at the start and keep the rest unchanged.

D. Use floating point division instead of integer division to handle negatives.

Solution

1. Step 1: Understand the problem change

   Allowing negative x means we must compute the integer square root of |x|.

2. Step 2: Modify input safely

   Adding x = abs(x) at the start ensures the algorithm works unchanged on non-negative inputs.

3. Step 3: Avoid incorrect changes

   Removing base cases or changing loop conditions breaks correctness; floating point division violates problem constraints.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Absolute value conversion preserves algorithm correctness [OK]

Hint: Convert input to non-negative before applying integer sqrt [OK]

Common Mistakes:

• Removing base cases
• Changing loop condition incorrectly
• Using float division causing precision errors