Problem Description

You start at height 0 (the floor) and need to climb a ladder to reach the last rung. The ladder has rungs at specific heights given in the array rungs, which is strictly increasing.

You have a maximum step distance dist - this means you can only climb from your current position to the next rung if the height difference is at most dist. If the next rung is too far (height difference > dist), you need to add intermediate rungs to make the climb possible.

The key rules are:

• You start at height 0 (the floor)
• You can only step up by at most dist units at a time
• You can insert new rungs at any positive integer height where a rung doesn't already exist
• You want to reach the last rung in the rungs array

The goal is to find the minimum number of rungs you need to add to make it possible to climb from the floor to the last rung.

For example, if rungs = [3, 6, 8, 10] and dist = 3:

• From floor (0) to rung at height 3: distance is 3, which equals dist, so no additional rungs needed
• From height 3 to height 6: distance is 3, which equals dist, so no additional rungs needed
• From height 6 to height 8: distance is 2, which is less than dist, so no additional rungs needed
• From height 8 to height 10: distance is 2, which is less than dist, so no additional rungs needed
• Total additional rungs needed: 0

The solution uses the formula ⌊(b - a - 1) / dist⌋ to calculate how many rungs need to be inserted between position a and position b when the distance exceeds dist.

Intuition

The key insight is that we want to minimize the number of rungs added, so we should only add rungs when absolutely necessary. When do we need to add rungs? Only when the gap between our current position and the next rung exceeds our maximum step distance dist.

Think about it greedily: if we're at position a and need to reach position b where b - a > dist, we want to take the largest possible steps of size dist each time. This means we should place new rungs at positions a + dist, a + 2*dist, a + 3*dist, and so on, until we can reach b.

How many such rungs do we need? If the gap is b - a, we need to divide this gap into chunks of size dist. However, we don't need to add a rung at position b since it already exists. So we're really asking: how many complete steps of size dist fit in the gap b - a - 1?

This gives us the formula ⌊(b - a - 1) / dist⌋:

• b - a is the total gap
• We subtract 1 because if the gap is exactly divisible by dist, we don't need an extra rung (we can reach b directly)
• Integer division gives us the number of complete dist-sized steps we need to insert

For example, if the gap is 10 and dist = 3:

• We need rungs at positions +3, +6, +9 from our current position
• That's 3 rungs, which equals ⌊(10 - 1) / 3⌋ = ⌊9 / 3⌋ = 3

We apply this calculation to each consecutive pair of positions (including from the floor at height 0 to the first rung), and sum up all the required rungs.

Learn more about Greedy patterns.

Solution Approach

The implementation follows a greedy simulation approach. We process each gap between consecutive positions and calculate how many rungs need to be inserted.

Step 1: Prepare the data

• We prepend 0 to the rungs array to represent the floor: rungs = [0] + rungs
• This allows us to treat the problem uniformly - every climb is from one position to the next, including the initial climb from the floor

Step 2: Process consecutive pairs

• We use pairwise(rungs) to iterate through consecutive pairs (a, b) where:
  • a is the current position
  • b is the next rung we want to reach

Step 3: Calculate rungs needed for each gap

• For each pair (a, b), we calculate the number of rungs to insert using the formula: (b - a - 1) // dist
• This formula works because:
  • If b - a <= dist, the formula gives 0 (no rungs needed)
  • If b - a > dist, we need ⌊(b - a - 1) / dist⌋ intermediate rungs

Step 4: Sum up all insertions

• We use Python's sum() function with a generator expression to calculate the total number of rungs needed across all gaps

Example walkthrough: If rungs = [3, 6, 8, 10] and dist = 2:

• After prepending: [0, 3, 6, 8, 10]
• Pairs: (0,3), (3,6), (6,8), (8,10)
• Calculations:
  • Gap 0→3: (3-0-1)//2 = 2//2 = 1 rung needed
  • Gap 3→6: (6-3-1)//2 = 2//2 = 1 rung needed
  • Gap 6→8: (8-6-1)//2 = 1//2 = 0 rungs needed
  • Gap 8→10: (10-8-1)//2 = 1//2 = 0 rungs needed
• Total: 1 + 1 + 0 + 0 = 2 rungs

The time complexity is O(n) where n is the length of the rungs array, and the space complexity is O(1) if we don't count the space used for the modified input.

Example Walkthrough

Let's walk through a concrete example with rungs = [4, 8, 12, 16] and dist = 3.

Initial Setup:

• We start at height 0 (the floor)
• We need to reach the last rung at height 16
• We can only step up by at most 3 units at a time

Step 1: Prepare the data

• Prepend 0 to represent the floor: [0, 4, 8, 12, 16]
• This gives us pairs: (0,4), (4,8), (8,12), (12,16)

Step 2: Process each gap

Gap 1: From floor (0) to first rung (4)

• Distance: 4 - 0 = 4
• Since 4 > 3 (our max step), we need intermediate rungs
• Apply formula: (4 - 0 - 1) // 3 = 3 // 3 = 1
• We need 1 additional rung (placed at height 3)
• Path: 0 → 3 (added) → 4

Gap 2: From height 4 to height 8

• Distance: 8 - 4 = 4
• Since 4 > 3, we need intermediate rungs
• Apply formula: (8 - 4 - 1) // 3 = 3 // 3 = 1
• We need 1 additional rung (placed at height 7)
• Path: 4 → 7 (added) → 8

Gap 3: From height 8 to height 12

• Distance: 12 - 8 = 4
• Since 4 > 3, we need intermediate rungs
• Apply formula: (12 - 8 - 1) // 3 = 3 // 3 = 1
• We need 1 additional rung (placed at height 11)
• Path: 8 → 11 (added) → 12

Gap 4: From height 12 to height 16

• Distance: 16 - 12 = 4
• Since 4 > 3, we need intermediate rungs
• Apply formula: (16 - 12 - 1) // 3 = 3 // 3 = 1
• We need 1 additional rung (placed at height 15)
• Path: 12 → 15 (added) → 16

Step 3: Calculate total

• Total rungs needed: 1 + 1 + 1 + 1 = 4

Final climbing path: 0 → 3 → 4 → 7 → 8 → 11 → 12 → 15 → 16

The formula (b - a - 1) // dist elegantly handles all cases:

• When the gap equals dist + 1 (like 4 in our example), we need exactly 1 rung
• When the gap is less than or equal to dist, the formula gives 0
• For larger gaps, it calculates the exact number of intermediate steps needed

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the rungs list. This is because:

• Creating [0] + rungs takes O(n) time to create a new list
• The pairwise(rungs) function iterates through the list once, generating n pairs
• The sum() function with the generator expression iterates through all pairs once, performing O(1) arithmetic operations for each pair

The space complexity is O(n), not O(1) as stated in the reference answer. This is because:

• The operation [0] + rungs creates a new list of size n + 1, which requires O(n) additional space
• The pairwise() function returns an iterator that doesn't require additional space beyond O(1)
• The generator expression (b - a - 1) // dist for a, b in pairwise(rungs) uses O(1) space

The reference answer's claim of O(1) space complexity would only be correct if the code modified the original list in-place or used a different approach that avoided creating a new list.

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

Common Pitfalls

1. Forgetting to Consider the Starting Position (Floor)

Many solutions incorrectly calculate the distance from the first rung without considering that you start at height 0. This leads to missing additional rungs needed for the initial climb.

Incorrect approach:

def addRungs(self, rungs: List[int], dist: int) -> int:
    total = 0
    # Wrong: only checking gaps between existing rungs
    for i in range(len(rungs) - 1):
        gap = rungs[i + 1] - rungs[i]
        total += (gap - 1) // dist
    return total

Why it fails: If rungs = [4, 8, 12] and dist = 2, this would miss that climbing from floor (0) to height 4 requires additional rungs.

Solution: Always include the starting position 0 in your calculations, either by prepending it to the array or handling the first rung separately.

2. Using the Wrong Formula for Calculating Additional Rungs

A common mistake is using gap // dist instead of (gap - 1) // dist.

Incorrect formula:

additional_rungs = gap // dist  # Wrong!

Why it fails: When the gap is exactly divisible by dist, this formula gives one extra rung. For example, if gap = 6 and dist = 3, we can climb in exactly 2 steps (0→3→6), needing only 1 additional rung. But 6 // 3 = 2 would suggest we need 2 additional rungs.

Correct formula:

additional_rungs = (gap - 1) // dist

This works because:

• If gap <= dist: (gap - 1) // dist = 0 (no additional rungs needed)
• If gap > dist: We need exactly ⌊(gap - 1) / dist⌋ intermediate rungs

3. Off-by-One Errors in Loop Iteration

When iterating through pairs, it's easy to create index out of bounds errors.

Incorrect approach:

for i in range(len(rungs)):  # Wrong: will cause index error
    gap = rungs[i + 1] - rungs[i]

Solution: Use range(len(rungs) - 1) or better yet, use Python's pairwise function from itertools for cleaner code:

from itertools import pairwise
for current, next_rung in pairwise([0] + rungs):
    gap = next_rung - current
    total += (gap - 1) // dist