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Add Minimum Number of Rungs

You are given a strictly increasing integer array rungs that represents the height of rungs on a ladder. You are currently on the floor at height 0, and you want to reach the last rung.

You are also given an integer dist. You can only climb to the next highest rung if the distance between where you are currently at (the floor or on a rung) and the next rung is at most dist. You are able to insert rungs at any positive integer height if a rung is not already there.

Return the minimum number of rungs that must be added to the ladder in order for you to climb to the last rung.

 

Example 1:

Input: rungs = [1,3,5,10], dist = 2
Output: 2
Explanation:
You currently cannot reach the last rung.
Add rungs at heights 7 and 8 to climb this ladder. 
The ladder will now have rungs at [1,3,5,7,8,10].

Example 2:

Input: rungs = [3,6,8,10], dist = 3
Output: 0
Explanation:
This ladder can be climbed without adding additional rungs.

Example 3:

Input: rungs = [3,4,6,7], dist = 2
Output: 1
Explanation:
You currently cannot reach the first rung from the ground.
Add a rung at height 1 to climb this ladder.
The ladder will now have rungs at [1,3,4,6,7].

 

Constraints:

  • 1 <= rungs.length <= 105
  • 1 <= rungs[i] <= 109
  • 1 <= dist <= 109
  • rungs is strictly increasing.

Solution Explanation for LeetCode 1936: Add Minimum Number of Rungs

This problem involves finding the minimum number of rungs to add to a ladder to make it climbable, given a maximum climbing distance (dist). The solution uses a greedy approach.

Approach:

The core idea is to iterate through the rungs array, keeping track of the current reachable height. For each rung, we calculate the gap between the current reachable height and the rung's height. If this gap exceeds dist, we need to add rungs. The number of rungs to add for each gap is determined by integer division of the gap minus 1 by dist. This is because we can place rungs at intervals of dist, and subtracting 1 accounts for the fact that the current height is already reachable.

Time Complexity: O(n), where n is the length of the rungs array. We iterate through the array once.

Space Complexity: O(1). The algorithm uses a constant amount of extra space.

Code Explanation with Examples:

Let's break down the Python code as an example:

from itertools import pairwise
 
class Solution:
    def addRungs(self, rungs: List[int], dist: int) -> int:
        rungs = [0] + rungs #Add 0 to represent starting at the ground level.
        return sum((b - a - 1) // dist for a, b in pairwise(rungs))

  1. from itertools import pairwise: This line imports the pairwise function, which efficiently generates pairs of consecutive elements from an iterable.

  2. rungs = [0] + rungs: We prepend a 0 to the rungs list to represent starting at the ground level (height 0).

  3. return sum((b - a - 1) // dist for a, b in pairwise(rungs)): This is the core logic:

    • pairwise(rungs): Generates pairs of consecutive rung heights. For example, if rungs is [0, 1, 3, 5, 10], pairwise yields (0, 1), (1, 3), (3, 5), (5, 10).
    • (b - a - 1) // dist: For each pair (a, b), this calculates the number of rungs to add. b - a is the gap between the rungs. We subtract 1 because a is already reachable. The integer division // gives the number of full intervals of length dist that fit within the gap, representing the minimum number of rungs needed.
    • sum(...): Finally, we sum up the number of rungs to add for all gaps.

Example Walkthrough:

Let's consider the example rungs = [1, 3, 5, 10], dist = 2.

  1. rungs becomes [0, 1, 3, 5, 10].
  2. pairwise(rungs) yields (0, 1), (1, 3), (3, 5), (5, 10).
  3. Let's calculate the number of rungs to add for each pair:
    • (1 - 0 - 1) // 2 = 0 (No rungs needed between 0 and 1)
    • (3 - 1 - 1) // 2 = 0 (No rungs needed between 1 and 3)
    • (5 - 3 - 1) // 2 = 0 (No rungs needed between 3 and 5)
    • (10 - 5 - 1) // 2 = 2 (2 rungs needed between 5 and 10: we can add them at 7 and 9).
  4. The total number of rungs to add is 0 + 0 + 0 + 2 = 2.

The other language solutions follow the same logic, adapting the syntax to their respective languages. The key is the efficient calculation of the number of rungs needed for each gap between consecutive rungs, leveraging integer division to find the minimum number of additional rungs.