LeetCode #1936 — MEDIUM

Add Minimum Number of Rungs

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[1,3,5,10]
2

Example 2

[3,6,8,10]
3

Example 3

[3,4,6,7]
2

Core Insight

What unlocks the optimal approach

Go as far as you can on the available rungs before adding new rungs.
If you have to add a new rung, add it as high up as possible.
Try using division to decrease the number of computations.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1936: Add Minimum Number of Rungs
class Solution {
    public int addRungs(int[] rungs, int dist) {
        int ans = 0, prev = 0;
        for (int x : rungs) {
            ans += (x - prev - 1) / dist;
            prev = x;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1936: Add Minimum Number of Rungs
func addRungs(rungs []int, dist int) (ans int) {
	prev := 0
	for _, x := range rungs {
		ans += (x - prev - 1) / dist
		prev = x
	}
	return
}

# Accepted solution for LeetCode #1936: Add Minimum Number of Rungs
class Solution:
    def addRungs(self, rungs: List[int], dist: int) -> int:
        rungs = [0] + rungs
        return sum((b - a - 1) // dist for a, b in pairwise(rungs))

// Accepted solution for LeetCode #1936: Add Minimum Number of Rungs
impl Solution {
    pub fn add_rungs(rungs: Vec<i32>, dist: i32) -> i32 {
        let mut ans = 0;
        let mut prev = 0;

        for &x in rungs.iter() {
            ans += (x - prev - 1) / dist;
            prev = x;
        }

        ans
    }
}

// Accepted solution for LeetCode #1936: Add Minimum Number of Rungs
function addRungs(rungs: number[], dist: number): number {
    let ans = 0;
    let prev = 0;
    for (const x of rungs) {
        ans += ((x - prev - 1) / dist) | 0;
        prev = x;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.