How do you solve Add Minimum Number of Rungs on LeetCode?

Add Minimum Number of Rungs (LeetCode #1936) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Always check the distance to the next rung before moving.

What is the time complexity of Add Minimum Number of Rungs?

The optimal solution for Add Minimum Number of Rungs runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the rungs array once, making it efficient even for larger inputs.

What is the brute force solution for Add Minimum Number of Rungs?

The brute force approach for Add Minimum Number of Rungs is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each rung one by one and determining if we can reach it from our current position. If we can't, we add a rung at the maximum possible height we can reach and continue. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Add Minimum Number of Rungs?

Add Minimum Number of Rungs uses the following concepts: Array, Greedy. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Add Minimum Number of Rungs?

Think about edge cases, such as the first rung being unreachable. Explain your thought process clearly while coding. Consider the efficiency of your solution and discuss potential optimizations.

What are common mistakes when solving Add Minimum Number of Rungs?

Failing to account for the distance from the ground to the first rung. Not updating the current height correctly after adding rungs.

#1936

Add Minimum Number of Rungs

Medium

ArrayGreedyGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution leverages a greedy approach. We only need to check the distance between the current position and the next rung, adding rungs only when necessary. This minimizes the number of rungs added by always filling the largest gaps first.

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Algorithm

6 steps

1Step 1: Initialize current_height to 0 and added_rungs to 0.
2Step 2: Iterate through each rung in the rungs array.
3Step 3: For each rung, check if the distance from current_height to the rung exceeds dist.
4Step 4: If it does, calculate how many rungs need to be added to bridge the gap and update added_rungs and current_height accordingly.
5Step 5: If the distance is within the limit, simply update current_height to the rung's height.
6Step 6: Return added_rungs after processing all rungs.

solution.py9 lines

1def addRungs(rungs, dist):
2    current_height = 0
3    added_rungs = 0
4    for rung in rungs:
5        if rung - current_height > dist:
6            gap = rung - current_height
7            added_rungs += (gap - 1) // dist
8        current_height = rung
9    return added_rungs

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Complexity note: The time complexity is O(n) because we only traverse the rungs array once, making it efficient even for larger inputs.

1Always check the distance to the next rung before moving.
2Adding rungs at the maximum height possible minimizes the number of rungs needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.