How do you solve Patching Array on LeetCode?

Patching Array (LeetCode #330) 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: Using a greedy approach minimizes the number of patches by always extending the range of reachable sums.

What is the brute force solution for Patching Array?

The brute force approach for Patching Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible sums from the existing array and checking if we can form all numbers from 1 to n. This method is straightforward but inefficient, as it can lead to exponential time complexity. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Patching Array?

Patching Array uses the following concepts: Array, Greedy. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Patching Array?

Clearly explain your thought process and the reasoning behind each step. Be prepared to discuss the trade-offs between different approaches. Practice explaining your solution as if teaching someone else, which can help clarify your understanding.

What are common mistakes when solving Patching Array?

Failing to recognize when to add a patch, often trying to add too many at once. Not keeping track of the maximum reachable sum correctly, leading to unnecessary patches.

#330

Patching Array

Hard

ArrayGreedyGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a greedy strategy to ensure that we can form all numbers from 1 to n with the minimum number of patches. We keep track of the maximum number we can form and add patches only when necessary.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to track the maximum number that can be formed (initially 0).
2Step 2: Iterate through the numbers from 1 to n.
3Step 3: If the current number is greater than the maximum number that can be formed, add it as a patch and update the maximum number.
4Step 4: If the current number is in the array, update the maximum number that can be formed by adding this number.

solution.py14 lines

1# Full working Python code
2def min_patches(nums, n):
3    patches = 0
4    max_reachable = 0
5    i = 0
6    while max_reachable < n:
7        if i < len(nums) and nums[i] <= max_reachable + 1:
8            max_reachable += nums[i]
9            i += 1
10        else:
11            patches += 1
12            max_reachable += (max_reachable + 1)
13    return patches
14

ℹ

Complexity note: The complexity is O(n) because we iterate through the array and the numbers we need to cover, ensuring we only make necessary patches.

1Using a greedy approach minimizes the number of patches by always extending the range of reachable sums.
2The maximum number that can be formed at any point is crucial for determining whether a patch is needed.

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