Patching Array

The given problem can be solved using a greedy algorithm. The idea is to iterate through the given array and maintain a variable reach, which represents the current range [1, reach] that we can cover with the numbers seen so far.

We’ll also keep a count of the number of patches we need to add to cover the range [1, n].

Here’s the step-by-step explanation:

1. Initialize Variables: Start with reach = 0 and patches = 0. reach represents the current range that we can cover without adding any patches.

2. Iterate through the Array: Iterate through the given array. If the current number is less than or equal to reach + 1, then you can extend your reach. Otherwise, you’ll need to add a patch, and the value to be patched is reach + 1.

3. Check for Patches: If num <= reach + 1, extend the reach by adding the current number to it: reach += num. If num > reach + 1, add a patch for reach + 1, increment patches, and double the reach: reach += reach + 1.

4. Final Check: Continue checking for patches until you have covered the entire range [1, n].

5. Return Result: Return the number of patches required.

Here’s the code that implements the above logic:

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class Solution:
    def minPatches(self, nums, n):
        patches, reach, i = 0, 0, 0
        while reach < n:
            if i < len(nums) and nums[i] <= reach + 1:
                reach += nums[i]
                i += 1
            else:
                patches += 1
                reach += reach + 1
        return patches

Key Takeaways

• The algorithm iteratively adds patches to the array to cover the range [1, n].
• By following a greedy approach, the algorithm ensures that it always picks the best immediate choice to extend the range that can be formed.
• The time complexity of the solution is (O(k)), where (k) is the length of the input array, and the space complexity is (O(1)).

Here’s the implementation of the minPatches method:

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class Solution:
    def minPatches(self, nums, n):
        patches = 0  # Count of patches added
        covered = 0  # Range covered so far
        index = 0

        while covered < n:
            if index < len(nums) and nums[index] <= covered + 1:
                covered += nums[index]  # Extend the covered range
                index += 1
            else:
                patches += 1  # Add a patch to cover the uncovered range
                covered = covered * 2 + 1

        return patches

This implementation calculates the minimum number of patches required to cover the range from 1 to n using the given sorted nums array. It iterates through the nums array and extends the covered range if the current number can be added to it. If not, it adds a patch to cover the range and doubles the covered range plus one. The loop continues until the entire range is covered. The patches variable keeps track of the added patches, which is the result returned by the method.

10 Prerequisite LeetCode Problems

“Patching Array” requires an understanding of the greedy algorithm approach and array manipulation. Here are some simpler problems to prepare for it:

1. 55. Jump Game: Helps understand the concept of reaching a target by jumping over elements in an array.

2. 45. Jump Game II: This is a more complex version of Jump Game and introduces the concept of minimal jumps.

3. 134. Gas Station: This problem introduces the concept of completing a circuit by greedy approach.

4. 605. Can Place Flowers: Helps to understand how to place elements in an array in the most optimal way.

5. 392. Is Subsequence: This problem teaches you how to determine whether an array is a subsequence of another.

6. 122. Best Time to Buy and Sell Stock II: This problem helps understand greedy approach in an array to maximize profit.

7. 406. Queue Reconstruction by Height: This problem introduces the concept of arranging elements in a specific order in an array.

8. 53. Maximum Subarray: Teaches the concept of finding a contiguous subarray which has the largest sum, which is fundamental for array problems.

9. 283. Move Zeroes: A simple problem which teaches how to rearrange elements in an array.

10. 121. Best Time to Buy and Sell Stock: This problem helps to understand how to maximize profit using the price array.

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class Solution:
  def minPatches(self, nums: List[int], n: int) -> int:
    miss, i, patches = 1, 0, 0
    while miss <= n:
        if i < len(nums) and nums[i] <= miss:
            miss += nums[i]
            i += 1
        else:
            miss *= 2
            patches += 1
    return patches

Problem Classification

Patching Array - This problem requires you to add elements to an array to make it possible to generate a sequence of numbers. This is a Array Modification Problem.

Language Agnostic Coding Drills

1. Understand and work with variables: This problem uses integers and lists in Python. Understanding how to work with these basic data types and manipulate their values is essential.

2. Control flow with conditionals and loops: The solution utilizes a while loop and an if-else condition within it. Understanding the concepts of loops, conditionals, and how to use them in various scenarios is a crucial step.

3. Working with lists: Understanding how to work with lists, especially accessing their elements and getting their length, is crucial for this problem.

4. Basic arithmetic operations: The solution uses basic arithmetic operations such as addition, comparison, and multiplication.

5. Understanding the problem-specific logic (Greedy algorithm): In this problem, we try to increase the range of covered numbers as much as possible in each step. This is the essence of a greedy algorithm: making the choice that seems to be the best at the moment.

Problem-solving approach:

1. Initialize the variables miss (the smallest sum in the [0,n] range that we might be missing), patches (number of patches used), and i (index to the given array nums).

2. Run a loop until miss is smaller or equal to n. Inside the loop, we have two main cases:

   • If the current array element is less than or equal to miss, that means we can add the current number to the range [0, miss) to create a new range [0, miss+nums[i]). After considering this number, move forward in the array.
   • If the current array element is greater than miss, we can’t create a continuous range. The best option is to patch miss itself to the array. This is because adding miss doubles the range that we’ve covered so far.

3. Return the number of patches used when the loop is over.

At the end of the process, the number of patches (patches) is the minimum number of patches we need to apply to ensure every number in the range [1, n] is covered.

Targeted Drills in Python

1. Understanding and working with variables Here’s a simple drill to practice variable assignments and basic arithmetic.

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   # Assign values to variables and perform arithmetic operations a = 10 b = 20 print(a + b) # Should print 30

2. Control flow with conditionals and loops Practice the usage of if-else statements and a while loop.

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   # Print the first 10 natural numbers using a while loop i = 1 while i <= 10: print(i) i += 1

3. Working with lists Learn to access elements in a list and find its length.

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   # Create a list and perform operations nums = [1, 2, 3, 4, 5] print(nums[0]) # Print the first element print(len(nums)) # Print the length of the list

4. Basic arithmetic operations Get comfortable with Python’s arithmetic operations.

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   # Perform arithmetic operations a = 15 b = 5 print(a * b) # Multiplication, should print 75 print(a + b) # Addition, should print 20

5. Understanding the problem-specific logic (Greedy algorithm) The core idea is to keep track of the range of numbers we can form and try to extend it as much as possible in each step. This drill simplifies the problem: you have a number n and you can either add 1 to it or double it, starting from 1. Find the smallest number of operations to reach n.

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   def min_operations(n): # We start with number 1 current = 1 operations = 0 while current < n: if current * 2 <= n: # It's safe to double the number current *= 2 else: # It's safe to add 1 current += 1 operations += 1 return operations print(min_operations(9)) # Should print 4: 1->2->4->8->9

By practicing these drills, you should get comfortable with the basic programming concepts and the problem-specific logic involved in the solution. These individual drills can then be combined to solve the original problem.