How do you solve Minimum Replacements to Sort the Array on LeetCode?

Minimum Replacements to Sort the Array (LeetCode #2366) 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: Breaking larger numbers into smaller parts can help maintain order.

What is the time complexity of Minimum Replacements to Sort the Array?

The optimal solution for Minimum Replacements to Sort the Array runs in O(n) time and uses O(1) space. This complexity is achieved because we only traverse the array once, making it efficient for large inputs.

What is the brute force solution for Minimum Replacements to Sort the Array?

The brute force approach for Minimum Replacements to Sort the Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible way to replace elements to sort the array. This is simple but inefficient, as it doesn't consider optimal replacements. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Replacements to Sort the Array?

Minimum Replacements to Sort the Array uses the following concepts: Array, Math, Greedy. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Minimum Replacements to Sort the Array?

Think about edge cases, such as already sorted arrays. Always clarify the problem requirements and constraints before jumping into coding. Discuss your thought process with the interviewer to demonstrate your understanding.

What are common mistakes when solving Minimum Replacements to Sort the Array?

Failing to recognize that the last element should not be replaced. Not calculating the number of replacements correctly when the current element exceeds the previous bound.

#2366

Minimum Replacements to Sort the Array

Hard

ArrayMathGreedyGreedyArray

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 solution iterates backward through the array, ensuring each element is less than or equal to the next. By breaking larger elements into smaller parts, we minimize operations.

⚙️

Algorithm

4 steps

1Step 1: Initialize a counter for operations and set a previous bound as infinity.
2Step 2: Iterate from the second last element to the first element of the array.
3Step 3: If the current element is greater than the previous bound, calculate how many replacements are needed to ensure it is less than or equal to the previous bound.
4Step 4: Update the operations counter accordingly.

solution.py14 lines

1# Full working Python code
2
3def min_replacements_optimal(nums):
4    operations = 0
5    prev_bound = float('inf')
6    for i in range(len(nums) - 2, -1, -1):
7        if nums[i] > prev_bound:
8            operations += (nums[i] - 1) // prev_bound
9            nums[i] = prev_bound
10        prev_bound = nums[i]
11    return operations
12
13# Example usage
14print(min_replacements_optimal([3, 9, 3]))

ℹ

Complexity note: This complexity is achieved because we only traverse the array once, making it efficient for large inputs.

1Breaking larger numbers into smaller parts can help maintain order.
2Always consider the previous element's value when deciding how to break down the current element.

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