#2009

Hard

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Minimum Number of Operations to Make Array Continuous

Find the minimum number of operations to make an array continuous by modifying elements using array scanning and hash lookup.

array hash table binary search sliding window

Problem Statement

You are given an integer array nums. In one operation, you can replace any element in nums with any integer. The goal is to make the array continuous. An array is continuous if it consists of consecutive integers, with no gaps in between. For example, [4, 2, 5, 3] is continuous, but [1, 2, 3, 5, 6] is not.

The task is to find the minimum number of operations required to make the array continuous. You may change any element to any integer, and the array should have consecutive integers. The input constraint allows the array to be as large as 100,000 elements, and each element can be as large as one billion.

Examples

Example 1

Input: nums = [4,2,5,3]

Output: 0

nums is already continuous.

Example 2

Input: nums = [1,2,3,5,6]

Output: 1

One possible solution is to change the last element to 4. The resulting array is [1,2,3,5,4], which is continuous.

Example 3

Input: nums = [1,10,100,1000]

Output: 3

One possible solution is to:

Change the second element to 2.
Change the third element to 3.
Change the fourth element to 4. The resulting array is [1,2,3,4], which is continuous.

Constraints

1 <= nums.length <= 105
1 <= nums[i] <= 109

Solution Approach

Sort the Array

The first step is to sort the array to facilitate easier identification of gaps between consecutive integers. By sorting, you can more easily find continuous subarrays. This allows us to focus on smaller, sorted windows instead of brute-forcing every possible arrangement.

Sliding Window Approach

Once the array is sorted, use a sliding window technique to find the longest subarray of consecutive integers. This approach reduces the complexity of checking each potential subarray, allowing for efficient searching of the largest continuous segment.

Hash Lookups for Gap Detection

To identify gaps between consecutive elements in the sorted array, hash tables can be employed. This enables quick lookups to verify which elements need to be modified and ensures that only the minimal number of operations are performed.

Complexity Analysis

Metric	Value
Time	O(n \cdot \log{}n)
Space	O(n)

The time complexity of this solution is O(n log n) due to the sorting step. The sliding window and hash lookups operate in O(n) time. Space complexity is O(n) as we store the array and any auxiliary data structures for tracking continuous subarrays.

What Interviewers Usually Probe

The candidate understands the need for sorting before applying any further optimizations.
The candidate efficiently identifies the correct subarray using sliding window logic.
The candidate uses hash tables effectively to minimize operations and detect gaps.

Common Pitfalls or Variants

Common pitfalls

Failing to sort the array initially, leading to unnecessary complexity in gap detection.
Misunderstanding the sliding window approach and applying it incorrectly, resulting in higher complexity.
Not considering the impact of large array sizes and struggling with time and space constraints.

Follow-up variants

What if the array contains only a single element? The problem becomes trivial as no operations are needed.
What if the array is already sorted or already continuous? The result will be 0 operations required.
If the array contains duplicate elements, they can still be part of the solution if they fit into the continuous subarray.

FAQ

What is the minimum number of operations to make an array continuous?

It is the fewest number of element replacements needed to transform the array into a continuous sequence of consecutive integers.

How do I solve the 'Minimum Number of Operations to Make Array Continuous' problem?

The solution involves sorting the array, using a sliding window to find the longest continuous subarray, and employing hash lookups to minimize the number of modifications.

What is the primary approach to solving the 'Minimum Number of Operations to Make Array Continuous' problem?

Array scanning combined with hash lookup is the primary approach, utilizing sorting and sliding windows to efficiently solve the problem.

What data structure is used for gap detection in this problem?

A hash table is used to quickly detect gaps between consecutive elements in the sorted array.

What is the time complexity of the solution to this problem?

The time complexity is O(n log n) due to the sorting step, with O(n) complexity for the sliding window and hash lookups.

terminal

Solution

Solution 1: Sorting + Deduplication + Binary Search

First, we sort the array and remove duplicates.

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class Solution:
    def minOperations(self, nums: List[int]) -> int:
        ans = n = len(nums)
        nums = sorted(set(nums))
        for i, v in enumerate(nums):
            j = bisect_right(nums, v + n - 1)
            ans = min(ans, n - (j - i))
        return ans

Solution 2: Sorting + Deduplication + Two Pointers

Similar to Solution 1, we first sort the array and remove duplicates.

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class Solution:
    def minOperations(self, nums: List[int]) -> int:
        ans = n = len(nums)
        nums = sorted(set(nums))
        for i, v in enumerate(nums):
            j = bisect_right(nums, v + n - 1)
            ans = min(ans, n - (j - i))
        return ans

Continue Practicing

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