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Minimum Number of Operations to Make Array Continuous

Difficulty: Hard

Problem Description

You are given an integer array nums. In one operation, you can replace any element in nums with any integer. nums is considered continuous if both of the following conditions are fulfilled:

  • All elements in nums are unique.
  • The difference between the maximum element and the minimum element in nums equals nums.length - 1.

Return the minimum number of operations to make nums continuous.

Key Insights

  • To have a continuous array, we need unique elements that form a consecutive range.
  • The maximum difference between the highest and lowest values in the array should equal the length of the array minus one.
  • Using a sliding window approach helps find the longest valid subarray of unique numbers that can be transformed into a continuous array with the least number of changes.

Space and Time Complexity

Time Complexity: O(n log n) - due to sorting the array. Space Complexity: O(n) - for storing the sorted array and any additional data structures used.

Solution

The problem can be approached using the following steps:

  1. Sort the array to easily find the maximum and minimum values in any subarray.
  2. Use a two-pointer (sliding window) technique to find the longest subarray where the conditions for continuity are satisfied.
  3. The number of operations needed would be the size of the array minus the size of the longest valid subarray.

The primary data structure used is a sorted array, and the algorithm leverages sorting and a sliding window approach to efficiently compute the result.

Code Solutions

def minOperations(nums):
    nums = sorted(set(nums))  # Sort and remove duplicates
    n = len(nums)
    max_length = 0
    
    left = 0
    for right in range(n):
        # Ensure the current window is valid
        while nums[right] - nums[left] >= n:
            left += 1
        max_length = max(max_length, right - left + 1)
    
    return n - max_length  # The number of operations needed

# Example usage:
print(minOperations([1, 2, 3, 5, 6]))  # Output: 1
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