LeetCode 2009: Minimum Number of Operations to Make Array Continuous Solution

Master LeetCode problem 2009 (Minimum Number of Operations to Make Array Continuous), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2009. Minimum Number of Operations to Make Array Continuous

Hard

Array Hash Table Binary Search Sliding Window

Problem Explanation

Explanation

To make the array continuous, we need to find the minimum number of operations required to ensure that all elements are unique and the difference between the maximum and minimum elements is equal to the length of the array minus 1.

We first sort the array.
For each possible subarray of length N (where N is the length of the input array), we find the valid range of numbers that can form a continuous array.
We iterate through each subarray, finding the count of elements that need to be replaced to form a continuous array.
We keep track of the minimum count of replacements needed.

Time complexity: O(NlogN) - sorting the array Space complexity: O(N) - to store the sorted array

Solution Code

import java.util.Arrays;

class Solution {
    public int minOperations(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        int minOps = n;
        int j = 0;

        for (int i = 0; i < n; i++) {
            while (j < n && nums[j] <= nums[i] + n - 1) {
                j++;
            }

            minOps = Math.min(minOps, n - (j - i));
        }

        return minOps;
    }
}

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