Minimum Number of Operations to Make Array Continuous

Problem Statement

Leetcode: 2009. Minimum Number of Operations to Make Array Continuous

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.

For example, nums = [4, 2, 5, 3] is continuous, but nums = [1, 2, 3, 5, 6] is not continuous.

Return the minimum number of operations to make nums continuous.

Example 1:

Input: nums = [4,2,5,3]  
Output: 0  
Explanation: nums is already continuous.  

Example 2:

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

class Solution {
public:

    int minOperations(vector<int>& nums) {

        vector<int> unique;
        map<int, int> seen;
        for(auto n: nums) {
            if (!seen[n]) {
                seen[n] = 1;
                unique.push_back(n);
            }
        }
        sort(unique.begin(), unique.end());

        // find the mininum number -> max = length - min
        // for each num, check the index for upper_bount of max element
        // for example - for array of length 7, if min = 2
        // and max = 8, and there are 4 unique elements [2, 8]

        vector<int>::iterator low, up;
        int ops = INT_MAX;
        int n = nums.size();

        for (low = unique.begin(); low < unique.end(); low++) {
            int val = *low;
            int maximum = nums.size() + val - 1;
            up = upper_bound(low, unique.end(), maximum);
            int present = up - low;

            ops = min(ops, n - present);

            if (ops < (low - unique.begin())) {
                break;
            }
        }

        return ops;
    }
};