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
equalsnums.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
class Solution:
def minOperations(self, nums: List[int]) -> int:
n = len(nums)
nums = sorted(set(nums))
ans = n
for i, start in enumerate(nums):
end = start + n - 1
j = bisect_right(nums, end)
remainLen = j - i
ans = min(ans, n - remainLen)
return ans
class Solution {
public int minOperations(int[] nums) {
int N = nums.length;
if (N == 1) return 0;
Arrays.sort(nums);
int M = 1;
for (int i = 1; i < N; i++) {
if (nums[i] != nums[i - 1])
nums[M++] = nums[i];
}
int j = 0;
int ans = N;
for (int i = 0; i < M; i++) {
while (j < M && nums[j] <= N + nums[i] - 1) j++;
ans = Math.min(ans, N - j + i);
}
return ans;
}
}
class Solution {
public:
int minOperations(vector<int>& nums) {
sort(nums.begin(), nums.end());
int End = unique(nums.begin(), nums.end()) - nums.begin();
int n = nums.size();
int len = 0;
for (int i = 0; i < End; ++i) {
int temp = upper_bound(nums.begin(), nums.begin() + End, n + nums[i] - 1) - nums.begin() - i;
len = max(len, temp);
}
return n - len;
}
};