**Minimize Amplitude After K Removals - Google Top Interview Questions**

### Problem Statement :

You are given a list of integers nums, and an integer k. Given that you must remove k elements in nums, return the minimum max(nums) - min(nums) value we can achieve. Constraints 1 ≤ n ≤ 100,000 where n is the length of nums k < n Example 1 Input nums = [2, 10, 14, 12, 30] k = 2 Output 4 Explanation If we remove 2 and 30 then we'd get [10, 14, 12] and 14 - 10 = 4.

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums, int k) {
int n = nums.size();
sort(nums.begin(), nums.end()); // O(nlogn)
int min = nums.back() - nums[0];
if (k == 0) return min;
for (int i = 0; i + n - k - 1 < nums.size(); ++i) { // O(n)
if (nums[i + n - k - 1] - nums[i] < min) min = nums[i + n - k - 1] - nums[i];
}
return min;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums, int k) {
Arrays.sort(nums);
int n = nums.length;
if (k >= n - 1)
return 0;
int res = Integer.MAX_VALUE;
for (int i = 0; i <= k; i++) {
int remove_other = k - i;
// remove i from left
res = Math.min(res, nums[n - 1 - remove_other] - nums[i]);
// remove i from right
res = Math.min(res, nums[n - 1 - i] - nums[remove_other]);
}
return res;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums, k):
keep = len(nums) - k
nums.sort()
return min(nums[i + keep - 1] - nums[i] for i in range(len(nums) - keep + 1))
```

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