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 :



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                        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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