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Formatted question description: https://leetcode.ca/all/908.html

908. Smallest Range I (Easy)

Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add x to A[i].

After this process, we have some array B.

Return the smallest possible difference between the maximum value of B and the minimum value of B.

 

Example 1:

Input: A = [1], K = 0
Output: 0
Explanation: B = [1]

Example 2:

Input: A = [0,10], K = 2
Output: 6
Explanation: B = [2,8]

Example 3:

Input: A = [1,3,6], K = 3
Output: 0
Explanation: B = [3,3,3] or B = [4,4,4]

 

Note:

  1. 1 <= A.length <= 10000
  2. 0 <= A[i] <= 10000
  3. 0 <= K <= 10000

Solution 1.

  • class Solution {
        public int smallestRangeI(int[] A, int K) {
            Arrays.sort(A);
            int length = A.length;
            int min = A[0], max = A[length - 1];
            int difference = max - min;
            return Math.max(0, difference - 2 * K);
        }
    }
    
    ############
    
    class Solution {
        public int smallestRangeI(int[] nums, int k) {
            int mx = 0;
            int mi = 10000;
            for (int v : nums) {
                mx = Math.max(mx, v);
                mi = Math.min(mi, v);
            }
            return Math.max(0, mx - mi - k * 2);
        }
    }
    
  • // OJ: https://leetcode.com/problems/smallest-range-i/
    // Time: O(N)
    // Space: O(1)
    class Solution {
    public:
        int smallestRangeI(vector<int>& A, int K) {
            int minVal = INT_MAX, maxVal = INT_MIN;
            for (int n : A) {
                minVal = min(minVal, n);
                maxVal = max(maxVal, n);
            }
            return max(maxVal - minVal - 2 * K, 0);
        }
    };
    
  • class Solution:
        def smallestRangeI(self, nums: List[int], k: int) -> int:
            mx, mi = max(nums), min(nums)
            return max(0, mx - mi - k * 2)
    
    ############
    
    class Solution(object):
        def smallestRangeI(self, nums, k):
            """
            :type nums: List[int]
            :type k: int
            :rtype: int
            """
            diff = max(nums) - min(nums)
            if diff > 2 * k:
                return diff - 2 * k
            return 0
    
  • func smallestRangeI(nums []int, k int) int {
    	mx, mi := 0, 10000
    	for _, v := range nums {
    		mx = max(mx, v)
    		mi = min(mi, v)
    	}
    	return max(0, mx-mi-k*2)
    }
    
    func max(a, b int) int {
    	if a > b {
    		return a
    	}
    	return b
    }
    
    func min(a, b int) int {
    	if a < b {
    		return a
    	}
    	return b
    }
    
  • function smallestRangeI(nums: number[], k: number): number {
        const max = nums.reduce((r, v) => Math.max(r, v));
        const min = nums.reduce((r, v) => Math.min(r, v));
        return Math.max(max - min - k * 2, 0);
    }
    
    
  • impl Solution {
        pub fn smallest_range_i(nums: Vec<i32>, k: i32) -> i32 {
            let max = nums.iter().max().unwrap();
            let min = nums.iter().min().unwrap();
            0.max(max - min - k * 2)
        }
    }
    
    

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