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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 <= A.length <= 10000
0 <= A[i] <= 10000
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); } }
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// 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); } };
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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
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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 }
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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); }
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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) } }