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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.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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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