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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0901_1000.s0908_smallest_range_i;

// #Easy #Array #Math #2022_03_28_Time_2_ms_(88.84%)_Space_41.9_MB_(99.76%)

/**
 * 908 - Smallest Range I\.
 *
 * Easy
 *
 * You are given an integer array `nums` and an integer `k`.
 *
 * In one operation, you can choose any index `i` where `0 <= i < nums.length` and change `nums[i]` to `nums[i] + x` where `x` is an integer from the range `[-k, k]`. You can apply this operation **at most once** for each index `i`.
 *
 * The **score** of `nums` is the difference between the maximum and minimum elements in `nums`.
 *
 * Return _the minimum **score** of_ `nums` _after applying the mentioned operation at most once for each index in it_.
 *
 * **Example 1:**
 *
 * **Input:** nums = [1], k = 0
 *
 * **Output:** 0
 *
 * **Explanation:** The score is max(nums) - min(nums) = 1 - 1 = 0.
 *
 * **Example 2:**
 *
 * **Input:** nums = [0,10], k = 2
 *
 * **Output:** 6
 *
 * **Explanation:** Change nums to be [2, 8]. The score is max(nums) - min(nums) = 8 - 2 = 6.
 *
 * **Example 3:**
 *
 * **Input:** nums = [1,3,6], k = 3
 *
 * **Output:** 0
 *
 * **Explanation:** Change nums to be [4, 4, 4]. The score is max(nums) - min(nums) = 4 - 4 = 0.
 *
 * **Constraints:**
 *
 * *   1 <= nums.length <= 10⁴
 * *   0 <= nums[i] <= 10⁴
 * *   0 <= k <= 10⁴
**/
public class Solution {
    public int smallestRangeI(int[] nums, int k) {
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        for (int num : nums) {
            min = Math.min(min, num);
            max = Math.max(max, num);
        }
        if (min + k >= max - k) {
            return 0;
        }
        return (max - k) - (min + k);
    }
}