LeetCode 908: Smallest Range I Solution

Problem Explanation

Explanation

To minimize the score of nums after applying the operation, we need to make the range of values in nums as small as possible. We can achieve this by adjusting the elements in nums to bring them closer together. Since we can only change each element once within the range [-k, k], the optimal strategy is to increase the minimum values and decrease the maximum values. This will reduce the overall difference between the maximum and minimum elements, thus minimizing the score.

To calculate the minimum score after applying the operation, we need to find the current range of values in nums and check if it can be reduced by at most 2k. If the current range is already smaller than 2k, the minimum score will be 0. Otherwise, we can reduce the range by 2*k to minimize the score.

Time complexity: O(n) where n is the length of the input array nums.
Space complexity: O(1)

Solution Code

class Solution {
    public int smallestRangeI(int[] nums, int k) {
        int minNum = nums[0];
        int maxNum = nums[0];
        
        for (int num : nums) {
            minNum = Math.min(minNum, num);
            maxNum = Math.max(maxNum, num);
        }
        
        int diff = maxNum - minNum - 2 * k;
        
        return diff > 0 ? diff : 0;
    }
}

Frequently Asked Questions

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908. Smallest Range I