LeetCode in Kotlin
1848. Minimum Distance to the Target Element
Easy
Given an integer array nums (0-indexed) and two integers target and start, find an index i such that nums[i] == target and abs(i - start) is minimized. Note that abs(x) is the absolute value of x.
Return abs(i - start).
It is guaranteed that target exists in nums.
Example 1:
Input: nums = [1,2,3,4,5], target = 5, start = 3
Output: 1
Explanation: nums[4] = 5 is the only value equal to target, so the answer is abs(4 - 3) = 1.
Example 2:
Input: nums = [1], target = 1, start = 0
Output: 0
Explanation: nums[0] = 1 is the only value equal to target, so the answer is abs(0 - 0) = 0.
Example 3:
Input: nums = [1,1,1,1,1,1,1,1,1,1], target = 1, start = 0
Output: 0
Explanation: Every value of nums is 1, but nums[0] minimizes abs(i - start), which is abs(0 - 0) = 0.
Constraints:
1 <= nums.length <= 10001 <= nums[i] <= 1040 <= start < nums.lengthtargetis innums.
Solution
import kotlin.math.abs
class Solution {
fun getMinDistance(nums: IntArray, target: Int, start: Int): Int {
var result = 0
var minDiff = Int.MAX_VALUE
for (i in nums.indices) {
if (nums[i] == target && abs(start - i) < minDiff) {
minDiff = abs(start - i)
result = minDiff
}
}
return result
}
}