carloscn
commented
1 year ago Analysis
pub fn get_min_distance(nums: Vec<i32>, target: i32, start: i32) -> i32
{
if nums.len() < 1 {
return -1;
}
if nums[start as usize] == target {
return 0;
}
let mut i:i32 = start + 1;
let mut j:i32 = start;
let len:i32 = nums.len() as i32;
while i <= len || j >= 0 {
if i < len && nums[i as usize] == target {
return i - start;
}
i += 1;
if j >= 0 && nums[j as usize] == target {
return start - j;
}
j -= 1;
}
return -1;
Description
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 <= 1000 1 <= nums[i] <= 104 0 <= start < nums.length target is in nums.