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.lengthtarget is in nums.struct Solution;
impl Solution {
fn get_min_distance(nums: Vec<i32>, target: i32, start: i32) -> i32 {
let n = nums.len();
let mut res = std::i32::MAX;
for i in 0..n {
if nums[i] == target {
res = res.min((i as i32 - start).abs());
}
}
res
}
}
#[test]
fn test() {
let nums = vec![1, 2, 3, 4, 5];
let target = 5;
let start = 3;
let res = 1;
assert_eq!(Solution::get_min_distance(nums, target, start), res);
let nums = vec![1];
let target = 1;
let start = 0;
let res = 0;
assert_eq!(Solution::get_min_distance(nums, target, start), res);
let nums = vec![1, 1, 1, 1, 1, 1, 1, 1, 1, 1];
let target = 1;
let start = 0;
let res = 0;
assert_eq!(Solution::get_min_distance(nums, target, start), res);
}