LeetCode-in-Java
3740. Minimum Distance Between Three Equal Elements I
Easy
You are given an integer array nums.
A tuple (i, j, k) of 3 distinct indices is good if nums[i] == nums[j] == nums[k].
The distance of a good tuple is abs(i - j) + abs(j - k) + abs(k - i), where abs(x) denotes the absolute value of x.
Return an integer denoting the minimum possible distance of a good tuple. If no good tuples exist, return -1.
Example 1:
Input: nums = [1,2,1,1,3]
Output: 6
Explanation:
The minimum distance is achieved by the good tuple (0, 2, 3).
(0, 2, 3) is a good tuple because nums[0] == nums[2] == nums[3] == 1. Its distance is abs(0 - 2) + abs(2 - 3) + abs(3 - 0) = 2 + 1 + 3 = 6.
Example 2:
Input: nums = [1,1,2,3,2,1,2]
Output: 8
Explanation:
The minimum distance is achieved by the good tuple (2, 4, 6).
(2, 4, 6) is a good tuple because nums[2] == nums[4] == nums[6] == 2. Its distance is abs(2 - 4) + abs(4 - 6) + abs(6 - 2) = 2 + 2 + 4 = 8.
Example 3:
Input: nums = [1]
Output: -1
Explanation:
There are no good tuples. Therefore, the answer is -1.
Constraints:
1 <= n == nums.length <= 1001 <= nums[i] <= n
Solution
public class Solution {
public int minimumDistance(int[] nums) {
int len = nums.length;
int[] last2 = new int[len];
int res = 200;
for (int i = 0; i < len; i++) {
int val = nums[i] - 1;
int pos = i + 1;
int pack = last2[val];
int old = pack & 255;
int cur = pack >> 8;
last2[val] = cur | (pos << 8);
if (old > 0) {
res = Math.min(res, (pos - old) << 1);
}
}
return res == 200 ? -1 : res;
}
}