Description
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
Solutions
This solution uses an array instead of a tuple to store indices, and checks arr.length >= 2 before accessing the second-to-last index. It iterates through the array, and for each duplicate value with at least two prior occurrences, it calculates the distance k - arr[arr.length - 2] and tracks the minimum. Finally, it returns -1 if no duplicates are found, or 2 * min for the minimum distance between any two occurrences.
/**
* @param {number[]} nums
* @return {number}
*/
var minimumDistance = function(nums) {
const map = new Map();
let min = Infinity;
for (let k = 0; k < nums.length; k++) {
const arr = map.get(nums[k]) || [];
if (arr.length === 0) {
map.set(nums[k], arr);
} else if (arr.length >= 2) {
min = Math.min(min, k - arr[arr.length - 2]);
}
arr.push(k);
}
if (min === Infinity) {
return -1;
}
return 2 * min;
};