4115 Minimum Distance Between Three Equal Elements I
Minimum Distance Between Three Equal Elements I 
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
2
1 <= n == nums.length <= 100
1 <= nums[i] <= n
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class Solution:
def minimumDistance(self, nums: List[int]) -> int:
# if i sort these on nums and then on index
# then i slide a window of len 3
# and calculate the dist
if len(nums) < 3:
return -1
arr = []
for (i, k) in enumerate(nums):
arr.append((k,i))
arr.sort(key = lambda x : (x[0], x[1]))
print(arr)
res = 1000000
left = 0
right = 2
while(right < len(arr)):
if arr[left][0] == arr[left + 1][0] and arr[left+1][0] == arr[right][0]:
res = min(res, abs(arr[left][1] - arr[left + 1][1]) + abs(arr[left + 1][1] - arr[left + 2][1]) + abs(arr[left + 2][1] - arr[left][1]))
right += 1
left += 1
return -1 if res == 1000000 else res