1200. Minimum Absolute Difference
Easy
Given an array of distinct integers arr, find all pairs of elements with the minimum absolute difference of any two elements.
Return a list of pairs in ascending order(with respect to pairs), each pair [a, b] follows
a, bare fromarra < bb - aequals to the minimum absolute difference of any two elements inarr
Example 1:
Input: arr = [4,2,1,3] Output: [[1,2],[2,3],[3,4]] Explanation: The minimum absolute difference is 1. List all pairs with difference equal to 1 in ascending order.
Example 2:
Input: arr = [1,3,6,10,15] Output: [[1,3]]
Example 3:
Input: arr = [3,8,-10,23,19,-4,-14,27] Output: [[-14,-10],[19,23],[23,27]]
Constraints:
2 <= arr.length <= 105-106 <= arr[i] <= 106
class Solution:
def minimumAbsDifference(self, nums: List[int]) -> List[List[int]]:
# nums.sort()
# diff_map = defaultdict(list)
# for i, num in enumerate(nums[1:]):
# diff = abs(num - nums[i])
# diff_map[diff].append([nums[i], num] )
# return diff_map[min(diff_map.keys())]
# nums.sort()
# min_diff = sys.maxsize
# diff_map = []
# for i, num in enumerate(nums[1:]):
# diff = abs(num - nums[i])
# if diff == min_diff:
# diff_map.append([nums[i], num])
# elif diff < min_diff:
# min_diff = diff
# diff_map = [[nums[i], num]]
# return diff_map
# ---------------Using Count sort ----------------------
mini, maxi = min(nums), max(nums)
counter = [0] * (maxi-mini+1)
shift = -mini
for num in nums:
counter[num+shift] = 1
min_abs_diff = maxi-mini
min_abs_list = []
prev = 0
for curr in range(1, maxi+shift+1):
if counter[curr] == 0:
continue
if curr-prev == min_abs_diff:
min_abs_list.append([prev-shift, curr-shift])
elif curr-prev < min_abs_diff:
min_abs_list = [[prev-shift, curr-shift]]
min_abs_diff = curr - prev
prev = curr
return min_abs_list