Minimum Absolute Difference - LeetCode 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, b are from arr
a < b
b - a equals to the minimum absolute difference of any two elements in arr
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 <= 10^5
-10^6 <= arr[i] <= 10^6
- code
class Solution:
def minimumAbsDifference(self, arr: List[int]) -> List[List[int]]:
arr.sort()
min_d = inf
res = []
for i in range(1, len(arr)):
cur_d = arr[i] - arr[i-1]
if cur_d < min_d:
res = [[arr[i-1], arr[i]]]
min_d = cur_d
elif cur_d == min_d:
res.append([arr[i-1], arr[i]])
return res
- code counting sort Counting sort - Wikipedia
class Solution {
public List<List<Integer>> minimumAbsDifference(int[] arr) {
int min = arr[0], max = arr[0], shift;
for (int num: arr){
min = Math.min(min, num);
max = Math.max(max, num);
}
shift = -min;
int[] line = new int[max - min + 1];
for (int num: arr){
line[num + shift] = 1;
}
List<List<Integer>> res = new ArrayList();
int last_i = 0;
int min_diff = max - min;
for (int i = 1; i < line.length; ++i){
if (line[i] == 1){
if (i - last_i < min_diff){
res = new ArrayList();
res.add(Arrays.asList(last_i - shift, i - shift));
min_diff = i - last_i;
}
else if (i - last_i == min_diff){
res.add(Arrays.asList(last_i - shift, i - shift));
}
last_i = i;
}
}
return res;
}
}