Description
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 arra < bb - 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 <= 105-106 <= arr[i] <= 106
Solutions
Solution 1: Sorting
According to the problem description, we need to find the minimum absolute difference between any two elements in the array arr. Therefore, we can first sort the array arr, then traverse the adjacent elements to get the minimum absolute difference mi.
Finally, we traverse the adjacent elements again to find all pairs of elements where the minimum absolute difference equals mi.
The time complexity is O(n × log n), and the space complexity is O(log n). Here, n is the length of the array arr.
PythonJavaC++GoTypeScriptRust
class Solution:
def minimumAbsDifference(self, arr: List[int]) -> List[List[int]]:
arr.sort()
mi = min(b - a for a, b in pairwise(arr))
return [[a, b] for a, b in pairwise(arr) if b - a == mi](code-box)
class Solution {
public List<List<Integer>> minimumAbsDifference(int[] arr) {
Arrays.sort(arr);
int n = arr.length;
int mi = 1 << 30;
for (int i = 0; i < n - 1; ++i) {
mi = Math.min(mi, arr[i + 1] - arr[i]);
}
List<List<Integer>> ans = new ArrayList<>();
for (int i = 0; i < n - 1; ++i) {
if (arr[i + 1] - arr[i] == mi) {
ans.add(List.of(arr[i], arr[i + 1]));
}
}
return ans;
}
}(code-box)
class Solution {
public:
vector<vector<int>> minimumAbsDifference(vector<int>& arr) {
sort(arr.begin(), arr.end());
int mi = 1 << 30;
int n = arr.size();
for (int i = 0; i < n - 1; ++i) {
mi = min(mi, arr[i + 1] - arr[i]);
}
vector<vector<int>> ans;
for (int i = 0; i < n - 1; ++i) {
if (arr[i + 1] - arr[i] == mi) {
ans.push_back({arr[i], arr[i + 1]});
}
}
return ans;
}
};(code-box)
func minimumAbsDifference(arr []int) (ans [][]int) {
sort.Ints(arr)
mi := 1 << 30
n := len(arr)
for i := 0; i < n-1; i++ {
if t := arr[i+1] - arr[i]; t < mi {
mi = t
}
}
for i := 0; i < n-1; i++ {
if arr[i+1]-arr[i] == mi {
ans = append(ans, []int{arr[i], arr[i+1]})
}
}
return
}(code-box)
function minimumAbsDifference(arr: number[]): number[][] {
arr.sort((a, b) => a - b);
let mi = 1 << 30;
const n = arr.length;
for (let i = 0; i < n - 1; ++i) {
mi = Math.min(mi, arr[i + 1] - arr[i]);
}
const ans: number[][] = [];
for (let i = 0; i < n - 1; ++i) {
if (arr[i + 1] - arr[i] === mi) {
ans.push([arr[i], arr[i + 1]]);
}
}
return ans;
}(code-box)
impl Solution {
pub fn minimum_abs_difference(mut arr: Vec<i32>) -> Vec<Vec<i32>> {
arr.sort();
let n = arr.len();
let mut mi = i32::MAX;
for i in 1..n {
mi = mi.min(arr[i] - arr[i - 1]);
}
let mut ans = Vec::new();
for i in 1..n {
if arr[i] - arr[i - 1] == mi {
ans.push(vec![arr[i - 1], arr[i]]);
}
}
ans
}
}(code-box)
