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1200. Minimum Absolute Difference

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 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 <= 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 $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $arr$.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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]
  
  

• 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
  }
  

• 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;
  }
  
  

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