# LeetCode 2340. Minimum Adjacent Swaps to Make a Valid Array

You are given a 0-indexed integer array nums.

Swaps of adjacent elements are able to be performed on nums.

A valid array meets the following conditions:

• The largest element (any of the largest elements if there are multiple) is at the rightmost position in the array.
• The smallest element (any of the smallest elements if there are multiple) is at the leftmost position in the array.

Return the minimum swaps required to make nums a valid array.

Example 1:

Input: nums = [3,4,5,5,3,1]
Output: 6
Explanation: Perform the following swaps:
- Swap 1: Swap the 3rd and 4th elements, nums is then [3,4,5,3,5,1].
- Swap 2: Swap the 4th and 5th elements, nums is then [3,4,5,3,1,5].
- Swap 3: Swap the 3rd and 4th elements, nums is then [3,4,5,1,3,5].
- Swap 4: Swap the 2nd and 3rd elements, nums is then [3,4,1,5,3,5].
- Swap 5: Swap the 1st and 2nd elements, nums is then [3,1,4,5,3,5].
- Swap 6: Swap the 0th and 1st elements, nums is then [1,3,4,5,3,5].
It can be shown that 6 swaps is the minimum swaps required to make a valid array.


Example 2:

Input: nums = [9]
Output: 0
Explanation: The array is already valid, so we return 0.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105

List some examples and we can find the routine.

Find the right most max num ind and left most min num ind in case there are duplicates.

min swap =  max ind to the right edge + min ind to the left edge.

If the min ind > max Ind, we need to minus one since when we max to the right, we already swap once for the min.

Time Complexity: O(n). n = nums.length.

Space: O(1).

AC Java:

 1 class Solution {
2     public int minimumSwaps(int[] nums) {
3         int n = nums.length;
4         int maxInd = n - 1;
5         int minInd = 0;
6         for(int i = 0; i < n; i++){
7             if(nums[minInd] > nums[i]){
8                 minInd = i;
9             }
10
11             if(nums[maxInd] < nums[n - 1 - i]){
12                 maxInd = n - 1 - i;
13             }
14         }
15
16         return n - 1 - maxInd + minInd - (minInd > maxInd ? 1 : 0);
17     }
18 }

posted @ 2024-06-17 13:49  Dylan_Java_NYC  阅读(5)  评论(0编辑  收藏  举报