2340. Minimum Adjacent Swaps to Make a Valid Array πŸ”’ 

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Description

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

Solutions

Solution 1: Maintain Index of Extremes + Case Analysis

We can use indices $i$ and $j$ to record the index of the first minimum value and the last maximum value in the array $\textit{nums}$, respectively. Traverse the array $\textit{nums}$ to update the values of $i$ and $j$.

Next, we need to consider the number of swaps.

  • If $i = j$, it means the array $\textit{nums}$ is already a valid array, and no swaps are needed. Return $0$.

  • If $i < j$, it means the minimum value in the array $\textit{nums}$ is to the left of the maximum value. The number of swaps needed is $i + n - 1 - j$, where $n$ is the length of the array $\textit{nums}$.

  • If $i > j$, it means the minimum value in the array $\textit{nums}$ is to the right of the maximum value. The number of swaps needed is $i + n - 1 - j - 1$.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.

Python3

class Solution:
    def minimumSwaps(self, nums: List[int]) -> int:
        i = j = 0
        for k, v in enumerate(nums):
            if v < nums[i] or (v == nums[i] and k < i):
                i = k
            if v >= nums[j] or (v == nums[j] and k > j):
                j = k
        return 0 if i == j else i + len(nums) - 1 - j - (i > j)

Java

class Solution {
    public int minimumSwaps(int[] nums) {
        int n = nums.length;
        int i = 0, j = 0;
        for (int k = 0; k < n; ++k) {
            if (nums[k] < nums[i] || (nums[k] == nums[i] && k < i)) {
                i = k;
            }
            if (nums[k] > nums[j] || (nums[k] == nums[j] && k > j)) {
                j = k;
            }
        }
        if (i == j) {
            return 0;
        }
        return i + n - 1 - j - (i > j ? 1 : 0);
    }
}

C++

class Solution {
public:
    int minimumSwaps(vector<int>& nums) {
        int n = nums.size();
        int i = 0, j = 0;
        for (int k = 0; k < n; ++k) {
            if (nums[k] < nums[i] || (nums[k] == nums[i] && k < i)) {
                i = k;
            }
            if (nums[k] > nums[j] || (nums[k] == nums[j] && k > j)) {
                j = k;
            }
        }
        if (i == j) {
            return 0;
        }
        return i + n - 1 - j - (i > j);
    }
};

Go

func minimumSwaps(nums []int) int {
	var i, j int
	for k, v := range nums {
		if v < nums[i] || (v == nums[i] && k < i) {
			i = k
		}
		if v > nums[j] || (v == nums[j] && k > j) {
			j = k
		}
	}
	if i == j {
		return 0
	}
	if i < j {
		return i + len(nums) - 1 - j
	}
	return i + len(nums) - 2 - j
}

TypeScript

function minimumSwaps(nums: number[]): number {
    let i = 0;
    let j = 0;
    const n = nums.length;
    for (let k = 0; k < n; ++k) {
        if (nums[k] < nums[i] || (nums[k] == nums[i] && k < i)) {
            i = k;
        }
        if (nums[k] > nums[j] || (nums[k] == nums[j] && k > j)) {
            j = k;
        }
    }
    return i == j ? 0 : i + n - 1 - j - (i > j ? 1 : 0);
}