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2340. Minimum Adjacent Swaps to Make a Valid Array

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

  • 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);
        }
    }
    
  • 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);
        }
    };
    
  • 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)
    
    
  • 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
    }
    
  • 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);
    }
    
    

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