LC 765. Couples Holding Hands - Weekly Contest 67 - LeetCode

N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum number of swaps so that every couple is sitting side by side. A swap consists of choosing any two people, then they stand up and switch seats.

The people and seats are represented by an integer from 0 to 2N-1, the couples are numbered in order, the first couple being (0, 1), the second couple being (2, 3), and so on with the last couple being (2N-2, 2N-1).

The couples' initial seating is given by row[i] being the value of the person who is initially sitting in the i-th seat.

Example 1:

Input: row = [0, 2, 1, 3]
Output: 1
Explanation: We only need to swap the second (row[1]) and third (row[2]) person.
Example 2:

Input: row = [3, 2, 0, 1]
Output: 0
Explanation: All couples are already seated side by side.
Note:

len(row) is even and in the range of [4, 60].
row is guaranteed to be a permutation of 0...len(row)-1.

	class Solution {
	/**
	* @param row
	* @return
	*/
	public int minSwapsCouples(int[] row) {
	// filter abnormal cases
	if (row == null || row.length == 0) {
	return 0;
	}
	// core logic
	int swapCounter = 0;
	HashMap<Integer, Integer> lookup = new HashMap<>();
	for (int i = 0; i < row.length; i++) {
	lookup.put(row[i], i);
	}
	for (int i = 0; i < row.length; i += 2) {
	if (!isPair(row[i], row[i + 1])) {
	swap(row, i + 1, lookup.get(getPartner(row[i])), lookup);
	swapCounter++;
	}
	}
	// return the final result
	return swapCounter;
	}
	private int getPartner(int i) {
	if (i % 2 == 0) {
	return i + 1;
	} else {
	return i - 1;
	}
	}
	private void swap(int[] row, int i, int j, HashMap<Integer, Integer> lookup) {
	int tmp = row[i];
	lookup.put(row[i], j);
	lookup.put(row[j], i);
	row[i] = row[j];
	row[j] = tmp;
	}
	private boolean isPair(int a, int b) {
	return a / 2 == b / 2;
	}
	}

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