LeetCode #765 — HARD

Couples Holding Hands

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

There are n couples sitting in 2n seats arranged in a row and want to hold hands.

The people and seats are represented by an integer array row where row[i] is the ID of the person sitting in the i^th seat. 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).

Return 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.

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.

Constraints:

2n == row.length
2 <= n <= 30
0 <= row[i] < 2n
All the elements of row are unique.

Step 01

Brute Force Baseline

Problem summary: There are n couples sitting in 2n seats arranged in a row and want to hold hands. The people and seats are represented by an integer array row where row[i] is the ID of the person sitting in the ith seat. 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). Return 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Greedy · Union-Find

Example 1

[0,2,1,3]

Example 2

[3,2,0,1]

Core Insight

What unlocks the optimal approach

Say there are N two-seat couches. For each couple, draw an edge from the couch of one partner to the couch of the other partner.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #765: Couples Holding Hands
class Solution {
    private int[] p;

    public int minSwapsCouples(int[] row) {
        int n = row.length >> 1;
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        for (int i = 0; i < n << 1; i += 2) {
            int a = row[i] >> 1, b = row[i + 1] >> 1;
            p[find(a)] = find(b);
        }
        int ans = n;
        for (int i = 0; i < n; ++i) {
            if (i == find(i)) {
                --ans;
            }
        }
        return ans;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

// Accepted solution for LeetCode #765: Couples Holding Hands
func minSwapsCouples(row []int) int {
	n := len(row) >> 1
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	for i := 0; i < n<<1; i += 2 {
		a, b := row[i]>>1, row[i+1]>>1
		p[find(a)] = find(b)
	}
	ans := n
	for i := range p {
		if find(i) == i {
			ans--
		}
	}
	return ans
}

# Accepted solution for LeetCode #765: Couples Holding Hands
class Solution:
    def minSwapsCouples(self, row: List[int]) -> int:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        n = len(row) >> 1
        p = list(range(n))
        for i in range(0, len(row), 2):
            a, b = row[i] >> 1, row[i + 1] >> 1
            p[find(a)] = find(b)
        return n - sum(i == find(i) for i in range(n))

// Accepted solution for LeetCode #765: Couples Holding Hands
struct Solution;

impl Solution {
    fn min_swaps_couples(mut row: Vec<i32>) -> i32 {
        let n = row.len();
        let mut res = 0;
        for i in 0..n {
            if i % 2 == 1 {
                continue;
            }
            if row[i] == row[i + 1] ^ 1 {
                continue;
            }
            res += 1;
            for j in i + 2..n {
                if row[i] == row[j] ^ 1 {
                    row.swap(i + 1, j);
                    break;
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let row = vec![0, 2, 1, 3];
    let res = 1;
    assert_eq!(Solution::min_swaps_couples(row), res);
    let row = vec![3, 2, 0, 1];
    let res = 0;
    assert_eq!(Solution::min_swaps_couples(row), res);
}

// Accepted solution for LeetCode #765: Couples Holding Hands
function minSwapsCouples(row: number[]): number {
    const n = row.length >> 1;
    const p: number[] = Array(n)
        .fill(0)
        .map((_, i) => i);
    const find = (x: number): number => {
        if (p[x] !== x) {
            p[x] = find(p[x]);
        }
        return p[x];
    };
    for (let i = 0; i < n << 1; i += 2) {
        const a = row[i] >> 1;
        const b = row[i + 1] >> 1;
        p[find(a)] = find(b);
    }
    let ans = n;
    for (let i = 0; i < n; ++i) {
        if (i === find(i)) {
            --ans;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × \alpha(n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.