LeetCode #3587 — MEDIUM

Minimum Adjacent Swaps to Alternate Parity

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array nums of distinct integers.

In one operation, you can swap any two adjacent elements in the array.

An arrangement of the array is considered valid if the parity of adjacent elements alternates, meaning every pair of neighboring elements consists of one even and one odd number.

Return the minimum number of adjacent swaps required to transform nums into any valid arrangement.

If it is impossible to rearrange nums such that no two adjacent elements have the same parity, return -1.

Example 1:

Input: nums = [2,4,6,5,7]

Output: 3

Explanation:

Swapping 5 and 6, the array becomes [2,4,5,6,7]

Swapping 5 and 4, the array becomes [2,5,4,6,7]

Swapping 6 and 7, the array becomes [2,5,4,7,6]. The array is now a valid arrangement. Thus, the answer is 3.

Example 2:

Input: nums = [2,4,5,7]

Output: 1

Explanation:

By swapping 4 and 5, the array becomes [2,5,4,7], which is a valid arrangement. Thus, the answer is 1.

Example 3:

Input: nums = [1,2,3]

Output: 0

Explanation:

The array is already a valid arrangement. Thus, no operations are needed.

Example 4:

Input: nums = [4,5,6,8]

Output: -1

Explanation:

No valid arrangement is possible. Thus, the answer is -1.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
All elements in nums are distinct.

Step 01

Brute Force Baseline

Problem summary: You are given an array nums of distinct integers. In one operation, you can swap any two adjacent elements in the array. An arrangement of the array is considered valid if the parity of adjacent elements alternates, meaning every pair of neighboring elements consists of one even and one odd number. Return the minimum number of adjacent swaps required to transform nums into any valid arrangement. If it is impossible to rearrange nums such that no two adjacent elements have the same parity, return -1.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[2,4,6,5,7]

Example 2

[2,4,5,7]

Example 3

[1,2,3]

Step 02

Core Insight

What unlocks the optimal approach

Compute <code>evenCnt</code> and <code>oddCnt</code> in <code>nums</code>. If abs(evenCnt - oddCnt) > 1, return -1 immediately.
Let <code>n</code> = len(nums). You’ll try at most two target parity‐patterns: one starting with even at index 0, one with odd at index 0.
If <code>n</code> is odd, only the pattern whose starting parity matches the larger of <code>evenCnt</code> or <code>oddCnt</code> is feasible. If <code>n</code> is even, both starting‐even and starting‐odd patterns are possible—compute both and take the minimum.
For a given target pattern, collect the indices of all even elements in <code>nums</code> (or odd elements) and the indices where an even (or odd) should go in the pattern.
The minimum adjacent‐swap cost to align those elements is the sum of absolute differences between each element’s current index and its target index.
Return the smallest cost over valid patterns.

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

Upper-end input sizes

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 #3587: Minimum Adjacent Swaps to Alternate Parity
class Solution {
    private List<Integer>[] pos = new List[2];
    private int[] nums;

    public int minSwaps(int[] nums) {
        this.nums = nums;
        Arrays.setAll(pos, k -> new ArrayList<>());
        for (int i = 0; i < nums.length; ++i) {
            pos[nums[i] & 1].add(i);
        }
        if (Math.abs(pos[0].size() - pos[1].size()) > 1) {
            return -1;
        }
        if (pos[0].size() > pos[1].size()) {
            return calc(0);
        }
        if (pos[0].size() < pos[1].size()) {
            return calc(1);
        }
        return Math.min(calc(0), calc(1));
    }

    private int calc(int k) {
        int res = 0;
        for (int i = 0; i < nums.length; i += 2) {
            res += Math.abs(pos[k].get(i / 2) - i);
        }
        return res;
    }
}

// Accepted solution for LeetCode #3587: Minimum Adjacent Swaps to Alternate Parity
func minSwaps(nums []int) int {
	pos := [2][]int{}
	for i, x := range nums {
		pos[x&1] = append(pos[x&1], i)
	}
	if abs(len(pos[0])-len(pos[1])) > 1 {
		return -1
	}
	calc := func(k int) int {
		res := 0
		for i := 0; i < len(nums); i += 2 {
			res += abs(pos[k][i/2] - i)
		}
		return res
	}
	if len(pos[0]) > len(pos[1]) {
		return calc(0)
	}
	if len(pos[0]) < len(pos[1]) {
		return calc(1)
	}
	return min(calc(0), calc(1))
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #3587: Minimum Adjacent Swaps to Alternate Parity
class Solution:
    def minSwaps(self, nums: List[int]) -> int:
        def calc(k: int) -> int:
            return sum(abs(i - j) for i, j in zip(range(0, len(nums), 2), pos[k]))

        pos = [[], []]
        for i, x in enumerate(nums):
            pos[x & 1].append(i)
        if abs(len(pos[0]) - len(pos[1])) > 1:
            return -1
        if len(pos[0]) > len(pos[1]):
            return calc(0)
        if len(pos[0]) < len(pos[1]):
            return calc(1)
        return min(calc(0), calc(1))

// Accepted solution for LeetCode #3587: Minimum Adjacent Swaps to Alternate Parity
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3587: Minimum Adjacent Swaps to Alternate Parity
// class Solution {
//     private List<Integer>[] pos = new List[2];
//     private int[] nums;
// 
//     public int minSwaps(int[] nums) {
//         this.nums = nums;
//         Arrays.setAll(pos, k -> new ArrayList<>());
//         for (int i = 0; i < nums.length; ++i) {
//             pos[nums[i] & 1].add(i);
//         }
//         if (Math.abs(pos[0].size() - pos[1].size()) > 1) {
//             return -1;
//         }
//         if (pos[0].size() > pos[1].size()) {
//             return calc(0);
//         }
//         if (pos[0].size() < pos[1].size()) {
//             return calc(1);
//         }
//         return Math.min(calc(0), calc(1));
//     }
// 
//     private int calc(int k) {
//         int res = 0;
//         for (int i = 0; i < nums.length; i += 2) {
//             res += Math.abs(pos[k].get(i / 2) - i);
//         }
//         return res;
//     }
// }

// Accepted solution for LeetCode #3587: Minimum Adjacent Swaps to Alternate Parity
function minSwaps(nums: number[]): number {
    const pos: number[][] = [[], []];
    for (let i = 0; i < nums.length; ++i) {
        pos[nums[i] & 1].push(i);
    }
    if (Math.abs(pos[0].length - pos[1].length) > 1) {
        return -1;
    }
    const calc = (k: number): number => {
        let res = 0;
        for (let i = 0; i < nums.length; i += 2) {
            res += Math.abs(pos[k][i >> 1] - i);
        }
        return res;
    };
    if (pos[0].length > pos[1].length) {
        return calc(0);
    }
    if (pos[0].length < pos[1].length) {
        return calc(1);
    }
    return Math.min(calc(0), calc(1));
}

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)

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.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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.