LeetCode #2206 — EASY

Divide Array Into Equal Pairs

Build confidence with an intuition-first walkthrough focused on array fundamentals.

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The Problem

Problem Statement

You are given an integer array nums consisting of 2 * n integers.

You need to divide nums into n pairs such that:

Each element belongs to exactly one pair.
The elements present in a pair are equal.

Return true if nums can be divided into n pairs, otherwise return false.

Example 1:

Input: nums = [3,2,3,2,2,2]
Output: true
Explanation: 
There are 6 elements in nums, so they should be divided into 6 / 2 = 3 pairs.
If nums is divided into the pairs (2, 2), (3, 3), and (2, 2), it will satisfy all the conditions.

Example 2:

Input: nums = [1,2,3,4]
Output: false
Explanation: 
There is no way to divide nums into 4 / 2 = 2 pairs such that the pairs satisfy every condition.

Constraints:

nums.length == 2 * n
1 <= n <= 500
1 <= nums[i] <= 500

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums consisting of 2 * n integers. You need to divide nums into n pairs such that: Each element belongs to exactly one pair. The elements present in a pair are equal. Return true if nums can be divided into n pairs, otherwise return false.

Baseline thinking

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

Pattern signal: Array · Hash Map · Bit Manipulation

Example 1

[3,2,3,2,2,2]

Example 2

[1,2,3,4]

Core Insight

What unlocks the optimal approach

For any number x in the range [1, 500], count the number of elements in nums whose values are equal to x.
The elements with equal value can be divided completely into pairs if and only if their count is even.

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 #2206: Divide Array Into Equal Pairs
class Solution {
    public boolean divideArray(int[] nums) {
        int[] cnt = new int[510];
        for (int v : nums) {
            ++cnt[v];
        }
        for (int v : cnt) {
            if (v % 2 != 0) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #2206: Divide Array Into Equal Pairs
func divideArray(nums []int) bool {
	cnt := [510]int{}
	for _, x := range nums {
		cnt[x]++
	}
	for _, v := range cnt {
		if v%2 != 0 {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #2206: Divide Array Into Equal Pairs
class Solution:
    def divideArray(self, nums: List[int]) -> bool:
        cnt = Counter(nums)
        return all(v % 2 == 0 for v in cnt.values())

// Accepted solution for LeetCode #2206: Divide Array Into Equal Pairs
use std::collections::HashMap;

impl Solution {
    pub fn divide_array(nums: Vec<i32>) -> bool {
        let mut cnt = HashMap::new();
        for x in nums {
            *cnt.entry(x).or_insert(0) += 1;
        }
        cnt.values().all(|&v| v % 2 == 0)
    }
}

// Accepted solution for LeetCode #2206: Divide Array Into Equal Pairs
function divideArray(nums: number[]): boolean {
    const cnt = Array(501).fill(0);

    for (const x of nums) {
        cnt[x]++;
    }

    for (const x of cnt) {
        if (x & 1) return false;
    }

    return true;
}

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

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.