LeetCode #2341 — EASY

Maximum Number of Pairs in Array

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

The Problem

Problem Statement

You are given a 0-indexed integer array nums. In one operation, you may do the following:

Choose two integers in nums that are equal.
Remove both integers from nums, forming a pair.

The operation is done on nums as many times as possible.

Return a 0-indexed integer array answer of size 2 where answer[0] is the number of pairs that are formed and answer[1] is the number of leftover integers in nums after doing the operation as many times as possible.

Example 1:

Input: nums = [1,3,2,1,3,2,2]
Output: [3,1]
Explanation:
Form a pair with nums[0] and nums[3] and remove them from nums. Now, nums = [3,2,3,2,2].
Form a pair with nums[0] and nums[2] and remove them from nums. Now, nums = [2,2,2].
Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [2].
No more pairs can be formed. A total of 3 pairs have been formed, and there is 1 number leftover in nums.

Example 2:

Input: nums = [1,1]
Output: [1,0]
Explanation: Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [].
No more pairs can be formed. A total of 1 pair has been formed, and there are 0 numbers leftover in nums.

Example 3:

Input: nums = [0]
Output: [0,1]
Explanation: No pairs can be formed, and there is 1 number leftover in nums.

Constraints:

1 <= nums.length <= 100
0 <= nums[i] <= 100

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums. In one operation, you may do the following: Choose two integers in nums that are equal. Remove both integers from nums, forming a pair. The operation is done on nums as many times as possible. Return a 0-indexed integer array answer of size 2 where answer[0] is the number of pairs that are formed and answer[1] is the number of leftover integers in nums after doing the operation as many times as possible.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[1,3,2,1,3,2,2]

Example 2

[1,1]

Example 3

[0]

Core Insight

What unlocks the optimal approach

What do we need to know to find how many pairs we can make? We need to know the frequency of each integer.
When will there be a leftover number? When the frequency of an integer is an odd number.

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 #2341: Maximum Number of Pairs in Array
class Solution {
    public int[] numberOfPairs(int[] nums) {
        int[] cnt = new int[101];
        for (int x : nums) {
            ++cnt[x];
        }
        int s = 0;
        for (int v : cnt) {
            s += v / 2;
        }
        return new int[] {s, nums.length - s * 2};
    }
}

// Accepted solution for LeetCode #2341: Maximum Number of Pairs in Array
func numberOfPairs(nums []int) []int {
	cnt := [101]int{}
	for _, x := range nums {
		cnt[x]++
	}
	s := 0
	for _, v := range cnt {
		s += v / 2
	}
	return []int{s, len(nums) - s*2}
}

# Accepted solution for LeetCode #2341: Maximum Number of Pairs in Array
class Solution:
    def numberOfPairs(self, nums: List[int]) -> List[int]:
        cnt = Counter(nums)
        s = sum(v // 2 for v in cnt.values())
        return [s, len(nums) - s * 2]

// Accepted solution for LeetCode #2341: Maximum Number of Pairs in Array
impl Solution {
    pub fn number_of_pairs(nums: Vec<i32>) -> Vec<i32> {
        let n = nums.len();
        let mut count = [0; 101];
        for &v in nums.iter() {
            count[v as usize] += 1;
        }
        let mut sum = 0;
        for v in count.iter() {
            sum += v >> 1;
        }
        vec![sum as i32, (n - sum * 2) as i32]
    }
}

// Accepted solution for LeetCode #2341: Maximum Number of Pairs in Array
function numberOfPairs(nums: number[]): number[] {
    const n = nums.length;
    const count = new Array(101).fill(0);
    for (const num of nums) {
        count[num]++;
    }
    const sum = count.reduce((r, v) => r + (v >> 1), 0);
    return [sum, n - sum * 2];
}

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(C)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.