Divide Array Into Equal Pairs

Divide Array Into Equal Pairs - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [3,2,3,2,2,2]

› Output: true

💡 Note: We can pair the elements as (3,3), (2,2), and (2,2). Each pair contains equal elements.

Example 2 — Impossible Pairing

$ Input: nums = [1,2,3,4]

› Output: false

💡 Note: All elements are different, so no equal pairs can be formed.

Example 3 — Odd Frequency

$ Input: nums = [1,1,2,2]

› Output: true

💡 Note: Element 1 appears 2 times (even) and element 2 appears 2 times (even). We can form pairs (1,1) and (2,2).

Constraints

1 ≤ nums.length ≤ 100
nums.length is even
-100 ≤ nums[i] ≤ 100

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that every element must appear an even number of times to form equal pairs. The frequency counting approach efficiently solves this by counting occurrences and checking divisibility by 2. Alternatively, XOR operation provides an O(1) space solution. Time: O(n), Space: O(1) for XOR or O(n) for frequency counting.

Common Approaches

✓ Bit Manipulation (XOR)

⏱️ Time: O(n) Space: O(1)

Leverage the XOR operation property where identical numbers cancel out (a ⊕ a = 0). If all elements can be paired, XORing all elements should result in 0.

Brute Force - Try All Combinations

⏱️ Time: O((2n)!) Space: O(n)

Generate all possible pairings of elements and check if any pairing consists entirely of equal pairs. This involves checking every combination of how to group the 2*n elements into n pairs.

Frequency Count

⏱️ Time: O(n) Space: O(n)

Count how many times each element appears in the array. For the array to be dividable into equal pairs, every element must appear an even number of times.

Bit Manipulation (XOR) — Algorithm Steps

XOR all elements together
If result is 0, all elements have pairs
If result is non-zero, some elements are unpaired

Visualization

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Step-by-Step Walkthrough

XOR Operation

XOR all elements together

Cancellation

Paired elements cancel out (a ⊕ a = 0)

Result

If final result is 0, all elements are paired

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(int* nums, int numsSize) {
    int xorResult = 0;
    for (int i = 0; i < numsSize; i++) {
        xorResult ^= nums[i];
    }
    return xorResult == 0;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int size = 0;
    char* token = strtok(line, "[],\n ");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, "[],\n ");
    }
    
    bool result = solution(nums, size);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array XORing each element

✓ Linear Growth

Space Complexity

O(1)

Only uses one variable to store XOR result

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Bit Manipulation (XOR) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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