Maximum Number of Pairs in Array - Practice Coding Problems

Maximum Number of Pairs in Array - Problem

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.

Input & Output

Example 1 — Basic Case

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

› Output: [3,1]

💡 Note: Form pairs: (1,1), (3,3), (2,2). This gives us 3 pairs. One element 2 is left over, so leftovers = 1.

Example 2 — All Pairs

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

› Output: [3,0]

💡 Note: Form pairs: (1,1), (2,2), (3,3). All elements are paired, so 3 pairs and 0 leftovers.

Example 3 — No Pairs Possible

$ Input: nums = [0]

› Output: [0,1]

💡 Note: Single element cannot form a pair. Result: 0 pairs, 1 leftover.

Constraints

1 ≤ nums.length ≤ 1000
0 ≤ nums[i] ≤ 1000

Visualization

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Understanding the Visualization

Input Array

Array with potentially duplicate elements

Pair Formation

Match identical elements to form pairs

Count Results

Return [pairs_formed, elements_leftover]

Key Takeaway

🎯 Key Insight: Count frequency of each element, then for f occurrences make f÷2 pairs with f%2 leftovers

Asked in

a Amazon 15 M Microsoft 12 G Google 8 🍎 Apple 6

The key insight is that for any element appearing f times, we can form f÷2 pairs with f%2 leftovers. Best approach is frequency counting using a hash map. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Frequency Counting - Hash Map	O(n)	O(n)	Count frequency of each element, then calculate pairs from frequencies
Single Pass Optimization	O(n)	O(n)	Count frequencies and calculate pairs in a single pass through the array
Brute Force - Nested Loop Pairing	O(n³)	O(1)	For each element, search for its pair in the rest of the array

Frequency Counting - Hash Map — Algorithm Steps

Count frequency of each element using hash map
For each unique element, calculate pairs = frequency ÷ 2
Sum all pairs and leftovers from all elements

Visualization

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Step-by-Step Walkthrough

Count frequencies

Build hash map with element frequencies

Calculate pairs

For each frequency f, pairs = f÷2, leftovers = f%2

Sum results

Total pairs and leftovers from all elements

Code -

solution.c — C

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

// Simple hash table for integers
#define HASH_SIZE 10007

typedef struct Node {
    int key;
    int count;
    struct Node* next;
} Node;

int hash(int key) {
    return abs(key) % HASH_SIZE;
}

void insert(Node** table, int key) {
    int h = hash(key);
    Node* current = table[h];
    
    while (current) {
        if (current->key == key) {
            current->count++;
            return;
        }
        current = current->next;
    }
    
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->key = key;
    newNode->count = 1;
    newNode->next = table[h];
    table[h] = newNode;
}

int* solution(int* nums, int numsSize, int* returnSize) {
    *returnSize = 2;
    int* result = (int*)malloc(2 * sizeof(int));
    
    // Initialize hash table
    Node* table[HASH_SIZE] = {NULL};
    
    // Count frequencies
    for (int i = 0; i < numsSize; i++) {
        insert(table, nums[i]);
    }
    
    int pairs = 0;
    int leftovers = 0;
    
    // Calculate pairs and leftovers
    for (int i = 0; i < HASH_SIZE; i++) {
        Node* current = table[i];
        while (current) {
            pairs += current->count / 2;
            leftovers += current->count % 2;
            Node* temp = current;
            current = current->next;
            free(temp);
        }
    }
    
    result[0] = pairs;
    result[1] = leftovers;
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple array parsing
    int nums[1000];
    int size = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, "[],\n");
    }
    
    int returnSize;
    int* result = solution(nums, size, &returnSize);
    printf("[%d,%d]\n", result[0], result[1]);
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array to count frequencies, then one pass through hash map

✓ Linear Growth

Space Complexity

O(n)

Hash map stores frequency for each unique element, at most n entries

⚡ Linearithmic Space

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Medium Frequency

~8 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Frequency Counting - Hash Map — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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