Array Intersection - Problem

Given two integer arrays nums1 and nums2, return an array of their intersection.

Each element in the result must be unique and you may return the result in any order.

Note: You cannot use built-in set operations or hash maps for this problem.

Input & Output

Example 1 — Basic Intersection

$ Input: nums1 = [1,2,2,1], nums2 = [2,2]

› Output: [2]

💡 Note: The only common element between both arrays is 2. Even though 2 appears multiple times, we return it only once.

Example 2 — Multiple Common Elements

$ Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]

› Output: [4,9]

💡 Note: Both 4 and 9 appear in both arrays. The result can be in any order.

Example 3 — No Intersection

$ Input: nums1 = [1,2,3], nums2 = [4,5,6]

› Output: []

💡 Note: No elements are common between the two arrays, so return empty array.

Constraints

1 ≤ nums1.length, nums2.length ≤ 1000
0 ≤ nums1[i], nums2[i] ≤ 1000

Visualization

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Asked in

G Google 25 a Amazon 18 M Microsoft 15 A Apple 12

Find common elements between two arrays without using hash maps. The key insight is to sort both arrays first, then use two pointers to find matches efficiently. Best approach is Sort and Two Pointers with Time: O(n log n), Space: O(1).

Common Approaches

✓ Brute Force - Check Every Pair

⏱️ Time: O(n×m) Space: O(1)

Compare every element in the first array with every element in the second array. When a match is found, add it to the result if it's not already there.

Sort and Two Pointers

⏱️ Time: O(n log n + m log m) Space: O(1)

Sort both input arrays, then use two pointers to traverse them simultaneously. When both pointers point to the same value, we found an intersection.

Brute Force - Check Every Pair — Algorithm Steps

Step 1: For each element in nums1, search through all of nums2
Step 2: When match found, check if already in result to avoid duplicates
Step 3: Add unique matches to result array

Visualization

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Step-by-Step Walkthrough

Take First Element

Take nums1[0] and search all of nums2

Linear Search

Compare against every element in nums2

Add If Match

If found and not duplicate, add to result

Code -

solution.c — C

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

int* solution(int* nums1, int nums1Size, int* nums2, int nums2Size, int* returnSize) {
    int* result = (int*)malloc(1000 * sizeof(int));
    *returnSize = 0;
    
    for (int i = 0; i < nums1Size; i++) {
        // Check if this number exists in nums2
        int found = 0;
        for (int j = 0; j < nums2Size; j++) {
            if (nums1[i] == nums2[j]) {
                found = 1;
                break;
            }
        }
        
        // If found and not already in result, add it
        if (found) {
            int exists = 0;
            for (int k = 0; k < *returnSize; k++) {
                if (result[k] == nums1[i]) {
                    exists = 1;
                    break;
                }
            }
            if (!exists) {
                result[*returnSize] = nums1[i];
                (*returnSize)++;
            }
        }
    }
    
    return result;
}

int main() {
    int nums1[1000], nums2[1000];
    int nums1Size = 0, nums2Size = 0;
    char line[10000];
    
    // Parse nums1
    fgets(line, sizeof(line), stdin);
    char* ptr = strchr(line, '[') + 1;
    char* token = strtok(ptr, ",]");
    while (token != NULL) {
        nums1[nums1Size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    // Parse nums2
    fgets(line, sizeof(line), stdin);
    ptr = strchr(line, '[') + 1;
    token = strtok(ptr, ",]");
    while (token != NULL) {
        nums2[nums2Size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int returnSize;
    int* result = solution(nums1, nums1Size, nums2, nums2Size, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×m)

For each of n elements in nums1, we search through all m elements in nums2

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Array Intersection - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Every Pair — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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