Find Common Elements Between Two Arrays

Find Common Elements Between Two Arrays - Problem

You are given two integer arrays nums1 and nums2 of sizes n and m, respectively.

Calculate the following values:

answer1: the number of indices i such that nums1[i] exists in nums2
answer2: the number of indices i such that nums2[i] exists in nums1

Return [answer1, answer2].

Input & Output

Example 1 — Basic Case

$ Input: nums1 = [4,3,2,3,1], nums2 = [2,3,1,4,7]

› Output: [5,4]

💡 Note: For nums1: all elements 4,3,2,3,1 exist in nums2, so answer1=5. For nums2: elements 2,3,1,4 exist in nums1 (7 doesn't), so answer2=4.

Example 2 — Partial Matches

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

› Output: [0,0]

💡 Note: No elements from nums1 exist in nums2, and no elements from nums2 exist in nums1, so both counts are 0.

Example 3 — All Different Sizes

$ Input: nums1 = [2,4,6,8], nums2 = [1,2,3,4,5]

› Output: [2,2]

💡 Note: From nums1: elements 2,4 exist in nums2 (answer1=2). From nums2: elements 2,4 exist in nums1 (answer2=2).

Constraints

1 ≤ nums1.length, nums2.length ≤ 1000
1 ≤ nums1[i], nums2[i] ≤ 100

Visualization

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Asked in

a Amazon 35 G Google 28

The key insight is using hash sets to convert O(n) linear searches into O(1) constant-time lookups. Best approach is the Hash Set method which creates sets from both arrays and counts elements efficiently. Time: O(n+m), Space: O(n+m).

Common Approaches

✓ Brute Force - Nested Loops

⏱️ Time: O(n×m) Space: O(1)

Check every element in nums1 against every element in nums2 to count matches, then repeat in the opposite direction. This requires nested loops for both counting operations.

Hash Set Optimization

⏱️ Time: O(n+m) Space: O(n+m)

Create hash sets from both arrays to enable constant-time lookups. Then iterate through each array and count how many elements exist in the other array's set.

Brute Force - Nested Loops — Algorithm Steps

Count elements in nums1 that exist in nums2 using nested loops
Count elements in nums2 that exist in nums1 using nested loops
Return both counts as an array

Visualization

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Step-by-Step Walkthrough

Check nums1 in nums2

For each element in nums1, scan all of nums2

Check nums2 in nums1

For each element in nums2, scan all of nums1

Count Results

Return counts as [answer1, answer2]

Code -

solution.c — C

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

int* solution(int* nums1, int nums1Size, int* nums2, int nums2Size) {
    int* result = (int*)malloc(2 * sizeof(int));
    int answer1 = 0;
    int answer2 = 0;
    
    // Count elements in nums1 that exist in nums2
    for (int i = 0; i < nums1Size; i++) {
        for (int j = 0; j < nums2Size; j++) {
            if (nums1[i] == nums2[j]) {
                answer1++;
                break;
            }
        }
    }
    
    // Count elements in nums2 that exist in nums1
    for (int i = 0; i < nums2Size; i++) {
        for (int j = 0; j < nums1Size; j++) {
            if (nums2[i] == nums1[j]) {
                answer2++;
                break;
            }
        }
    }
    
    result[0] = answer1;
    result[1] = answer2;
    return result;
}

int main() {
    char line[1000];
    int nums1[100], nums2[100];
    int nums1Size = 0, nums2Size = 0;
    
    // Parse first array
    fgets(line, sizeof(line), stdin);
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (token[0] >= '0' && token[0] <= '9' || token[0] == '-') {
            nums1[nums1Size++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    // Parse second array
    fgets(line, sizeof(line), stdin);
    token = strtok(line, "[,]");
    while (token != NULL) {
        if (token[0] >= '0' && token[0] <= '9' || token[0] == '-') {
            nums2[nums2Size++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int* result = solution(nums1, nums1Size, nums2, nums2Size);
    printf("[%d,%d]\n", result[0], result[1]);
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×m)

For each element in first array (n), we scan the entire second array (m)

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for counting, no extra data structures

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Nested Loops — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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