Two Out of Three

Two Out of Three - Problem

Given three integer arrays nums1, nums2, and nums3, return a distinct array containing all the values that are present in at least two out of the three arrays. You may return the values in any order.

Input & Output

Example 1 — Basic Case

$ Input: nums1 = [1,1,3,2], nums2 = [2,3], nums3 = [3]

› Output: [2,3]

💡 Note: Value 2 appears in nums1 and nums2. Value 3 appears in all three arrays. Value 1 only appears in nums1.

Example 2 — No Overlap

$ Input: nums1 = [3,1], nums2 = [2,3], nums3 = [1,2]

› Output: [1,2,3]

💡 Note: Value 1 appears in nums1 and nums3. Value 2 appears in nums2 and nums3. Value 3 appears in nums1 and nums2.

Example 3 — Single Array Only

$ Input: nums1 = [1,2,2], nums2 = [4,3,3], nums3 = [5]

› Output: []

💡 Note: No value appears in at least two arrays, so the result is empty.

Constraints

1 ≤ nums1.length, nums2.length, nums3.length ≤ 100
1 ≤ nums1[i], nums2[j], nums3[k] ≤ 100

Visualization

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Asked in

G Google 12 a Amazon 8 M Meta 6 M Microsoft 4

The key insight is to track which arrays contain each value rather than counting occurrences. Best approach uses a frequency count map that maps each value to the set of arrays containing it. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force with Nested Loops

⏱️ Time: O(n²) Space: O(n)

For each unique value across all arrays, manually count how many arrays contain it. If the count is 2 or more, include it in the result.

Frequency Count Map

⏱️ Time: O(n) Space: O(n)

Use a hash map to track which arrays each value appears in, then collect values that appear in at least 2 arrays. This avoids redundant counting.

Two-Pass Hash Map

⏱️ Time: O(n) Space: O(n)

Convert arrays to sets for O(1) lookups, then iterate through unique values once to count array memberships efficiently.

Brute Force with Nested Loops — Algorithm Steps

Step 1: Collect all unique values from the three arrays
Step 2: For each unique value, count in how many arrays it appears
Step 3: Include values that appear in 2 or more arrays

Visualization

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Step-by-Step Walkthrough

Collect All

Get unique values from all arrays

Count Each

For each value, count array appearances

Filter

Keep values appearing in 2+ arrays

Code -

solution.c — C

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

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int contains(int* arr, int size, int val) {
    for (int i = 0; i < size; i++) {
        if (arr[i] == val) return 1;
    }
    return 0;
}

int* solution(int* nums1, int size1, int* nums2, int size2, int* nums3, int size3, int* returnSize) {
    int* allValues = malloc(1000 * sizeof(int));
    int allCount = 0;
    
    // Collect all unique values
    for (int i = 0; i < size1; i++) {
        int found = 0;
        for (int j = 0; j < allCount; j++) {
            if (allValues[j] == nums1[i]) {
                found = 1;
                break;
            }
        }
        if (!found) allValues[allCount++] = nums1[i];
    }
    
    for (int i = 0; i < size2; i++) {
        int found = 0;
        for (int j = 0; j < allCount; j++) {
            if (allValues[j] == nums2[i]) {
                found = 1;
                break;
            }
        }
        if (!found) allValues[allCount++] = nums2[i];
    }
    
    for (int i = 0; i < size3; i++) {
        int found = 0;
        for (int j = 0; j < allCount; j++) {
            if (allValues[j] == nums3[i]) {
                found = 1;
                break;
            }
        }
        if (!found) allValues[allCount++] = nums3[i];
    }
    
    int* result = malloc(1000 * sizeof(int));
    *returnSize = 0;
    
    for (int i = 0; i < allCount; i++) {
        int count = 0;
        if (contains(nums1, size1, allValues[i])) {
            count += 1;
        }
        if (contains(nums2, size2, allValues[i])) {
            count += 1;
        }
        if (contains(nums3, size3, allValues[i])) {
            count += 1;
        }
        
        if (count >= 2) {
            result[(*returnSize)++] = allValues[i];
        }
    }
    
    free(allValues);
    return result;
}

int main() {
    char line1[10000], line2[10000], line3[10000];
    fgets(line1, sizeof(line1), stdin);
    fgets(line2, sizeof(line2), stdin);
    fgets(line3, sizeof(line3), stdin);
    
    int nums1[1000], nums2[1000], nums3[1000];
    int size1, size2, size3;
    
    parseArray(line1, nums1, &size1);
    parseArray(line2, nums2, &size2);
    parseArray(line3, nums3, &size3);
    
    int returnSize;
    int* result = solution(nums1, size1, nums2, size2, nums3, size3, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        if (i > 0) printf(",");
        printf("%d", result[i]);
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each unique value, we search through all three arrays

✓ Linear Growth

Space Complexity

O(n)

Set to store unique values and result array

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force with Nested Loops — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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