Set Mismatch - Practice Coding Problems

Set Mismatch - Problem

You're given an array nums that was supposed to contain all integers from 1 to n exactly once, but something went wrong! One number got duplicated, which means another number went missing.

Your mission: Find both the duplicate number (appears twice) and the missing number (doesn't appear at all).

Example: If nums = [1,2,2,4], then:

The duplicate is 2 (appears twice)
The missing number is 3 (should be there but isn't)
Return [2,3]

Goal: Return an array [duplicate, missing] where the first element is the number that appears twice, and the second is the number that's missing.

Input & Output

example_1.py — Basic Case

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

› Output: [2,3]

💡 Note: The number 2 appears twice (duplicate) and the number 3 is missing from the sequence 1,2,3,4.

example_2.py — Different Duplicate

$ Input: nums = [1,1]

› Output: [1,2]

💡 Note: For array of length 2, we should have [1,2]. The number 1 is duplicated and 2 is missing.

example_3.py — Larger Array

$ Input: nums = [3,2,3,4,6,5]

› Output: [3,1]

💡 Note: Array should be [1,2,3,4,5,6]. The number 3 appears twice and 1 is missing.

Constraints

2 ≤ nums.length ≤ 10⁴
1 ≤ nums[i] ≤ 10⁴
nums contains n distinct numbers plus one duplicate
The array represents numbers from 1 to n with exactly one duplicate and one missing

Visualization

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Understanding the Visualization

Take Attendance

Call out each student number and mark who responds

Notice Duplicate

Student #2 responds twice - suspicious!

Find Missing

Use math: expected total responses minus actual responses plus duplicate

Solve Mystery

Student #2 answered twice, Student #3 is absent

Key Takeaway

🎯 Key Insight: Combine hash table duplicate detection with mathematical sum relationships to solve both parts of the problem efficiently in one pass!

Asked in

a Amazon 45 G Google 38 ⊞ Microsoft 32 f Meta 28

The optimal solution uses a one-pass hash table approach combined with mathematical insight. As we iterate through the array once, we detect the duplicate when a number is already in our seen set. We also calculate the array sum. The missing number is then found using the formula: missing = expectedSum - arraySum + duplicate. This achieves O(n) time and O(n) space complexity, making it the most efficient approach for this problem.

Common Approaches

Approach	Time	Space	Notes
✓ Two-Pass Hash Table	O(n)	O(n)	First pass counts frequencies, second pass finds duplicate and missing
Brute Force (Nested Loops)	O(n²)	O(1)	Check each number 1 to n by scanning the entire array
One-Pass Hash Table (Optimal)	O(n)	O(n)	Build frequency map and identify duplicate in one pass, calculate missing using math

Two-Pass Hash Table — Algorithm Steps

Create a hash map to store frequency of each number
First pass: iterate through array and count frequencies
Second pass: for each number from 1 to n, check its frequency
Frequency 2 = duplicate, frequency 0 = missing
Return [duplicate, missing]

Visualization

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Step-by-Step Walkthrough

First Pass

Build frequency map: {1:1, 2:2, 4:1}

Second Pass

Check frequencies for numbers 1 to n

Identify Issues

freq[2]=2 (duplicate), freq[3]=0 (missing)

Code -

solution.c — C

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

int* findErrorNums(int* nums, int numsSize, int* returnSize) {
    *returnSize = 2;
    int* result = (int*)malloc(2 * sizeof(int));
    int* freq = (int*)calloc(numsSize + 1, sizeof(int));
    
    // First pass: count frequencies
    for (int i = 0; i < numsSize; i++) {
        freq[nums[i]]++;
    }
    
    // Second pass: find duplicate and missing
    int duplicate = -1, missing = -1;
    for (int i = 1; i <= numsSize; i++) {
        if (freq[i] == 2) duplicate = i;
        else if (freq[i] == 0) missing = i;
    }
    
    result[0] = duplicate;
    result[1] = missing;
    free(freq);
    return result;
}

int main() {
    int nums[1000];
    int n = 0;
    char c;
    scanf("%c", &c); // skip '['
    
    while (scanf("%d%c", &nums[n], &c)) {
        n++;
        if (c == ']') break;
    }
    
    int returnSize;
    int* result = findErrorNums(nums, n, &returnSize);
    printf("[%d,%d]\n", result[0], result[1]);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

First pass O(n) to build frequency map, second pass O(n) to check 1 to n

✓ Linear Growth

Space Complexity

O(n)

Hash map stores frequency for up to n different numbers

⚡ Linearithmic Space

89.2K Views

High Frequency

~15 min Avg. Time

1.8K Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Two-Pass Hash Table — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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