Missing Number - Practice Coding Problems

Missing Number - Problem

Array Hash Table Math Binary Search Bit Manipulation Sorting Easy

You're given an array nums containing n distinct numbers taken from the range [0, n]. This means if the array has 3 elements, the numbers should come from [0, 1, 2, 3].

Here's the catch: exactly one number from this range is missing from the array. Your task is to find and return that missing number.

Example: If nums = [3, 0, 1], then n = 3 and the complete range should be [0, 1, 2, 3]. The missing number is 2.

Can you solve this efficiently? There are multiple clever approaches using math, bit manipulation, and more!

Input & Output

example_1.py — Basic Case

$ Input: nums = [3,0,1]

› Output: 2

💡 Note: Array has 3 elements, so complete range should be [0,1,2,3]. Numbers 0, 1, 3 are present, but 2 is missing.

example_2.py — Missing First

$ Input: nums = [1]

› Output: 0

💡 Note: Array has 1 element, so complete range should be [0,1]. Number 1 is present, but 0 is missing.

example_3.py — Missing Last

$ Input: nums = [0,1]

› Output: 2

💡 Note: Array has 2 elements, so complete range should be [0,1,2]. Numbers 0 and 1 are present, but 2 is missing.

Constraints

n == nums.length
1 ≤ n ≤ 10⁴
0 ≤ nums[i] ≤ n
All the numbers of nums are unique

Visualization

Tap to expand

Understanding the Visualization

Crime Scene Analysis

You know there should be n+1 pieces of evidence numbered 0 through n, but you only found n pieces.

The Brute Force Method

Check each possible evidence number by searching through your entire collection - slow but thorough.

The Hash Table Method

Create a quick-reference catalog of all evidence you have, then check which number isn't in your catalog.

The Mathematical Breakthrough

Use Gauss's discovery: the sum of numbers 0+1+2+...+n equals n(n+1)/2. The missing evidence number equals the difference between expected and actual sums!

Key Takeaway

🎯 Key Insight: Sometimes the most elegant solution comes from mathematical properties rather than complex algorithms. Gauss's sum formula transforms this from a search problem into a simple arithmetic calculation!

Asked in

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

The optimal solution uses Gauss's sum formula: calculate the expected sum of numbers 0 to n using n(n+1)/2, then subtract the actual sum of array elements. The difference is the missing number. This achieves O(n) time and O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Math Formula (Optimal)	O(n)	O(1)	Use Gauss's formula: expected sum minus actual sum equals missing number
Hash Set (Optimal for Lookup)	O(n)	O(n)	Store all numbers in a hash set, then check which number from 0 to n is missing
Brute Force (Nested Loops)	O(n²)	O(1)	For each number 0 to n, scan the entire array to check if it exists

Math Formula (Optimal) — Algorithm Steps

Calculate expected sum using formula: n(n+1)/2 where n is array length
Calculate actual sum by adding all elements in the array
The difference (expected sum - actual sum) is the missing number
Return the missing number

Visualization

Tap to expand

Step-by-Step Walkthrough

Calculate Expected Sum

For range [0,1,2,3]: sum = 3×4/2 = 6

Calculate Actual Sum

Array [3,0,1]: sum = 3+0+1 = 4

Find Difference

Expected - Actual = 6 - 4 = 2

Missing Number Found

The difference 2 is our missing number!

Code -

solution.c — C

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

int missingNumber(int* nums, int numsSize) {
    int n = numsSize;
    // Expected sum of numbers from 0 to n
    int expectedSum = n * (n + 1) / 2;
    // Actual sum of array elements
    int actualSum = 0;
    for (int i = 0; i < numsSize; i++) {
        actualSum += nums[i];
    }
    // Missing number is the difference
    return expectedSum - actualSum;
}

int main() {
    int nums[1000];
    int size = 0;
    char c;
    int num = 0;
    bool inNumber = false;
    
    while ((c = getchar()) != '\n' && c != EOF) {
        if (c >= '0' && c <= '9') {
            num = num * 10 + (c - '0');
            inNumber = true;
        } else if (inNumber) {
            nums[size++] = num;
            num = 0;
            inNumber = false;
        }
    }
    if (inNumber) nums[size++] = num;
    
    printf("%d\n", missingNumber(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array to calculate the sum of all elements

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to store sums, no additional data structures

✓ Linear Space

89.5K Views

Very High Frequency

~15 min Avg. Time

2.3K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Math Formula (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler