(Coding Clutch Challenge Series 101) – Coding Challenge #1
Imagine you’re playing detective and trying to find a missing piece of a puzzle. That’s precisely what we do in the “Find the Missing Number” challenge. It’s a fun yet practical problem that helps build problem-solving skills. Let’s dive in!
Coding Clutch Challenge Series 101 – All Coding Challenges
Approach to Solving the Problem
Understanding the Problem Statement
You’re given an array of n
unique integers ranging from 1
to n+1
. One number is missing. Your job is to find that missing number efficiently.
Methods to Solve the Problem
- Sum Formula: Use the mathematical formula for the sum of the first
n
natural numbers. - XOR Approach: Leverage the properties of XOR for a neat and efficient solution.
- Iterative Search: Check each number systematically (less efficient).
Coding Clutch Challenge Series 101 – All Coding Challenges
Implementation in Different Languages
Python Solution
def find_missing_number(arr): n = len(arr) + 1 total_sum = n * (n + 1) // 2 return total_sum - sum(arr) # Example numbers = [1, 2, 4, 5, 6] print("Missing number is:", find_missing_number(numbers))
Explanation:
- Calculate the total sum of numbers from
1
ton
. - Subtract the sum of the array from the total sum.
Output:Missing number is: 3
C Solution
#include <stdio.h> int findMissingNumber(int arr[], int n) { int total = (n + 1) * (n + 2) / 2; int sum = 0; for (int i = 0; i < n; i++) { sum += arr[i]; } return total - sum; } int main() { int arr[] = {1, 2, 4, 5, 6}; int n = sizeof(arr) / sizeof(arr[0]); printf("Missing number is: %d\n", findMissingNumber(arr, n)); return 0; }
Explanation:
- Use a loop to calculate the array sum.
- Find the difference between the total and array sums.
Output:Missing number is: 3
C++ Solution
#include <iostream> using namespace std; int findMissingNumber(int arr[], int n) { int total = (n + 1) * (n + 2) / 2; int sum = 0; for (int i = 0; i < n; i++) { sum += arr[i]; } return total - sum; } int main() { int arr[] = {1, 2, 4, 5, 6}; int n = sizeof(arr) / sizeof(arr[0]); cout << "Missing number is: " << findMissingNumber(arr, n) << endl; return 0; }
Output:Missing number is: 3
Coding Clutch Challenge Series 101 – All Coding Challenges
Java Solution
public class MissingNumber { public static int findMissingNumber(int[] arr, int n) { int total = (n + 1) * (n + 2) / 2; int sum = 0; for (int num : arr) { sum += num; } return total - sum; } public static void main(String[] args) { int[] arr = {1, 2, 4, 5, 6}; System.out.println("Missing number is: " + findMissingNumber(arr, arr.length)); } }
Output:Missing number is: 3
Coding Clutch Challenge Series 101 – All Coding Challenges
Optimizing the Code
Using the Sum Formula
The formula n(n+1)/2
reduces time complexity to O(n).
Using the XOR Approach
This method involves XORing all elements in the array and numbers from 1
to n+1
.
Comparative Analysis of Solutions
Time Complexity
- Sum Formula: O(n)
- XOR Approach: O(n)
Space Complexity
- All methods: O(1)
Common Mistakes and Debugging Tips
- Forgetting to include all numbers in the range.
- Miscalculating the total sum.
- Errors in loop indexing.
Coding Clutch Challenge Series 101 – All Coding Challenges
FAQs
- Can this be solved with recursion?
Yes, but it might not be as efficient due to stack overhead. - How does the XOR approach work?
It cancels out all paired numbers, leaving only the missing number. - Is there a difference in runtime between the languages?
No, the algorithm dictates the runtime, not the language. - What happens if the input array is unsorted?
The solution still works as the order doesn’t matter. - Can we use libraries to simplify the task?
Yes, libraries likeNumPy
in Python can help, but it’s good to understand the core logic.