Given an array containing n
distinct numbers taken from the range 0
to n
, find the one number that is missing from the array.
Input: An array of n
distinct integers.
Output: The missing integer.
Constraints:
0
to n
.Example:
Input: [3, 0, 1]
Output: 2
The core challenge of this problem is to identify the missing number in a sequence of distinct integers. This problem is significant in various applications such as data validation, error detection, and ensuring data integrity. A common pitfall is assuming the array is sorted or contains duplicates, which is not the case here.
To solve this problem, we can consider several approaches:
A naive solution would involve sorting the array and then checking for the missing number. However, this approach is not optimal due to the sorting step, which has a time complexity of O(n log n).
We can improve upon the naive solution with the following approaches:
We can use the formula for the sum of the first n
natural numbers: sum = n * (n + 1) / 2
. By calculating the expected sum and subtracting the actual sum of the array, we can find the missing number.
Using the properties of XOR, we can find the missing number by XORing all the array elements with all the numbers from 0
to n
. This approach leverages the fact that a ^ a = 0
and a ^ 0 = a
.
0
to n
using the formula n * (n + 1) / 2
.xor1
for the array elements and xor2
for the numbers from 0
to n
.xor1
.0
to n
and store the result in xor2
.xor1
and xor2
.// Sum Formula Approach
function findMissingNumber(arr) {
const n = arr.length;
const expectedSum = (n * (n + 1)) / 2;
const actualSum = arr.reduce((acc, num) => acc + num, 0);
return expectedSum - actualSum;
}
// Example usage:
const arr = [3, 0, 1];
console.log(findMissingNumber(arr)); // Output: 2
// XOR Approach
function findMissingNumberXOR(arr) {
const n = arr.length;
let xor1 = 0;
let xor2 = 0;
// XOR all array elements
for (let num of arr) {
xor1 ^= num;
}
// XOR all numbers from 0 to n
for (let i = 0; i <= n; i++) {
xor2 ^= i;
}
// The missing number is the XOR of xor1 and xor2
return xor1 ^ xor2;
}
// Example usage:
const arr = [3, 0, 1];
console.log(findMissingNumberXOR(arr)); // Output: 2
Time Complexity: O(n) - We iterate through the array once to calculate the sum.
Space Complexity: O(1) - We use a constant amount of extra space.
Time Complexity: O(n) - We iterate through the array and the range 0
to n
once.
Space Complexity: O(1) - We use a constant amount of extra space.
Consider the following edge cases:
[0]
or [1]
.[1, 2, 3]
or [0, 1, 2]
.Both approaches handle these edge cases effectively by following the same logic.
To test the solution comprehensively, consider the following test cases:
[3, 0, 1]
- Expected output: 2
.[0]
- Expected output: 1
.[0, 1, 2]
- Expected output: 3
.[1, 2, 3]
- Expected output: 0
.Use testing frameworks like Jest or Mocha for automated testing.
When approaching such problems:
In this blog post, we explored the "Find the Missing Number" problem and discussed various approaches to solve it. We covered the naive solution, optimized solutions using the sum formula and XOR, and provided detailed explanations and code implementations. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.
Keep practicing and exploring different algorithms to enhance your understanding and proficiency.