## Monday, August 24, 2015

### LeetCode OJ - Missing Number

Problem:

Solution:

Classic interview question. Just compare the sum and the expected sum for all numbers would solve the problem. To generalize it, any group would work just as well (e.g. xor), it is just harder to compute the expected value, but with the advantage that we can forget about overflows.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/missing-number/

#include "LEET_MISSING_NUMBER.h"
#include <stack>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_MISSING_NUMBER
{
class Solution
{
public:
int missingNumber(vector<int>& nums)
{
int expected_sum = (nums.size()) * (nums.size() + 1 ) / 2;
int actual_sum = 0;
for (int i = 0; i < nums.size(); i++)
{
actual_sum += nums[i];
}
return expected_sum - actual_sum;
}
};
};

using namespace _LEET_MISSING_NUMBER;

int LEET_MISSING_NUMBER()
{
Solution solution;
int case1[] = { 0, 1, 3 };
cout << (solution.missingNumber(vector<int>(case1, case1 + _countof(case1))) == 2) << endl;
return 0;
}