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Problem: Given an array of size n, which contains numbers in the range 1 to n. Each number is present at least once except for 2 numbers. Find the missing numbers.
Solution:read

#include <iostream>
#include <bitset>
using namespace std;

//Given an array of size n. It contains numbers in the range 1 to n. 
//Each number is present at least once except for 2 numbers. 
//Find the missing numbers.

static const int N = 10;

int main()
{
    int numArray[N] = { 1, 7, 3, 4, 6, 2, 9, 9, 10, 10 };
    bitset<N> bits;

    bits.set();

    for (int i = 0; i < N; i++)
    {
        // Clears bit that corresponds to the number; bit array is zero based
        bits.set(numArray[i] - 1, false);
    }

    for (int i = 0; i < N; i++)
    {
        if (bits.test(i)) {
            printf("Found unused value %d\n", i + 1);
        }
    }
    return 0;
}
//This solution is kindly provided by Jeff Fore.
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