C++ Fill an array with random numbers without duplicates

Question

I need to find a way to fill an array with random numbers without having duplicates,so i wrote this code and it works.My question is,is this code efficient and will it really have no duplicates? Thanks in advance!

#include <iostream>
#include <time.h>
#include <stdlib.h>

int main(void) {
    srand(time(NULL));
    std::size_t array_size=100;
    int array[array_size];
    for(int i=0;i<array_size;i++) {
        array[i]=rand()%105+1;
            for(int k=0;k<array_size;k++) {         // Checks if there is a duplicate in the array //
                    if(i!=k) {                      // Don't check for the same array position //
                        if(array[i]==array[k]) {    // If a duplicate is found,repeat the check process//
                        array[i]=rand()%105+1;
                        k=-1;                       // -1 so the for loop starts from zero //
                }
            }
        }
    }
    return 0;
}

Answer 1

That approach works fine when the number of desired values is much less than the number of possible values. Most of time it won't produce a duplicate value, so it just keeps the one it produced. But when there isn't a lot of slack there are lots of duplicates; when this code gets close to the end it's generating a value between 1 and 106 when there are only six or seven or so acceptable values. So it ends up spinning its wheels.

Instead of doing all that looping, create an array that holds all of the possible values, randomly shuffle it, and throw out the extra ones:

int array[105];
for (int i = 0; i < 105; ++i)
    array[i] = i + 1;
std::mt19937_64 mt;
std::shuffle(std::begin(array), std::end(array), mt);
for (int i = 0; i < 100; ++i)
    std::cout << array[i] << '\n';