replaces duplicate values in an array with -1 [closed] - c++

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Write a function that replaces duplicate values in an array with -1.
Assume that an array of positive numbers such as {1, 2, 5, 4, 2, 7, 1, 2} is passed to your function.
You must check duplicate values in this array and replace them with -1 except the first appearance.
Your final array of {1, 2, 5, 4, 2, 7, 1, 2} should look like {1,2,5,4,-1, 7, -1,-1}
this is my question can someone figure out ?
{
for(j=i+1; j<size; j++)
{
/* If any duplicate found */
if(arr[i] == arr[j])
{
/* Delete the current duplicate element */
for(k=j; k < size - 1; k++)
{
arr[k] = arr[k + 1];
}
/* Decrement size after removing duplicate element */
size--;
/* If shifting of elements occur then don't increment j */
j--;
}
}
}

Answer 1

In C++, you would do this exactly as you would do manually, with your brain.
Iterate over all elements in the array
Count the number of occurences of an element
If this element is more than one time present, then replace it.
For counting we will use a hash map, so an std::unordered_map. Please read here about that.
The very simple code would then look like this:
#include <iostream>
#include <vector>
#include <unordered_map>
int main() {
std::vector testData{ {1, 2, 5, 4, 2, 7, 1, 2} };
std::unordered_map<int, unsigned int> counter{};
for (int& i : testData) {
counter[i]++;
if (counter[i] > 1)
i = -1;
}
for (const int k : testData)
std::cout << k << ' ';
}

Answer 2

Maybe something like this:
#include <stdio.h>
#include <stdlib.h>
void replaceDuplicates(int *arr, int n)
{
int *differents = malloc(n * sizeof(n));
int iterator = 0;
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
{
if(arr[i] == differents[j])
{
arr[i] = -1;
}
}
if(arr[i] != -1)
{
differents[iterator++] = arr[i];
}
}
}
int main()
{
int arr[8] = {1, 2, 5, 4, 2, 7, 1, 2};
replaceDuplicates(arr, 8);
for(int i = 0; i < 8; i++)
{
printf("%d ", arr[i]);
}
return 0;
}
Note that I haven't tested this code that much. You should probably improve it
This code outputs:
1 2 5 4 -1 7 -1 -1