Understanding the time complexity

In this blog I have started learning the DSA in c++ so , 

to understand the time complexity

Let's take a example to see weather the pairs of (x, y) in the  given array is equal Z , 

where Z = sum of the pair. 

to execute this I wrote the following code;----->

        : a[ ] is an array which I will declare while creating a function

int n is the size of an array in bytes and int z is the sum that is the sum of the pairs

the at should be equal to z;

#include<iostream>

using namespace std;

/*a is an array which I will declare while creating a function

int n is the size of an array in bytes and int z is the sum that is the sum of the pairs

the at should be equal to z;

*/

bool Findpair(int a[], int n , int z)

{

    //ireting all the pairs of an array in a for loop

    for(int i = 0; i<n; i++){

        for(int j = 0; j<i; j++){

            if(i!=j && a[i] + a[j] == z){

                return true;

            }

        }

    }

    return false;

}

int main(){

   

    int a[] = {7, 8, 9, 4, 5, 12, 30};//if we increase the size of the array the time of exicution also increases

    int z = 15;

    int n = sizeof(a) / sizeof(a[0]);

    cout<<n;

    if (Findpair(a, n, z))

    {

        cout<<"True";

    }

    else{

        cout<<"False";

    }

    return 0;

}

/*

Note:-

Order of growth is how the time of execution depends on the length of the input

from the above example,

it is clearly that the time of execution quadratically depends on the length of the array.

Big oh = O(n^2)

*/