Answer to Question #273077 in C++ for Swale

Question

a. Given the following array of 8 integers {2,3,2,1,2,5,8,2} you are required to write a

C++ program that will loop through it and re-arrange it into the following final output

{5,3,1,8,2,2,2,2} and display it . The idea is to position all values equal to 2 in the

array to come after the other values not equal to 2. The order of the other values i.e 5,1,3

and 8 does not matter as long as they appear positioned before all 2s as illustrated in the hint

below:

Initial array: {2,3,2,1,2,5,8,2}

NumbertoMove = 2

Final Array: {5,3,1,8,2,2,2,2}

Write an algorithm as a pseudocode to solve the problem above. [7 Marks]

b. Transform the pseudocode in part (a) above into an actual C++ program that filters all

values equal to 2 in the array to come after the other values not equal to 2. The order of the

other values i.e 5,1,3 and 8 does not matter as long as they appear positioned before all 2s.

Answer 1

Pseudocode

1 Set insert_i to array length
2 If elemen at position insert_i is not equal 2 go to step 5
3 Decrease insert_i by 1
4 Go to step 2
5 Set check_i to 1
6 If element at position check_i equal 2
7     Swap elements at positions check_i and insert_i
8     Decrement insert_i by 1
9 Increment check_i by 1
10 If check_i is less than insert_i go to step 6

C++ code:

#include <iostream>
using namespace std;

void PrintArray(int a[], int n) {
    for (int i=0; i<n; i++) {
        cout << a[i] << " ";
    }
    cout << endl;
}

void Swap(int&a, int& b) {
    int tmp = a;
    a = b;
    b = tmp;
}

int main() {
    int arr[] = {2,3,2,1,2,5,8,2};
    int n = sizeof(arr) / sizeof(arr[0]);

    PrintArray(arr, n);

    int j=n-1, i=0;
    while (arr[j] == 2) {
        j--;
    }

    while ( i < j ) {
        if (arr[i] == 2) {
            Swap(arr[i], arr[j]);
            j--;
        }
        i++;
    }

    PrintArray(arr, n);
}