Question

Write function in C++:

int removeDups (string a[], int n); For every sequence of consecutive identical items in a, retain only one item of that sequence. Suppose we call the number of retained items r. Then when this function returns, elements through of a must contain the retained items (in the same relative order they were in originally), and the remaining elements may have whatever values you want. Return the number of retained items. Here's an example: string [9] = { "jon", "daenerys", "samwell", "samwell", "margaery", "margaery", "margaery", "samwell", "samwell" }; int p = removeDups(d, 9); // returns 5 // d[@] through d[4] now contain "jon" "daenerys" "samwell" "margaery" "samwell" // We no longer care what strings are in d[5] and beyond.

Answer 1

We can do it as follows,

1. Define a function removeDups() that accepts an integer array and an integer n, and returns an integer
2. Declare and initialize non_dup_index to 0
3. Use for-loop to iterate over each item in string array
4. Use while-loop to skip to next index if consecutive items are same
5. Update a[non_dup_index] with non duplicate item
6. Increment non_dup_index by 1
7. return non duplicate items count

Program:

#include <iostream>
#include <string.h>
using namespace std;

int removeDups(string a[], int n){  /* function removeDups() */
    int non_dup_index =0;           /* initialize non_dup_index to 0  */
    for(int i=0;i<n;i++){           /* for each item in string array */
        while(i<n-1&&a[i].compare(a[i+1])==0){ /* skip to next item if consecutive items are same */
            i++;                    /* skip to next item */
        }
        a[non_dup_index] = a[i];    /* update non_dup_index with non_dup_index item */
        non_dup_index++;            /* increment non_dup_index */
    }
    return non_dup_index;           /* return non duplicate count */
}

int main()
{
    int n=9;
    string d[9] = {
        "jon","daenerys","samwell","samwell","margaery","margaery", "margaery", "samwell", "samwell"
        
    };
    cout<<"Before : "<<endl;
    for(int i=0;i<n;i++){ /* print items before function call */
        cout<<d[i]<<endl;
    }
    
    n = removeDups(d, n); /* call the function removeDups() */
    
    cout<<endl<<endl<<"After : "<<endl; 
    for(int i=0;i<n;i++){ /* print items after function call */
        cout<<d[i]<<endl;
    }
    return 0;
}

Screenshot:

#include <iostream> #include <string.h> using namespace std; int removeDups (string a[], int n){ /* function removeDups() */ int non_dup_index =0; /* initialize non_dup_index to @ */ for(int i=0;i<n;i++){ /* for each item in string array */ while(i<n-1&&a[i].compare(a[i+1])==0){ /* skip to next item if consecutive items are same /* skip to next item */ i++; } a[non_dup_index] = a[i]; non_dup_index++; /* update non_dup_index with non_dup_index item */ /* increment non_dup_index */ } return non_dup_index; /* return non duplicate count */ بها int main() { int n=9; string d[9] = { "jon", "daenerys", "samwell", "samwell", "margaery", "margaery", "margaery", "samwell", "samwell" }; cout<<"Before : "<<endl; for(int i=0;i<n;i++){ /* print items before function call */ cout<<d[i]<<endl; } n = removeDups(d, n); /* call the function removeDups() */ cout<<endl<<endl<<"After : "<<endl; for(int i=0;i<n;i++){ /* print items after function call */ cout<<d[i]<<endl; } return 0; }

Output:

Before jon daenerys samwell samwell margaery margaery margaery samwell samwell After : jon daenerys samwell margaery samwell ..Program finished with exit code 0

Please don't forget to give a Thumbs Up.