Typo searching for at least 1 character - c++

Question

The original prompt was:
Write a program that keeps track of a speakers bureau. The program should use a
structure to store the following data about a speaker:
Name
Telephone Number
Speaking Topic
Fee Required
The program should use an array of at least 10 structures. It should let the user enter
data into the array, change the contents of any element, and display all the data stored
in the array. The program should have a menu-driven user interface.
Input Validation: When the data for a new speaker is entered, be sure the user enters
data for all the fields. No negative amounts should be entered for a speaker s fee.
The added prompt was:
I need this to expand the search pattern with the potential one character of letter or digit typos. Only one character maybe a typo, in any position Try these test patterns should get the following results:
0-9 is 0x30-0x39
a-z is 0x41-0x5A
A-Z is 0x61-0x7A (or lower case it)
And I can't get the added prompt to work with my current program.
No other characters in the search pattern may be changed.
#include <iostream>
#include <cstring>
#include<string>
using namespace std;
bool print_one_typo(string input, string people[11], bool is_found[11])
{
bool found = false;
if (input[0] == '?')
{
char *strPtr = NULL;
for (int i = 0; i < 11; i++)
{
strPtr = strstr(people[i].c_str(), input.substr(1).c_str());
if (strPtr != NULL)
{
cout << "\t" << people[i] << endl;
found = true;
is_found[i] = true;
}
}
}
else
{
for (int i = 0; i < 11; i++)
{
bool match = true;
string str = people[i];
int value = str.find(input[0]);
for (int k = 0; k < input.length(); k++)
{
if (input[k] != '?' && input[k] != str[value++])
{
match = false;
break;
}
}
if (match && !is_found[i])
{
found = true;
cout << "\t" << people[i] << endl;
}
}
}
return found;
}
int main()
{
string people[11] = { "Becky Warren, 555-1223",
"Joe Looney, 555-0097",
"Geri Palmer, 555-8787",
"Lynn Presnell, 555-1225",
"Holly Gaddis, 555-8878",
"Sam Wiggins, 555-0998",
"Bob Kain, 555-8712",
"Tim Haynes, 555-7676",
"Warren Gaddis, 555-9037",
"Jean James, 555-9223",
"Ron Palmer, 555-7227" };
bool is_found[11] = { false };
string lookUp;
int i;
cout << "\t People and Phone numbers" << endl;
cout << "Enter name or phone number: ";
cin >> lookUp;
cout << "result: " << endl;
bool found = false;
bool output = false;
for (int i = 0; i < lookUp.length(); i++)
{
string local = lookUp;
found = print_one_typo(local.replace(i, 1, 1, '?'), people, is_found);
if (found) output = true;
}
if (!output)
cout << "No matching product was found" << endl;
return 0;
}

Answer 1

I think your code overthinks the problem. Also, you didn't specify if "typo" just means "wrong character", or a fuller range that includes dropped characters or inserted characters.
If you are only looking for matches with zero or one incorrect characters, but otherwise the same length, I think this code should do:
bool print_one_typo(string input, string people[11], bool is_found[11])
{
for (int i = 0; i < 11; i++)
{
if ( is_found[i] ) // Skip if it had already been found?
continue;
if ( input.length() != people[i].length() ) // Are they same length?
continue; // No: Skip it.
int typos = 0;
size_t len = input.length();
for (size_t j = 0; j != len && typos < 2; j++)
if ( input[j] != people[i][j] )
typos++;
if (typos < 2) // Fewer than 2 typos: We have a winner! Return it.
{
is_found[i] = true;
return true;
}
}
return false;
}
This code skips strings that differ in length, or which you're filtering out via the is_found[] array. Not sure why you're doing that, but I preserved that bit of your original code.
If it finds two strings that are the same length, it just compares them character by character, counting up typos. If it sees 2 or more typos, it skips to the next one. Otherwise, it takes the first string that's the same length but fewer than 2 typos.