I Am Able To Go Outside Array Bounds - c++

Question

Given two strings, write a method to decide if one is an anagram/permutation of the other. This is my approach:
I wrote this function to check if 2 strings are anagrams (such as dog and god).
In ascii, a to z is 97 - 122.
Basically I have an array of bools that are all initially false. Everytime I encounter a char in string1, it marks it as true.
To check if its an anagram, I check if any chars of string2 are false (should be true if encountered in string1).
I'm not sure how but this works too: arr[num] = true; (shouldnt work because I dont take into account that ascii starts at 97 and thus goes out of bounds).
(Side note: is there a better approach than mine?)
EDIT: Thanks for all the responses! Will be carefully reading each one. By the way: not an assignment. This is a problem from a coding interview practice book
bool permutation(const string &str1, const string &str2)
{
// Cannot be anagrams if sizes are different
if (str1.size() != str2.size())
return false;
bool arr[25] = { false };
for (int i = 0; i < str1.size(); i++) // string 1
{
char ch = (char)tolower(str1[i]); // convert each char to lower
int num = ch; // get ascii
arr[num-97] = true;
}
for (int i = 0; i < str2.size(); i++) // string 2
{
char ch = (char)tolower(str2[i]); // convert char to lower
int num = ch; // get ascii
if (arr[num-97] == false)
return false;
}
return true;
}

Answer 1

There is nothing inherent in C++ arrays that prevents you from writing beyond the end of them. But, in doing so, you violate the contract you have with the compiler and it is therefore free to do what it wishes (undefined behaviour).
You can get bounds checking on "arrays" by using the vector class, if that's what you need.
As for a better approach, it's probably better if your array is big enough to cover every possible character (so you don't have to worry about bounds checking) and it shouldn't so much be a truth value as a count, so as to handle duplicate characters within the strings. If it's just a truth value, then here and her would be considered anagrams.
Even though you state it's not an assignment, you'll still learn more if you implement it yourself, so it's pseudo-code only from me. The basic idea would be:
def isAnagram (str1, str2):
# Different lengths means no anagram.
if len(str1) not equal to len(str2):
return false
# Initialise character counts to zero.
create array[0..255] (assumes 8-bit char)
for each index 0..255:
set count[index] to zero
# Add 1 for all characters in string 1.
for each char in string1:
increment array[char]
# Subtract 1 for all characters in string 2.
for each char in string2:
decrement array[char]
# Counts will be all zero for an anagram.
for each index 0..255:
if count[index] not equal to 0:
return false
return true

Answer 2

Working approach : with zero additional cost.
bool permutation(const std::string &str1, const std::string &str2)
{
// Cannot be anagrams if sizes are different
if (str1.size() != str2.size())
return false;
int arr[25] = {0 };
for (int i = 0; i < str1.size(); i++) // string 1
{
char ch = (char)tolower(str1[i]); // convert each char to lower
int num = ch; // get ascii
arr[num-97] = arr[num-97] + 1 ;
}
for (int i = 0; i < str2.size(); i++) // string 2
{
char ch = (char)tolower(str2[i]); // convert char to lower
int num = ch; // get ascii
arr[num-97] = arr[num-97] - 1 ;
}
for (int i =0; i< 25; i++) {
if (arr[i] != 0) {
return false;
}
}
return true;
}

Answer 3

Yes, C and C++ both doesn't carry out the index-out-of-bounds.
It is the duty of the programmer to make sure that the program logic doesn't cross the legitimate limits. It is the programmer who need to make checks for the violations.
Improved Code:
bool permutation(const string &str1, const string &str2)
{
// Cannot be anagrams if sizes are different
if (str1.size() != str2.size())
return false;
int arr[25] = { 0 }; //<-------- Changed
for (int i = 0; i < str1.size(); i++) // string 1
{
char ch = (char)tolower(str1[i]); // convert each char to lower
int num = ch; // get ascii
arr[num-97] += 1; //<-------- Changed
}
for (int i = 0; i < str2.size(); i++) // string 2
{
char ch = (char)tolower(str2[i]); // convert char to lower
int num = ch; // get ascii
arr[num-97] = arr[num-97] - 1 ; //<-------- Changed
}
for (int i =0; i< 25; i++) { //<-------- Changed
if (arr[i] != 0) { //<-------- Changed
return false; //<-------- Changed
}
}
return true;
}