What Method in the String Class Returns Only the First N Characters

What method in the String class returns only the first N characters?

public static string TruncateLongString(this string str, int maxLength)
{
    if (string.IsNullOrEmpty(str)) return str;

    return str.Substring(0, Math.Min(str.Length, maxLength));
}

In C# 8 or later it is also possible to use a Range to make this a bit terser:

public static string TruncateLongString(this string str, int maxLength)
{
    return str?[0..Math.Min(str.Length, maxLength)];
}

Which can be further reduced using an expression body:

public static string TruncateLongString(this string str, int maxLength) =>
    str?[0..Math.Min(str.Length, maxLength)];

Note null-conditional operator (?) is there to handle the case where str is null. This replaces the need for an explict null check.

How do I get the first n characters of a string without checking the size or going out of bounds?

Here's a neat solution:

String upToNCharacters = s.substring(0, Math.min(s.length(), n));

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^{Opinion: while this solution is "neat", I think it is actually less readable than a solution that uses if / else in the obvious way. If the reader hasn't seen this trick, he/she has to think harder to understand the code. IMO, the code's meaning is more obvious in the if / else version. For a cleaner / more readable solution, see @paxdiablo's answer.}

Keep only first 60 characters of a string

Use Substring

Note: you need to check you input's length

string data = Model.Products[i].Description;
string temp = string.Empty;
if(!string.IsNullOrEmpty(data) && data.Length >= 60)
   temp = data.Substring(0,60);
else
   temp = data;

First n characters of string or whole string; without SubscriptOutOfBounds

String>>truncateTo:

'abcdef' truncateTo: 30. "'abcdef'"
'abcdef' truncateTo: 3. "'abc'"

String-matching first n letters of two strings

I think the fastest approach would be to use Binaray Search, which will give you O(logn) complexity instead of O(n).
Here n is the length of smallest string.

The approach is simple in binary search. Look for similarity end for the index character in both the strings. For example if i is your index then check i+1 for dis-similarity character where character at i index is similar. And if that is the case break, return i as your answer. Or else keep on searching in sub-scope.

Edit

Adding function for better understanding.

int lengthOfFirstSimilarCharacters(String str1, String str2) {
    int strlen1 = str1.length();
    int strlen2 = str2.length();
    if(strlen1 > strlen2){
        return lengthOfFirstSimilarCharacters(str2,str1);
    }
    int i = 0;
    int j = strlen1-1;
    while(i<=j){
        int mid = i + (j-i)/2;
        if(str1.charAt(mid) == str2.charAt(mid)) {
            if(mid+1<strlen1 && str1.charAt(mid+1) != str2.charAt(mid+1)){
                return mid+1;
            }
            i = mid+1;
        }else{
            j = mid-1;
        }
    }
    return i;
}

Write a public method firstFifteen() that returns a String consisting of the first fifteen characters in the longNumber

I'm not entirely sure you need firstFifteen as a field. You can simply grab the first 15 characters

return longNumber.substring(0, 15);

You also don't need a parameter for that method, since longNumber is a field

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