Adding numbers which are stored in string variables

The purpose of this question is to add the numbers in the string form. You should not try to convert the strings to integers. The description says the length of the numbers could be up to 5100 digits. So the numbers are simply too big to be stored in integers and doubles. For instance In the following line:

double multiplier = Math.pow(10, num1.length() - 1);

You are trying to store 10^5100 in a double. In IEEE 754 binary floating point standard a double can a store number from ±4.94065645841246544e-324 to ±1.79769313486231570e+308. So your number won't fit. It will instead turn into Infinity. Even if it fits in double it won't be exact and you will encounter some errors in your follow up calculations.

Because the question specifies not to use BigInteger or similar libraries you should try and implement string addition yourself.

This is pretty straightforward just implement the exact algorithm you follow when you add two numbers on paper.

Here is working example of adding two strings without using BigInteger using char array as intermediate container. The point why double can't be used has been explained on @Tempux answer. Here the logic is similar to how adding two numbers on paper works.

public String addStrings(String num1, String num2) {
    int carry = 0;
    int m = num1.length(), n = num2.length();
    int len = m < n ? n : m;
    char[] res = new char[len + 1]; // length is maxLen + 1 incase of carry in adding most significant digits
    for(int i = 0; i <= len ; i++) {
        int a = i < m ? (num1.charAt(m - i - 1) - '0') : 0;
        int b = i < n ? (num2.charAt(n - i - 1) - '0') : 0;
        res[len - i] = (char)((a + b + carry) % 10 + '0');
        carry = (a + b + carry) / 10;
    }
    return res[0] == '0' ? new String(res, 1, len) : new String(res, 0, len + 1);
}

This snippet is relatively small and precise because here I didn't play with immutable String which is complicated/messy and yield larger code. Also one intuition is - there is no way of getting larger output than max(num1_length, num2_length) + 1 which makes the implementation simple.

You have to addition as you do on paper

you can't use BigInteger and the String Length is 5100, so you can not use int or long for addition. You have to use simple addition as we do on paper.

class AddString
{
public static void main (String[] args) throws java.lang.Exception
{
    String s1 = "98799932345";
    String s2 = "99998783456";
    //long n1 = Long.parseLong(s1);
    //long n2 = Long.parseLong(s2);
    System.out.println(addStrings(s1,s2));
    //System.out.println(n1+n2);
}
public static String addStrings(String num1, String num2) {
    StringBuilder ans = new StringBuilder("");
    int n = num1.length();
    int m = num2.length();
    int carry = 0,sum;
    int i, j;
    for(i = n-1,j=m-1; i>=0&&j>=0;i--,j--){
        int a = Integer.parseInt(""+num1.charAt(i));
        int b = Integer.parseInt(""+num2.charAt(j));
        //System.out.println(a+" "+b);
        sum = carry + a + b;
        ans.append(""+(sum%10));
        carry = sum/10;
    }
    if(i>=0){
        for(;i>=0;i--){
            int a = Integer.parseInt(""+num1.charAt(i));
            sum = carry + a;
            ans.append(""+(sum%10));
            carry = sum/10;
        }
    }
    if(j>=0){
        for(;j>=0;j--){
            int a = Integer.parseInt(""+num2.charAt(j));
            sum = carry + a;
            ans.append(""+(sum%10));
            carry = sum/10;
        }
    }
    if(carry!=0)ans.append(""+carry);
    return ans.reverse().toString();
}   
}

You can run the above code and see it works in all cases, this could be written in more compact way, but that would have been difficult to understand for you.

Hope it helps!

you can use this one that is independent of Integer or BigInteger methods

public String addStrings(String num1, String num2) {
    int l1 = num1.length();
    int l2 = num2.length();
    if(l1==0){
        return num2;
    }
    if(l2==0){
        return num1;
    }
    StringBuffer sb = new StringBuffer();
    int minLen = Math.min(l1, l2);
    int carry = 0;
    for(int i=0;i<minLen;i++){
        int ind = l1-i-1;
        int c1 = num1.charAt(ind)-48;
        ind = l2-i-1;
        int c2 = num2.charAt(ind)-48;
        int add = c1+c2+carry;
        carry = add/10;
        add = add%10;
        sb.append(add);
    }

    String longer = null;
    if(l1<l2){
        longer = num2;
    }
    else if(l1>l2){
        longer = num1;
    }
    if(longer!=null){
        int l = longer.length();
        for(int i=minLen;i<l;i++){
            int c1 = longer.charAt(l-i-1)-48;
            int add = c1+carry;
            carry = add/10;
            add = add%10;
            sb.append(add);
        }
    }
    return sb.reverse().toString();
}

The method takes two string inputs representing non-negative integers and returns the sum of the integers as a string. The algorithm works by iterating through the digits of the input strings from right to left, adding each digit and any carryover from the previous addition, and appending the resulting sum to a StringBuilder. Once both input strings have been fully processed, any remaining carryover is appended to the output string. Finally, the string is reversed to produce the correct output order.

Hope this will solve the issue.!

public string AddStrings(string num1, string num2) 
{
    int i = num1.Length - 1, j = num2.Length - 1, carry = 0;
    StringBuilder sb = new StringBuilder();
    while (i >= 0 || j >= 0 || carry != 0) {
        int x = i >= 0 ? num1[i--] - '0' : 0;
        int y = j >= 0 ? num2[j--] - '0' : 0;
        int sum = x + y + carry;
        sb.Append(sum % 10);
        carry = sum / 10;
    }
    char[] chars = sb.ToString().ToCharArray();
    Array.Reverse(chars);
    return new string(chars);
}

Previous solutions have excess code. This is all you need.

class ShortStringSolution {
    static String add(String num1Str, String num2Str) {
        return Long.toString(convert(num1Str) + convert(num2Str));
    }

    static long convert(String numStr) {
        long num = 0;
        for(int i = 0; i < numStr.length(); i++) {
            num = num * 10 + (numStr.charAt(i) - '0');
        }
        return num;
    }
}

class LongStringSolution {

    static String add(String numStr1, String numStr2) {
        StringBuilder result = new StringBuilder();
        int i = numStr1.length() - 1, j = numStr2.length() - 1, carry = 0;
        while(i >= 0 || j >= 0) {
            if(i >= 0) {
                carry += numStr1.charAt(i--) - '0';
            }
            if(j >= 0) {
                carry += numStr2.charAt(j--) - '0';
            }
            if(carry > 9) {
                result.append(carry - 10);
                carry = 1;
            } else {
                result.append(carry);
                carry = 0;
            }
        }
        if(carry > 0) {
            result.append(carry);
        }
        return result.reverse().toString();
    }
}

public class Solution {

    static String add(String numStr1, String numStr2) {
        if(numStr1.length() < 19 && numStr2.length() < 19) {
            return ShortStringSolution.add(numStr1, numStr2);
        }
        return LongStringSolution.add(numStr1, numStr2);
    }
}

For the sake of comprehension of the question

your method's name is addition

you are trying to do a power operation but the result is stored in a variable named multiplication...

there is more than one reason why that code doesnt work...

You need to do something like

Integer.parseInt(string)

in order to parse strings to integers

here the oficial doc