Given two strings representing integer numbers ("123" , "30") return a string representing the sum of the two numbers ("153")

Question

Luca Lupo · Accepted Answer

public static String sumStrings(String a, String b){
	    char[] num1 = a.toCharArray();
	    char[] num2 = b.toCharArray();

	    int i = num1.length - 1;
	    int j = num2.length - 1;
	    
	    StringBuilder sumString = new StringBuilder();
	    int carry = 0;
	        
	    while(i >= 0 || j >= 0){
	        int d1 = 0;
	        int d2 = 0;
	        
	        if (i >= 0) d1 = num1[i--] - '0';
	        if (j >= 0) d2 = num2[j--] - '0';
	        
	        int sum = d1 + d2 + carry;
	        if (sum >= 10){
	            carry = sum / 10;
	            sum = sum % 10;
	        }else carry = 0;
	        
	        sumString.insert(0, sum);
	    }
	    
	    return sumString.toString();
	}

Sigiloso · Answer

The interviewer wanted a loop through the digits starting form right to left, adding them one by one, and keeping track of the carriage.

Sigiloso · Answer

It's not stupid a stupid question. What if the strings have 10000 characters?

Matteo Gobbi · Answer

I don't understand...it's a very stupid question!

return Integer.toString(Integer.parseInt("123") + Integer.parseInt("30));

Adding Two Strings which are valid numbers. Without converting it to Integers using internal method · Answer

public static String addTwoStrings(String s1, String s2) {
        int len1 = s1.length();
        int len2 = s2.length();

        char[] s1Chars = s1.toCharArray();
        char[] s2Chars = s2.toCharArray();

        StringBuilder sb = new StringBuilder();
        int pointerA = len1 -1;
        int pointerB = len2 -1;

        int carry = 0;

        while (pointerA >= 0 || pointerB >= 0){

            int a = pointerA  10) {
              carry = sumTemp / 10;
              sumTemp =  sumTemp % 10;
            } else {
              carry = 0;
            }

            sb.insert(0, sumTemp);
        }
    
        if(carry >0) sb.insert(0, carry);

        return sb.toString();
    }

Ker · Answer

import java.util.Arrays;
import java.util.Scanner;

public class AddNumericStrings {

	public static void main(String[] args) {
		final Scanner in = new Scanner(System.in);
		while (true) {
			System.out.println("Enter 2 numeric strings : ");
			String x = in.nextLine();
			String y = in.nextLine();

			System.out.println(add(x.toCharArray(), y.toCharArray()));
		}
	}

	private static char[] add(char[] big, char[] small) {
		char[] result = new char[big.length + 1];
		Arrays.fill(result, '0');

		for (int i = big.length - 1, j = small.length - 1; i >= 0 || j >= 0; i--, j--) {
			char x = big[i];
			char y = '0';
			if (j >= 0) {
				y = small[j];
			}
			int val = x - '0';
			val += (y - '0');
			result[i+1] += val % 10;
			if (val > 10) {
				result[i] += (val/10);
			}
		}
		return result;
	}
}

Sigiloso · Answer

You all know that negative integers exist, right? The question does not specify if the integers are non-negative. One just assume, therefore, that negative integers are possible. It would not be called subtraction. Subtraction does not exist. It would just be addition of the additive inverse.

Sigiloso · Answer

Index the tuple to find the 1st & 2nd element. Convert the elements into ints and then convert the answer back to a string.

Code: str(int(inp[0]) + int(inp[1]))

Conner · Answer

public class StringToInt {
	public int stringToInt(String str) {
		int tens = 1;
		int num = 0;
		for(int i = 0; i < str.length(); ++i)
		{
			num += (str.charAt(str.length() - 1 - i) - '0') * tens;
			tens *= 10;
		}

		return num;
	}

	public int addStrings(String str1, String str2) {
		return stringToInt(str1) + stringToInt(str2);
	}

	public static void main(String [] args) {
		StringToInt s = new StringToInt();
		System.out.println(s.addStrings("145", "23"));
	}
}

Gabriel · Answer

@Conner

What if the strings are 1000 characters long? does your int tens and int num variables support that?

@staticaza · Answer

int stringToNumber(char *a){
    char *end = a;
    int it = 1;
    int acum = 0;
    while (*end != NULL){
        end++; //move pointer to last char of string
    }
    while (&end != &a){ 
        acum+=((*end - '0') * it);
        it *= 10;
        end--;
    }
    return acum;
}
int sum (char *a, char *b){
    return stringToNumber(a) + stringToNumber(b);
}

Sigiloso · Answer

It is basic but yet not stupid.
I assume that the interviewer asked to implement atoi and itoa (in case the interview was in C/C++).

Sigiloso · Answer

lol it is a stupid question i agree.
All you have to do is parse the strings add em parse em again and return em

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