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I am designing a problem in which I have to use an int array to add or subtract values. For example instead of changing 100 to 101 by adding 1, I want to do the same thing using the int array. It work like this:

int[] val = new int[3];

val[0] = 1;
val[1] = 0;
val[2] = 0;

val[2] += 1;

so, If I have to get a value of 101, I will add 1 to val[2].

The only problem I have is finding a way to make int array work like how adding and subtracting from an ordinary integer data set works.

Is this possible using a for loop or a while loop? Any help will be appreciated!

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  • Could you add a more complete example for clarifying the problem? Commented Sep 15, 2015 at 23:15
  • It is a bit tricky to explain what I want. What I trying to achieve is having an array of integer storing each single digit from the integer. For example: 100 will be stored in a array called val like: val[0] = 1, val[1] = 0 and val[2] = 0. Then, When I add something to the hundred, I wanna do it using the array that represents the integer. I hope it explains it! Commented Sep 15, 2015 at 23:27

5 Answers 5

Reset to default
3

Here's your homework:

public static int[] increment(int[] val) {
    for (int i = val.length - 1; i >= 0; i--) {
        if (++val[i] < 10)
            return val;
        val[i] = 0;
    }
    val = new int[val.length + 1];
    val[0] = 1;
    return val;
}

Make sure you understand how and why it works before submitting it as your own work.

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5 Comments

Could you please explain me the purpose of the three lines of code at the end of the for loop? This doesn't even get executed.
Actually never mind. I got it now. So, you don't have to explain the purpose of the lines of code.
Anyway, since the question arised: The last 3 lines handle the overflow. When incrementing an array with length n filled with all 9s, a new array with length n+1 is created to fit the result. Incrementing {9, 9} results in {1, 0, 0}.
@adityajain019 how so? Hint: The “last 3 lines” are executed if, and only if, the input is all 9’s in which case it return 1 followed by zeros where there were 9’s
My bad. removing my comment.
1

Solution of this problem is designed by using String
You can refer to this method which will return sum of 2 nos having input in String format.

Input String should contain only digits.

class Demo {
public static String add(String a1, String b1) {
    int[] a = String_to_int_Array(a1);
    int[] b = String_to_int_Array(b1);
    int l = a.length - 1;
    int m = b.length - 1;
    int sum = 0;
    int carry = 0;
    int rem = 0;
    String temp = "";
    if (a.length > b.length) {
        while (m >= 0) {
            sum = a[l] + b[m] + carry;
            carry = sum / 10;
            rem = sum % 10;
            temp = rem + temp;
            m--;
            l--;
        }
        while (l >= 0) {
            sum = a[l] + carry;
            carry = sum / 10;
            rem = sum % 10;
            temp = rem + temp;
            l--;
        }
        if (carry > 0) {
            temp = carry + temp;
        }
    } else {
        while (l >= 0) {
            sum = a[l] + b[m] + carry;
            carry = sum / 10;
            rem = sum % 10;
            temp = rem + temp;
            m--;
            l--;
        }
        while (m >= 0) {
            sum = b[m] + carry;
            carry = sum / 10;
            rem = sum % 10;
            temp = rem + temp;
            m--;
        }
        if (carry > 0) {
            temp = carry + temp;
        }
    }
    return temp;

}

public static int[] String_to_int_Array(String s) {
    int arr[] = new int[s.length()], i;
    for (i = 0; i < s.length(); i++)
    arr[i] = Character.digit(s.charAt(i), 10);
    return arr;
}

public static void main(String a[]) {
    System.out.println(add("222", "111"));
}
}
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1

Quick & dirty:

static void increment(int[] array){
    int i = array.length-1;
    do{
        array[i]=(array[i]+1)%10;
    }while(array[i--]==0 && i>=0);
}

Note the overflow when incementing e.g. {9, 9}. Result is {0, 0} here.

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enter image description here

public static void increment() {    
    int[] acc = {9,9,9,9};
    String s="";
    for (int i = 0; i < acc.length; i++)
        s += (acc[i] + "");
    int i = Integer.parseInt(s);
    i++;
    System.out.println("\n"+i);
    String temp = Integer.toString(i);
    int[] newGuess = new int[temp.length()];
    for (i = 0; i < temp.length(); i++)
    {
        newGuess[i] = temp.charAt(i) - '0';
    }

    printNumbers(newGuess);
}

public static void printNumbers(int[] input) {      
    for (int i = 0; i < input.length; i++) {
        System.out.print(input[i] + ", ");
    }
    System.out.println("\n");
}
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0

If someone is looking for this solution using JavaScript or if you can translate it to java, here's your optimum solution:

function incrementArr(arr) {
  let toBeIncrementedFlag = 1,   // carry over logic
    i = arr.length - 1;
  while (toBeIncrementedFlag) {
    if (arr[i] === 9) {
      arr[i] = 0;    // setting the digit as 0 and using carry over
      toBeIncrementedFlag = 1;
    } else {
      toBeIncrementedFlag = 0;
      arr[i] += 1;
      break;    // Breaking loop once no carry over is left
    }
    if (i === 0) {    // handling case of [9,9] [9,9,9] and so on
      arr.unshift(1);
      break;
    }
    i--;    // going left to right because of carry over
  }

  return arr;
}
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