**Contents**hide

In this post, we will discuss how to count the number of carry operations required to add two arrays of numbers in Java. This is an extension of the “carry forward” approach

**The Algorithm**

To count the number of carry operations required to add two arrays of numbers, we can use the following algorithm:

- Initialize a variable
**count****0**to keep track of the number of carry operations. - Convert both arrays to strings.
- Starting from the rightmost element of both arrays, add the corresponding elements.
- If the sum is greater than 9, increment
**count** - Repeat steps 3-4 until all elements have been added.
- Return
**count**.

## Pseudo Code

Here is some pseudo code that implements the above algorithm:

```
function countCarryOperations(a, b):
count = 0
carry = 0
aStr = convert a to string
bStr = convert b to string
for i from length of aStr - 1 to 0:
sum = aStr[i] + bStr[i] + carry
if sum > 9:
count++
carry = 1
else:
carry = 0
return count
```

**Java Code with Comments**

Here is some Java code that implements the above algorithm:

```
public class CarryForward {
public static int countCarryOperations(int[] a, int[] b) {
int count = 0;
```*// Initialize count to 0*
int carry = 0; *// Initialize carry to 0*
String[] aStr = new String[a.length]; *// Convert a to string array*
String[] bStr = new String[b.length]; *// Convert b to string array*
for (int i = 0; i < a.length; i++) {
aStr[i] = Integer.toString(a[i]);
}
for (int i = 0; i < b.length; i++) {
bStr[i] = Integer.toString(b[i]);
}
// Start from rightmost element and add corresponding elements
for (int i = aStr.length - 1; i >= 0; i--) {
int sum=Integer.parseInt(aStr[i])+Integer.parseInt(bStr[i])+
carry;
if (sum > 9) {
*// If sum is greater than 9, increment count and carry the 1*
count++;
carry = 1;
} else {
carry = 0;
}
}
return count;
}
}

**Dry Run**

Let’s do a dry run of the above code with the input **a = [1,2,3]** and **b = [4,5,6]**.

**count****0****carry**is initialized to**0**.**aStr****[“1″,”2″,”3”]**and**bStr****[“4″,”5″,”6”]****.**- The for loop starts at index 2 (the rightmost element).
- The first iteration of the loop calculates
**sum****3 + 6 + 0 = 9**. Since**sum <= 9**,**count****0****carry****0**. - The second iteration of the loop calculates
**sum****2 + 5 + 0 = 7****.**Since**sum <= 9**, count remains**0**and**carry**is set to**0**. - The third iteration of the loop calculates
**sum****1 + 4 + 0 = 5****.**Since**sum <= 9**, count remains**0****carry****0**. - The for loop ends and the function returns count, which is
**0**.

Therefore, for the input **a = [1,2,3]** and **b = [4,5,6]**, the function returns 0, indicating that no carry operations are required.

**Time and Space Complexity**

The time complexity of this algorithm is **O(n)**, where n is the number of elements in the larger array. This is because we need to iterate over each element once.

The space complexity of this algorithm is **O(n)**, since we need to create two new string arrays with n elements each.

**Main Method**

Here is an example main method that uses the above code:

```
public static void main(String[] args) {
int[] a = {1, 2, 3};
int[] b = {4, 5, 6};
int result = countCarryOperations(a, b);
System.out.println("Number of carry operations: " + result);
}
```

When run, this code will output **Number of carry operations: 0****.**