## Question

Given a list of bytes `a`

, each representing one byte of a larger integer (ie. `{0x12, 0x34, 0x56, 0x78}`

represents the integer `0x12345678`

), and an integer `b`

, find `a % b`

.

eg.

`mod({0x03, 0xED}, 10) = 5`

Once you think that you’ve solved the problem, click below to see the solution.

*As always, remember that practicing coding interview questions is as much about how you practice as the question itself. Make sure that you give the question a solid go before skipping to the solution. Ideally if you have time, write out the solution first by hand and then only type it into your computer to verify your work once you've verified it manually. To learn more about how to practice, check out this blog post.*

## Solution

How was that problem? You can check out the solution in the video below.

Here is the source code for the solution shown in the video (Github):

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
// Compute the mod. We use a char array as it is equivalent to an array of // unsigned bytes public static int mod(char[] a, int b) { // If input is null, let's just return 0 if (a == null) return 0; int m = 0; // Start with modding the most significant byte, then repeatedly shift // left. This way our value never gets larger than an int for (int i = 0; i < a.length; i++) { m <<= 8; m += (a[i] & 0xFF); m %= b; } return m; } |

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