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LESSON 52 OF 409 · CALCULATIONS - CHALLENGES

Sum of the Last Two Digits in C++ with % and /

Practice Read the full explanation

Given a non-negative integer n with at least two digits, compute and return the sum of the last two digits

Example:

Input: n = 2379

Output: 16

Explanation:
Last two digits of n are 7 and 9, so our answer is 7 + 9 = 16
Hints:

You can make use of modulo operator

You can get the last digit of a number using modulo 10:
2379 % 10 = 9

You can get a number consisting of last 2 digits using modulo 100:
2379 % 100 = 79

If you have a 2 digit number, you can get the first digit by using integer division

Divide a 2 digit number to 10 to get the first digit:
79 / 10 = 7 (integer division)
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    C++ · CALCULATIONS - CHALLENGES

    Last Two Digit Sum in C++, explained

    Problem overview

    The method lastTwoDigitSum(int n) receives a non-negative integer with at least two digits and returns the sum of its last two digits. For n = 2379 the last two digits are 7 and 9, so the answer is 16.

    The task does not need a loop or a string. Two integer operators, remainder (%) and integer division (/), pull the digits apart.

    Approach: remainders and integer division

    In base 10, dividing by 10 or 100 and looking at the remainder picks out the low digits:

    • n % 10 is the last digit. 2379 % 10 = 9.
    • n % 100 is the number formed by the last two digits. 2379 % 100 = 79.
    • For a two-digit number x, x / 10 is its first digit. 79 / 10 = 7, because dividing two int values in C++ throws away the fractional part.

    So first reduce n to its last two digits with % 100, then split that two-digit number into a ones digit (% 10) and a tens digit (/ 10), and add them.

    The same two operators extend to any number of digits. Repeating "take n % 10, then replace n with n / 10" in a loop peels off the digits one at a time, which is how you would sum all the digits of a number. Here only two digits are needed, so no loop is required.

    Step by step

    1. Take the last two digits. int lastTwoDigits = n % 100; gives 79 for 2379.
    2. Take the last digit of that number: lastTwoDigits % 10, which is 9.
    3. Take the second-to-last digit of n, which is the first digit of lastTwoDigits: lastTwoDigits / 10, which is 7. Both operands are int, so the division gives a whole number.
    4. Return the sum of the two, 9 + 7 = 16.

    The solution in C++

    class Solution {
public:
    int lastTwoDigitSum(int n) {
        int lastTwoDigits = n % 100;
        int lastDigit = lastTwoDigits % 10;
        int secondLastDigit = lastTwoDigits / 10;
        return lastDigit + secondLastDigit;
    }
};

    Two C++ points to keep in mind:

    • Integer division truncates. If either operand is a double, the result is a double: 79 / 10 is 7, but 79 / 10.0 is 7.9. Keeping both sides as int is what makes this work.
    • The sign of %. In C++ the result of % takes the sign of the left operand, and division truncates toward zero. For -2379, % 100 gives -79 and the digit sum comes out negative. The statement promises a non-negative n, so this does not arise here.

    Complexity

    Time and space are both O(1): a fixed number of arithmetic operations, whatever the size of n.

    Edge cases and common mistakes

    • A zero in the tens place. For n = 105, n % 100 is 5, so lastTwoDigits / 10 is 0 and the sum is 5. That is right: the last two digits are 0 and 5.
    • Exactly two digits. For n = 99, n % 100 is 99 and the answer is 18. For n = 10 it is 1.
    • Multiples of 100. For n = 100 the last two digits are 0 and 0, and the answer is 0.
    • Dividing instead of taking the remainder. n / 100 removes the last two digits and keeps the rest, the opposite of what is needed.
    • Converting to a string. It works, but it is slower and needs extra code to turn characters back into digits. The arithmetic version is shorter.

    Testing

    • 2379 gives 16.
    • 10 gives 1.
    • 99 gives 18.
    • 100 gives 0.
    • 105 gives 5.
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