digitSumDifference

with Carmen Salas • 2024/11/04

Code Challenge

Using methods...

/*
## PEDAC
**Problem**
Parse a number into individual digits and output the difference between the sum of all evens and all odds.
Number -> Number indicating the difference (as described above)

**Examples/Edge Cases:** 
2134 -> 6 (evens) - 4 (odds) == 2

**Data Structure**
- Number
- String
- Array

**Algorithm** 
1. Convert number into string.
2. Split string into array.
3. Convert each item into a number.
4. Look at each number in the array: if even, add value to sum, else subtract value from sum.
*/
function solution(n) {
  // method approach
  return [...String(n)] // step 1 and 2
           .reduce((sum, val) => // step 3 and 4
             // loose equality for type conversion
             (val%2 == 0) ? sum+Number(val) : sum-Number(val), 0);
}

Simple no-method approach

/*
## PEDAC
**Problem**
- Given a Number, look at each digit individually and sum all the evens and all the odds. The result will be the difference between sumEvens-sumOdds.
- We can skip the need for two sums because we notice that:
    all evens are additive, while
    all odds are subtractive.
- Meaning, we can just add all evens and subtract all odds to get the sum.

**Examples/Edge Cases:** 
input: 1234 => 1 , 2 , 3 , 4 => sum(2,4) - sum(1,3) => 2
               ^   ^   ^   ^    ^          ^           ^
               look at each     evens      odds        result
               
**Data Structures**
- Number
- While loops
- Conditional (ternary in this case)

**Algorithm** 
While n > 0, for each iteration:
  - do n%10 (this takes the value of the `ones` place)
  - if n%10 is even, ADD it to sum
                odd, SUBTRACT it from sum
  - do n // 10 (integer division)
Return the resulting sum.

Walkthrough for input = 1234
----------------------------
While (n > 0) loop:
  1st loop: sum = 0
    1234 % 10 => 4
    4 is even, sum += 4   
    1234 / 10 => Math.floor(123.4) => 123
  2nd loop: sum = 4
    123 % 10 => 3
    3 is odd, sum -= 3
    123 / 10 => Math.floor(12.3) => 12
  3rd loop: sum = 1
    12 % 10 => 2
    2 is odd, sum += 2
    12 / 10 => Math.floor(1.2) => 1
  4th loop: sum = 3
    1 % 10 => 1
    1 is odd, sum -= 1
    1 / 10 => Math.floor(0.1) => 0
Exit loop with sum = 2.
*/

function solution(n) {
  let sum = 0; // initialize the sum
  while (n > 0) {
    const rem = n%10; // remainder by 10 is effectively the number in the `ones` place
    (rem%2 === 0) ? sum += rem : sum -= rem; // add if even, subtract if odd
    n = Math.floor(n/10); // integer division, to remove the `ones` number we just processed
  }
  return sum; // return the sum
}