Java > Logic-2 > loneSum (CodingBat Solution)

Problem:

Given 3 int values, a b c, return their sum. However, if one of the values is the same as another of the values, it does not count towards the sum.

loneSum(1, 2, 3) → 6
loneSum(3, 2, 3) → 2
loneSum(3, 3, 3) → 0


Solution:

public int loneSum(int a, int b, int c) {
  if ( a == b && b == c)
    return 0;
  if (a == b)
    return c;
  if (b == c)
    return a;
  if (a == c)
    return b;
  else
    return a + b + c;
}

13 comments:

  1. AnonymousDecember 17, 2017 at 5:36 AM

    ReplyDelete
  2. AnonymousJanuary 11, 2018 at 9:55 AM

    public int loneSum(int a, int b, int c) {
    if(a == b && b == c) {
    return 0;
    } else if(b == c) {
    return a;
    } else if(a == c) {
    return b;
    } else if(a == b) {
    return c;
    } else {
    return a + b + c;
    }
    }

    ReplyDelete
  3. AnonymousJanuary 11, 2018 at 9:57 AM

    https://codingbatanswers.wordpress.com/logic-2/

    ReplyDelete
  4. French_SoleNovember 5, 2018 at 9:29 AM

    return(a==b&&b==c)?0:(a==b)?c:(a==c)?b:(b==c)?a:a+b+c;

    ReplyDelete
  5. UnknownFebruary 22, 2019 at 1:35 PM

    /*int sum = 0;
    if (a !=b && b!=c && c!=a) {return sum = sum + a+b+c;}
    if (a == b && b == c && c == a) {return sum;}
    else if (a == b && b !=c) {return sum = sum + c;}
    else if (b == c && c !=a) {return sum = sum + a;}
    else if (c == a && b != c) {return sum = sum + b;}
    return sum;
    */

    ReplyDelete
  6. kendallSeptember 20, 2019 at 11:51 AM

    Trying to refactor for arbitrary number of args. Is there are way to initialize the array to include all args, however many?
    --------------------------------------
    int[] arr = {a,b,c}; //how to do this programatically?
    int sum = 0;
    for (int i = 0; i < arr.length; i++){
    boolean flag = false;
    for (int j=0; j<arr.length; j++){
    if(i != j){
    if (arr[i] == arr[j]){
    flag = true;
    }
    }
    }
    if (!flag) sum += arr[i];
    }

    return sum;

    ReplyDelete
    Replies
    1. AnonApril 10, 2021 at 3:13 PM

      i like this solution because it's dynamic

      Delete
    2. AnonMarch 18, 2022 at 12:11 AM

      Came across this problem and decided to solve it in a functional way.

      public int loneSum(int...numbers) {
          return Arrays.stream(numbers).map(n - > {
              return Arrays.stream(numbers).filter(el - > {
                  return el == n;
              }).count() > 1 ? 0 : n;
          }).sum();
      }

      Delete
    3. AnonMarch 18, 2022 at 12:16 AM

      For my previous reply I used a formatter, which decided to space out the - and > in the lambdas. Sorry about that.

      Delete
  7. AnonymousSeptember 18, 2020 at 3:01 AM

    public int loneSum(int a, int b, int c) {
    if(a == b && b == c){
    return 0;
    }
    if(b == c){
    return a;
    }
    if(c == a){
    return b;
    }
    if(a == b){
    return c;
    }
    return a + b + c;
    }

    ReplyDelete
  8. ScientisTJanuary 22, 2022 at 10:35 PM

    int sum=a+b+c;
    if (a==b&&b==c){
    return 0;
    }
    if (a==b){
    return c;
    }
    if (b==c){
    return a;
    }
    if (a==c){
    return b;
    }
    else
    return sum;

    ReplyDelete
  9. rrAugust 19, 2022 at 10:19 AM

    public int loneSum(int a, int b, int c) {

    int num=0;
    if(a==b && b==c)
    return 0;
    num =(a==b?c:b==c?a:a==c?b:a+b+c);
    return num;
    }

    ReplyDelete
  10. unknownJune 7, 2023 at 11:07 AM

    public int loneSum(int a, int b, int c) {
    int sum = 0;
    if (a == b && b == c) {
    sum = 0;
    } else if (a == b) {
    sum = c;
    } else if (b == c) {
    sum = a;
    } else if (a == c) {
    sum = b;
    } else {
    sum = a + b + c;
    }

    return sum;
    }

    ReplyDelete