CODEWITHSUNDEEP
javaarrays
Harshad Kapoor
Harshad Kapoor

Reputation: 17

Return the sum of the numbers in the array, returning 0 for an empty array. Except the number 13 is very unlucky

I am working on practice coding problems and i came across this task.

Return the sum of the numbers in the array, returning 0 for an empty array. Except the number 13 is very unlucky. so it does not count and numbers that come immediately after a 13 also do not count

I came up with the right solution but i am not sure about one thing.

public int sum13(int[] nums) 
{
    int sum = 0;
    for(int i = 0; i < nums.length; i++)
    {
        if(nums[i] == 13)
            i++;
        else
            sum += nums[i];
    }
    return sum;
}

In one of the test cases output is 4, sum13([1, 2, 13, 2, 1, 13]) → 4 .

My question is ,in my "if" statement i incremented the counter by 1. Why did it not add the next element to the sum? Why did it skip the next element even though I'm only incrementing by 1 ?

Upvotes: 1

Views: 2221

Answers (3)

ritesh_ratn
ritesh_ratn

Reputation: 157

This code works fine:-

public int sum13(int[] nums) {
  int sum=0;
  for(int i=0;i<=nums.length-1;i++){
    if(nums[i]!=13){
      sum+=nums[i];
      if(i>0 && nums[i-1] == 13)
        sum -= nums[i];

    }
  }
  return sum;
}

Upvotes: 0

GhostCat
GhostCat

Reputation: 140523

Your problem is here:

if(nums[i] == 13)
  i++;

There is no point in increasing the index; that just makes you increase your loop counter twice in one iteration!

So, just go with

if(nums[i] != 13) {
  sum += nums[i];
}

instead (no else here; and: better always use { braces }; even when it is just a one-liner!)

And as you actually do not need that index at all, you could use the for-each loop style:

for (int num : nums ) {
  if (num != ...

Even less chance for messing up your index!

Upvotes: 2

djechlin
djechlin

Reputation: 60788

You are skipping 13 and the number after. You are incrementing i twice.

Upvotes: 1

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