There's a logical error. Please help me solve it

Question

I am trying to solve it. But some of tests cases just don't work right.
Is there anything wrong in my code?! Please tell me.

HERE IS THE QUESTION:

The snail climbs up 7 feet each day and slips back 2 feet each night.
How many days will it take the snail to get out of a well with the given depth?

Sample Input:
31

Sample Output:
6

Explanation: Let's break down the distance the snail covers each day:
Day 1: 7-2=5
Day 2: 5+7-2=10
Day 3: 10+7-2=15
Day 4: 15+7-2=20
Day 5: 20+7-2=25
Day 6: 25+7=32
So, on Day 6 the snail will reach 32 feet and get out of the well at day, without slipping back that night.

AND HERE IS MY ATTEMPT:

function main() {
var depth = parseInt(readLine(), 10);
//your code goes here
var total = 0
while (depth > 0) {
depth = (depth-7)+2
total ++
}
console.log(total)
}

Answer 1

The issue with your current implementation is that it doesn't account for the case where the snail can climb out of the well on the last day, without slipping back down.

For example, if the depth of the well is exactly 7 feet, the snail will climb out of the well on the first day and the loop won't even run.

To fix this issue, you can modify your while loop condition to check if the depth of the well is greater than or equal to 7 feet, and then add a separate condition to check if the snail can climb out on the last day without slipping back:
function main() {
var depth = parseInt(readLine(), 10);
// your code goes here
var total = 0;
while (depth >= 7) {
depth -= 5; // each day, the snail climbs up 7 feet and slips back 2 feet, for a net gain of 5 feet
total++;
}
if (depth > 0) { // if the snail can climb out on the last day without slipping back
total++;
}
console.log(total);
}

Answer 2

The snail 🐌 only slips down while it is not out of the well yet. Once depth == 0, you can stop (instead of adding 2).

Answer 3

Lisa,
How do I stop it?

Answer 4

You could check the depth before you add 2 and on add them, when the snail is still in the well.

Answer 5

In this modified implementation, the while loop continues as long as the depth of the well is greater than or equal to 7 feet, and on each iteration, the snail climbs up 7 feet and slips back 2 feet, resulting in a net gain of 5 feet each day. Once the depth of the well becomes less than 7 feet, we check if the snail can climb out on the last day without slipping back (i.e. the remaining depth is less than or equal to 7 feet). If it can, we increment the total number of days by 1.

With this modification, your implementation should now correctly handle all possible cases.

Answer 6

Sourav Chettri

Thanks a lot!! I got it!