How do I lessen the memory usage of my solution to LeetCode 495: Teemo Attacking

Question

How to make changes in my code so that it can take the least memory possible.? This is a leetcode program , I have tested all the testcases and it has passed, I tried to reduce the memory usage, but by far now, I have not been able to understand.

I want to look into this so that I can understand the usage of memory spaces better. Any input would be helpful. Thank you very much. I will also post the link to the question, but since the logic is already clear to me, The only question is that is this a good solution or memory heavy?

   class Solution {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        
        int i =0,old_goalpost=0,new_goalpost=0 ,diff=0;        
        int numberOfSeconds = duration;
       
        if (timeSeries.length >0)
            old_goalpost = timeSeries[0]+duration-1;
        else
            numberOfSeconds=0;
        
        if (timeSeries.length >0)
        for (i=1;i<timeSeries.length;i++)            
        {
            new_goalpost =  timeSeries[i]+duration-1; 
            
            diff = (new_goalpost - old_goalpost)>duration?duration:((new_goalpost - old_goalpost));
            
            
            numberOfSeconds = numberOfSeconds +diff;
            old_goalpost = new_goalpost;
            
        }
        return numberOfSeconds;
        
    }
}

the question is this

And it has passed all the test cases.

If you want, this is the link to the question https://leetcode.com/problems/teemo-attacking/

this code is 99.97% faster but memory heavy

class Solution {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        
        int i =0,old_goalpost=0,new_goalpost=0 ,diff=0;        
        int numberOfSeconds = duration;
       
        if (timeSeries.length >0)
            old_goalpost = timeSeries[0]+duration-1;
        else
            numberOfSeconds=0;
        
        if (timeSeries.length >0)
        for (i=1;i<timeSeries.length;i++)            
        {
            new_goalpost =  timeSeries[i]+duration-1;                 
            diff = (new_goalpost - old_goalpost)>duration?duration:((new_goalpost - old_goalpost)); 
            numberOfSeconds = numberOfSeconds +diff;
            old_goalpost = new_goalpost;
            
        }
        return numberOfSeconds;
        
    }
}

Answer 1

Disclaimer: Not a Code Reviewer

• Just tested your code, it has a 1ms runtime with 99.97% faster than other solutions.

  

• Don't pay too much attention to these runtime/memory data provided by LeetCode especially when:

  • it comes to memory; or
  • test cases are too limited (such as Teemo) because their benchmarkings are not so accurate.

• Maybe instead we'd format the code, just a bit making things based on Java's conventions:

public class Solution {
    public static final int findPoisonedDuration(
        final int[] timeSeries,
        final int duration
    ) {

        int i = 0;
        int oldGoalpost = 0;
        int newGoalpost = 0;
        int diff = 0;
        int numberOfSeconds = duration;

        if (timeSeries.length > 0) {
            oldGoalpost = timeSeries[0] + duration - 1;

        } else {
            numberOfSeconds = 0;
        }

        if (timeSeries.length > 0)
            for (i = 1; i < timeSeries.length; i++) {
                newGoalpost =  timeSeries[i] + duration - 1;

                diff = (newGoalpost - oldGoalpost) > duration ? duration : ((newGoalpost - oldGoalpost));


                numberOfSeconds = numberOfSeconds + diff;
                oldGoalpost = newGoalpost;

            }

        return numberOfSeconds;

    }
}

---

• Your solution has an order of N runtime with constant memory O(1).
• Algorithmically speaking, which is a key point on solving LeetCode questions, I don't think much can be done to make it any further efficient, I might be wrong though.

public class Solution {
    public static final int findPoisonedDuration(
        final int[] series,
        final int duration
    ) {
        if (series.length == 0 || duration == 0) {
            return 0;
        }

        int totalTime = duration;

        for (int i = 1; i < series.length; i++) {
            totalTime += Math.min(series[i] - series[i - 1], duration);
        }

        return totalTime;
    }
}