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I understand what you are doing here. But suppose you find j as the smallest starting point, your program makes sure there is enough gas to finish the remaining routes from j to N-1, since the route is a circle, but how can you guarantee your gas can cover the route from 0 to j-1 ?

Hey, Thanks for pointing it out!!
I think this serves it,

public class Solution {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int N =in.nextInt();
int sum = 0;
int position = 0;
int residue = 0;
int a,b;
int preA=0;
int preB=0;
for(int i=0;i<N;i++){
a = in.nextInt();
b = in.nextInt();
sum =sum +(a-b);
if(sum<0){
residue+=sum;
sum=0;
position = i+1;
preA=a;
preB=b;
}
if(i==N-1){
if(sum+residue>=0){
System.out.println(position);
}
else{
if(position<N-1){
i=position+1;
position=i;
sum=0;
residue+=preA+preB;
}
else{
System.out.println(-1);
}
}
}
}
}

Hi maximshen. There is no point going back to 0 to j-1 because you already know they can't be the starting point as they already have failed. However, it will be useful to know if they cant have a starting point since there is not enough fuel to complete the circle no matter where you start. But I believe all the test cases have solutions. So, there is really no need to bother about 0 to j-1.

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I understand what you are doing here. But suppose you find j as the smallest starting point, your program makes sure there is enough gas to finish the remaining routes from j to N-1, since the route is a circle, but how can you guarantee your gas can cover the route from 0 to j-1 ?

Hey, Thanks for pointing it out!! I think this serves it,

public class Solution {

}

Hey Vinay, can u try for this test case?

7

5 2

1 2

3 3

2 6

6 2

2 2

3 6

Its not mentioned. We can assume there is always a solution for this problem

Hi maximshen. There is no point going back to 0 to j-1 because you already know they can't be the starting point as they already have failed. However, it will be useful to know if they cant have a starting point since there is not enough fuel to complete the circle no matter where you start. But I believe all the test cases have solutions. So, there is really no need to bother about 0 to j-1.