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int i = 0;
int j = array.length - 1;
int sum = 0;
while (i != j) {
if(sum >= 0) {
sum -= array[j];
j--;
} else {
sum +=array[i];
i++;
}
}
return sum == 0 ? true : false;
Yep, I realized that too! I was looking to see if someone got that and you stated it more succinctly than I would have. It's much faster than the binary search method I originally thought of when I first read the problem.
Sherlock and Array
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From my point of view everithing pretty obvious imagine that one element it's a middle of array sum on left side = 0 as well as on right side = 0.
for the last line, you don't need to use ternary operator if all you want is to return a boolean
will be suffice
Yep, I realized that too! I was looking to see if someone got that and you stated it more succinctly than I would have. It's much faster than the binary search method I originally thought of when I first read the problem.
Great and Simple Solution!
fantastic algorithm
Awesome , Salute..
This fails following: n=4: 0 0 2 0
It passes all test cases, including n=4: 0 0 2 0 !!!! Just copy and paste the whole code, and do not forget to import java.util.Scanner !!!
You are right This is not also not work for 0 2 0 0
it fails for test input 0,0,2,0.
CHEERS