Print the second largest number on the list - dart

Question

I need to print the second largest number on the list, the output from the below code is all elements in the list except the first and the last one.
What is the mistake?
void main () {
List a = [9,6,4,10,13,2,3,5];
a.sort;
for(int x in a){
for (int max in a){
for (int second_last in a){
if (x > max) {
second_last = max;
max = x;
} else if (x > second_last && x != max) {
second_last = x;
print(second_last);
}
}
}
}
}

Answer 1

There are a few things wrong with your code:
You're not actually sorting the list. The sort method returns a new sorted list, it doesn't sort the existing list. So you need to do:
a = a.sort;
You're iterating over the list 3 times when you only need to iterate over it once.
You're not keeping track of the second largest number, you're just printing it out as you find it.
You're not checking for duplicates. If there are duplicate numbers in the list, your code will print them out multiple times.
Here's a corrected pseudo-code
void main() {
List a = [9, 6, 4, 10, 13, 2, 3, 5];
a.sort;
int max = a[0];
int second_last = a[0];
for (int x in a) {
if (x > max) {
second_last = max;
max = x;
} else if (x > second_last && x != max) {
second_last = x;
}
}
print(second_last);
}

Answer 2

I need to print the second largest number on the list
Sort the array (desc).
Access the second element.
List a = [9, 6, 4, 10, 13, 2, 3, 5];
a.sort((a, z) => z - a);
// final b = a.toList()..sort(...); if you do not want to modify the original array.
print(a.take(2)); // [13, 10]
print(a.take(2)[1]); // [10]
print(a.take(2).skip(1)); // [10]
print(a.skip(1).first); // [10]

Answer 3

You are missing () on sort. You can do
void main() {
List a = [9, 6, 4, 10, 13, 2, 3, 5];
a.sort();
print(a[a.length - 2]); //get second large item
List b = [9, 6, 4, 10, 13, 2, 3, 5];
//or like revese sort
b.sort((a, b) => b.compareTo(a));
print(b[1]);
}

Answer 4

If you find a solution that is without using the sort function, and also works with all possible conditions,
so you can use this solution as well,
void main(){
var list =[9,6,4,10,13,2,3,5,13];
var secondLargest = findSecLargest(list);
print(secondLargest);
}
findSecLargest(List<int> list) {
var secLargest =-1;
var largest=list[0];
for(int i = 1 ; i < list.length ; i ++){
if(largest<list[i]){
secLargest = largest;
largest = list[i];
}else if(secLargest<list[i] && list[i]!=largest){
secLargest = list[i];
}
}
return secLargest;
}