Question

Design and implement a Θ(n) algorithm that will simultaneously find the largest and second-largest elements (integers) in an array. Note that the second largest element should be
distinctly smaller than the largest element. You should also adhere to the following restrictions.
(1) You should traverse through the array ONLY ONCE.
(2) The array could have both positive and negative elements. So, you should NOT initialize any
temporary variables to very small negative values as part of your algorithm.
(3) You should NOT make any assumption regarding the values of the elements and their
distribution in the array.

Show the working of the algorithm with the following two types of example arrays:
(a) An example array of 10 integers that need to have the largest integer appearing at least twice.
(b) An example array with the largest integer and second-largest integer as respectively the first

Please put comments in the code (C++) and prove time complexity is theta(n)

Answer 1

#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
cin>>arr[i];
int lar, secLar; //variables to store the largest and second largest elements
lar=arr[0]; //initialising lar with the first element of array
int i=1;
while(arr[i]==lar){
i++; //getting past all the elements equal to lar
}   
if(arr[i]<lar) //getting the first element which is different from lar and comparing it to lar to get
secLar=arr[i]; //initial values for both lar and secLar
else{
secLar=lar;
lar=arr[i];
}
i++;
while(i<n){ //checking the rest of array for any modification in lar and secLar
if(arr[i]==lar || arr[i]==secLar){

} //do nothing if the next element is already lar or second lar
else{
if(arr[i]>lar){
secLar=lar; // if the next element is greater than lar , then secLar would get updated to lar and
lar=arr[i]; // lar would get updated to that element
}
else if(arr[i]>secLar){ //if the next element is smaller than lar, but greater than secLar , secLar would get
secLar=arr[i]; // updated to that element
}
else{ //do nothing if the element is smaller than both lar and secLar

}
}
i++;
}
cout<<"The largest and second largest elements of the array are: "<<lar<<" and "<<secLar<<" respectively"<<endl;
return 0;
}

for example, in case of array: 7 -2 4 6 7 3 2 0 -9 5

the result that comes is : lar: 7 and secLar: 6

the algorithm has the time complexity of o(n) and does the whole work in one scan.