### Thread: Algorithm to find second largest element in an array

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#### Algorithm to find second largest element in an array

Hello friends,
I need to find second largest number in an array of n elements. I have tried a c code for that...

#include <stdio.h>
int main(void) {
int pri, sec, i, v;
int arr[] = {4,10,3,8,6,7,2,7,9,2,0};
pri = sec = 0;
for (i = 0; arr[i]; ++i) {
v = arr[i];
if (v > pri) sec = pri, pri = v;
if (v > sec && v < pri) sec = v;
}
printf("pri is %d, sec is %d\n", pri, sec);
return 0;
}

but i want to know how to find complexity of the algorithm in worst case. and is there any algorithm whose complexity is 2n-3 which cud do the same task?

thnx,
raniiii
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If by "2n-3" you are referring to the number of comparisons, then yes: go through and find the largest element (n-1 comparisons), get it out of the way, and then go through and find the largest remaining element (n-2 comparisons).

As for analyzing worst-case complexity in an algorithm like this: the loop body does 3 comparisons and some number of assignments. In the worst case, you will do 3 assignments on each iteration (if the array is sorted). You do all of that n times, so you do 3n comparisons and 3n assignments in the worst case in the loop part of the code. Add to that the overhead for setting things up, initializing structures, etc.
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#### thnx

thnx for the reply but in a for loop u consider 0(n) complexity no matter how many comparisons u make rite ?
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Hi rani,
u r correct in a loop what does matters is how many times it runs and not the number of comparisons inside it.

so for your algorithm the complexity is O(n) as the loop is running for n times.

now if u want to have an algorithm giving you O(2n -3). then we could think of bubble short to find the second largest element in the array.

ok the complexity of bubble sort is O(n^2) to short complete array of elements.
as we need the second largest number, so we can run the loop only 2 times.
in the first time loop will run for (n-1) times and second time it will run for (n -2).
So all total we have (n-1) + (n-2) = 2n -3.

bye,
Sandeep Patra
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public void FindSec()
{
int[] arr = { 1,2,0,-1,8,26,1,26};
int high, secHigh, temp, i;

high = secHigh = arr[0];

for (i = 1; i < arr.Length; ++i)
{
temp = arr[i];
if(temp > high)
{
secHigh = high;
high = temp;
}
else if(temp >secHigh && temp != high)
{
secHigh = temp;
}
}

Console.WriteLine(secHigh);
}
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Originally Posted by sathish_leo
public void FindSec()
{
int[] arr = { 1,2,0,-1,8,26,1,26};
int high, secHigh, temp, i;

high = secHigh = arr[0];

for (i = 1; i < arr.Length; ++i)
{
temp = arr[i];
if(temp > high)
{
secHigh = high;
high = temp;
}
else if(temp >secHigh && temp != high)
{
secHigh = temp;
}
}

Console.WriteLine(secHigh);
}

I think this code will not work , when 1st element of the array will greatest elemnt .....
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#### This method is incorrect,it will not run at these type of arrays exp: a={10,3,4}

Error
This method is incorrect,it will not run at these type of arrays exp: a={10,3,4}

Originally Posted by sathish_leo
public void FindSec()
{
int[] arr = { 1,2,0,-1,8,26,1,26};
int high, secHigh, temp, i;

high = secHigh = arr[0];

for (i = 1; i < arr.Length; ++i)
{
temp = arr[i];
if(temp > high)
{
secHigh = high;
high = temp;
}
else if(temp >secHigh && temp != high)
{
secHigh = temp;
}
}

Console.WriteLine(secHigh);
}