CareerCup

Correct me if I am wrong . What if we try to develop an algorithm similar to quick sort ?
1. Set two pointers , one at index(0) as i and other at index(n) as j.
2. If a(i) > a(j) , Increment i else decrement j
3. Stop when i=j
4. No assignments needed right ?

public void FinMin(int[] array)
{
int minVal = array[0];
int cnt = 0;
for(int i=1; i<array.length; i++)
{
if(array[i] < min)
{
min = array[i];
cnt++;
}
}
Console.WriteLine("Min Value: {0}, number of assignments: {1}", minVal, cnt);

}

I would assume that this is the solution they get most often, is there a better way? if we were first to sort the array via mergeSort, would we get a better efficiency?

The simplest(yet efficient) solution for the above problem would be the one that most guys are posting here :-

int min(int nArray[], int nLength)
//
int nMin <- infinite
for int i <- 0 to nLength - 1
do
- if nArray[i] < nMin
then
- - nMin <- nArray[i]

return nMin


Time Complexity :- O(n)
Assignment operations :-
Worst case :- n, where n is the length of array (Array sorted in decreasing order)
Best case :- 1, Only for the first time (Array sorted in increasing order)

#include<stdio.h>
int main()
{
	int a[20],small,n,i;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	scanf("%d",&a[i]);
	small=a[0];
	for(i=1;i<n;i++)
	{
		if(a[i]<small)
		{
			small=a[i];
		}
	}
	printf("smallest=%d",small);
}
as the array is unsorted,so we have to go for linear search
worst case=n assignments within the loop
best case=1 assignment within the loop