How do you find the third smallest number in an array?

Find Third Smallest elements in a given array

1. Take three variables; let’s call them first, second and third and mark them as +∞.
2. Iterate through the array and for each element (let’s call it current),
3. At the end print the third, it will third-smallest element in the array.

How do you find the 3 Max elements in an array?

a) Let current array element be x. b) If (x > first) { // This order of assignment is important third = second second = first first = x } c) Else if (x > second) { third = second second = x } d) Else if (x > third) { third = x } 3) Print first, second and third. Below is the implementation of the above algorithm.

How many comparisons are required to find the maximum and minimum?

Lower bounds for the MIN and MAX Claim: Every comparison-based algorithm for finding both the minimum and the maximum of n elements requires at least (3n/2)-2 comparisons.

How do you find the third largest number in Python?

Find the third largest number in a list in python

1. +1. l=[5,8,2,9,4,10,7] print(“3rd largest in”,l,”is”,sorted(l)[-3]) 28th May 2018, 3:28 AM.
2. +13. Convert the list into a set.
3. +9. 2nd variant without using sort(): https://code.sololearn.com/cxKFjv7q5PQ8/#py.
4. +8. #If you are not allowed to use sort, try this.

How do you find the third largest number in a list in Python?

The commented numbers in the above program denote the step numbers below: First, create one number list holding few random numbers. We will find the third largest among these numbers. Create three variables to hold the largest number,second largest number, and third largest number of the list.

Now in order to get the k’th largest element, we need to add the frequencies till it becomes greater than or equal to 3. It is clear from the above that the frequency of 0 + frequency of 6 will become equal to 3 so the third smallest number in the array will be 6.

How do you reduce the number of comparisons in an array?

You need to decrease the number of comparisons as much as you can. 1. Searching linearly: Increment the loop by 1 We initialize both minimum and maximum element to the first element and then traverse the array, comparing each element and update minimum and maximum whenever necessary.

How many comparisons are there in a pair of arrays?

For each pair, there are a total of three comparisons, first among the elements of the pair and the other two with min and max. Critical ideas to think! Why min and max are initialized differently for even and odd sized arrays?

What to do if array size is odd or even?

If the array size is odd, we initialize the first element as both min and max, and if it’s even, we compare the first two elements and initialize min and max accordingly. Create max and min variables. 3. Traverse the array in pairs