How many comparisons is needed to find smallest number?

The key observation is that every element except the winner must lose at least one match. Hence, n − 1 comparisons are necessary to determine the minimum, and the algorithm MINIMUM is optimal with respect to the number of comparisons performed.

What is the smallest number of comparisons needed to find the minimum and maximum?

The formula for the minimum number of comparisons to find the minimum and the maximum elements is ( 3n / 2 ) – 2 where n is the size of the array.

How many minimum numbers are possible to find the second smallest element in an unsorted array?

In short, the Minimum Comparisons to find Second Largest Element or Second Smallest Element is N + logN – 2 comparisons . Hence, if there are 1024 elements, then we need at least 1024 + 10 – 2 = 1032 comparisons.

How do you find the second minimum of an array?

Java program to find the 2nd smallest number in an array

1. Compare the first two elements of the array.
2. If the first element is greater than the second swap them.
3. Then, compare 2nd and 3rd elements if the second element is greater than the 3rd swap them.
4. Repeat this till the end of the array.

How do you find second minimum?

Find 2nd Smallest Number in Array using Arrays

1. import java.util.*;
2. public class SecondSmallestInArrayExample1{
3. public static int getSecondSmallest(int[] a, int total){
4. Arrays.sort(a);
5. return a[1];
6. }
7. public static void main(String args[]){
8. int a[]={1,2,5,6,3,2};

Which is the smallest element in periodic table?

helium
As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.

How do you find the second lowest number?

Method 2 to find the second smallest element in an array

1. Declare an array and input the array elements.
2. Find the smallest element (first_smallest) in the array in the first traversal.
3. Find the smallest element (second_smallest) by skipping the first_smallest element.
4. Display second_smallest.

How do you find the second largest and smallest number in an array?

Algorithm

1. Step 1 − Declare and read the number of elements.
2. Step 2 − Declare and read the array size at runtime.
3. Step 3 − Input the array elements.
4. Step 4 − Arrange numbers in descending order.
5. Step 5 − Then, find the second largest and second smallest numbers by using an index.

What is the least number of comparisons which are required to find the minimum and maximum of 100 numbers?

The minimum number of comparisons required to find the minimum and the maximum of 100 numbers is ______________. Therefore, we need 3 comparisons for each 2 elements, so total number of required comparisons will be (3n)/2 – 2, because we do not need to update min or max in the very first step.

How many comparisons does it take to find the smallest number?

It will take 31 comparisons to find the smallest number as each comparison eliminates one possibility. Once this has been complete, make a list of the numbers the smallest number has been compared to. There should be 5. It will take 4 comparisons to find the smallest of these 5 numbers, for a total of 35 comparisons.

How do you find the smallest number in a list?

Take your list of 32 numbers and try to find the smallest number. Compare the first and second numbers, the third and fourth, etc. until you reduce the list by half and repeat. It will take 31 comparisons to find the smallest number as each comparison eliminates one possibility.

How many comparisons are needed to find the maximal element?

There should be 5. It will take 4 comparisons to find the smallest of these 5 numbers, for a total of 35 comparisons. 35 comparisons are optimal. Pair off the numbers and test them against each other. Take the winners, the winners’ winners, and so forth to find the maximal element; this takes 31 comparisons in all.

What is the total number of the rest of 30 numbers?

Assign largest and second largest to them based on comparison. Now compare rest 30 numbers with two of them and assign largest and second largest based on comparison. So for rest of 30 numbers we have to do 2 comparison for each number . this answer is 30*2 + 1 = 61.