How do you find the second largest element in C?

How do you find the second largest element in C?

Logic to find second largest element Declare two variables max1 and max2 to store first and second largest elements. Store minimum integer value in both i.e. max1 = max2 = INT_MIN . Iterate though all array elements, run a loop from 0 to size – 1 . Loop structure should look like for(i=0; i

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

The program output is also shown below.

1. /*
2. * C program to read elements into an array and find the.
3. * largest two elements in a given array.
4. #include
5. int main ()
6. {
7. int n = 0, i = 0, largest1 = 0, largest2 = 0, temp = 0;
8. printf (“Enter the size of the array\n”);

How can get second highest salary in SQL?

How To Find Second Highest Salary Using a Sub-Query

1. SELECT TOP 1 SALARY.
2. FROM (
3. SELECT DISTINCT TOP 2 SALARY.
4. FROM tbl_Employees.
5. ORDER BY SALARY DESC.
6. ) RESULT.
7. ORDER BY SALARY.

How do I find the second highest value in SQL?

SELECT MAX (column_name) FROM table_name WHERE column_name NOT IN (SELECT Max (column_name) FROM table_name); First we selected the max from that column in the table then we searched for the max value again in that column with excluding the max value which has already been found, so it results in the 2nd maximum value.

How to find the second largest number in an array?

We can find the second largest number in an array in java by sorting the array and returning the 2nd largest number. Let’s see the full example to find the second largest number in java array. public class SecondLargestInArrayExample {. public static int getSecondLargest (int[] a, int total) {. int temp;

How to find the second minimum in an array?

Idea to get second minimum is that it will be beaten by minimum candidate in one of previous round. So, we need to find minimum in potential candidates (beaten by minimum). So, no. of comparison to find minimum using tournament tree is n-1 and for second minimum is log n -1 sums up = n + ceil (log n) – 2

Do you need log n comparisons to find the second greatest number?

Where the X’s are your numbers , then you require n-1 comparisons to find the greatest. Then the second greatest could have only been beaten by the greatest number. So it is one of the log n numbers that greatest has beaten. Therefore you need log n comparisons to find second greatest, overall, approx n+log n instead of approx 2n.

How do you sort an array in Java?

You can also sort the elements of the given array using the sort method of the java.util.Arrays class then, print the second element from the end of the array.