Third Biggest Number
April 13, 2018
Today’s exercise is for all the beginning programming students who read this blog:
Write a program to read ten numbers input by the user and write the third largest of those ten numbers.
Your task is to write the program described above. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
Alternate solution keeps track of the three largest values using an array:
#!perl use strict; use warnings; use feature qw(say); my @biggest3 = (0, 0, 0); while(my $n = readline(*DATA)) { chomp $n; foreach my $i (0 .. $#biggest3) { if ($n >= $biggest3[$i]) { my $tmp = $biggest3[$i]; $biggest3[$i] = $n; foreach my $k ($i + 1 .. $#biggest3) { ($biggest3[$k], $tmp) = ($tmp, $biggest3[$k]); } last; } } } say $biggest3[-1]; __DATA__ 14 23 102 97 29 4 19 85 53 111int third_biggest(int arr[], int size){ int third = -1; int second = -1; int biggest = -1; printf("size: %d\n", size); int i = 0; while (size > i){ if (arr[i] > biggest) { third = second; second = biggest; biggest = arr[i]; } else if (arr[i] > second){ third = second; second = arr[i]; } else if (arr[i] > third) third = arr[i]; i++; printf("third: %d\n", third); } return third; } int main() { int input={100, 7, 50, 1, 2, 3, 4, 5, 6, 40}; int size = sizeof(input) / sizeof(int); printf("third largest %d\n", third_biggest(input, size)); // 40 is the answer return 0; }Mumps solutions
zjsg ; q 3rd ; k arr f i=1:1:10 r !,"Enter number: ",x s arr(x)=1 w !!,"3rd number is ",$o(arr($o(arr($o(arr("")))))) qUSER>k
USER>d 3rd^zjsg
Enter number: 999999
Enter number: 10000
Enter number: 2345
Enter number: 781
Enter number: 56
Enter number: 1
Enter number: 2
Enter number: 99
Enter number: 678
Enter number: 3456
3rd number is 56
A Haskell version.
-- Given a series of integers on standard input first remove duplicates, then -- print the 3'rd biggest number of those remaining, if it exists. -- -- Why remove duplicates? Consider the list [1, 1, 2, 2, 3, 3]. The third -- biggest number in the list is 1, not 2 (the third element after sorting). import Data.List.Ordered (nubSortOn) import Data.Ord (Down(Down)) -- Just the n'th element of a list (starting at 1) or Nothing if n < 1 or if the -- list has fewer than n elements. nth :: Int -> [a] -> Maybe a nth n xs = if n < 1 then Nothing else go n xs where go _ [] = Nothing go 1 (y:_) = Just y go m (_:ys) = go (m-1) ys -- Just the n'th biggest element of the list after duplicates have been removed, -- or Nothing if there are fewer than n elements left. nthBiggest :: Ord a => Int -> [a] -> Maybe a nthBiggest n = nth n . nubSortOn Down main :: IO () main = do x <- nthBiggest 3 . map read . words <$> getContents print (x :: Maybe Int)Python 3
import random def third_biggest(seq): uniq = sorted(set(seq)) return uniq[-3:-2] if len(uniq) > 2 else None if __name__ == "__main__": data = [random.randint(1, 10**6) for k in range(10)] print (data) print (third_biggest(data))Sample output:
[76631, 719928, 63284, 348692, 521266, 687164, 663331, 845208, 466387, 33721]
[687164]
Ruby. Assumes clean input of positive integers. This is a trick question right?
puts "#{10.times.map { print "Enter a number: "; Integer(gets) }.uniq.sort[-3] || 'Nop'}"My previous entry is invalid Ruby. This one does the trick.
puts 10.times.map { print "Enter a number: "; Integer(gets) }.uniq.sort[-3] || 'Nop'(1..10 | % { (Read-Host -Prompt “Enter a number”) -as [int] } | Sort -Unique)[-3]
A translation of my Ruby version in Go.
package main import ( "fmt" "sort" ) func main() { numbers := make([]int, 10) // Get numbers from user for i := range numbers { fmt.Print("Enter a number: ") fmt.Scan(&numbers[i]) } // Remove duplicates uniqNumbers := []int{} m := make(map[int]bool) for _, x := range numbers { m[x] = true } for k, _ := range m { uniqNumbers = append(uniqNumbers, k) } if len(uniqNumbers) < 3 { fmt.Println("Nop") return } // Sort in descending order sort.Sort(sort.Reverse(sort.IntSlice(numbers))) fmt.Println("Third lagest is:", numbers[2]) }A classic interview answer is to use a min-heap of fixed-size:
import java.util.PriorityQueue; import java.util.Scanner; public class ThirdBiggest { public static void main(String[] args) { PriorityQueue<Integer> numbers = new PriorityQueue<>(); System.out.println("Enter 10 numbers:"); Scanner sc = null; try { sc = new Scanner(System.in); for (int i = 0; i < 10; i++) { int num = sc.nextInt(); if (numbers.size() < 3) { numbers.add(num); } else if (num >= numbers.peek()){ numbers.remove(); numbers.add(num); } } } finally { if (sc != null) { sc.close(); } } System.out.println("The third biggest is " + numbers.peek()); } }Kotlin with a recursion approach. Give the first parameter
fun main(args: Array) {
var myList = mutableListOf(1,65,89,45,32,75,11,32,82,11)
println(recThirdBig(myList,null,null,null))
}
tailrec fun recThirdBig(testList: List, oneBig: Int?, twoBig: Int?, threeBig: Int?): Int? {
if (testList.isEmpty()) return threeBig
if (oneBig == null || testList[0] > oneBig) return recThirdBig(testList.drop(1), testList[0], oneBig, twoBig)
if (twoBig == null || testList[0] > twoBig) return recThirdBig(testList.drop(1), oneBig, testList[0], twoBig)
if (threeBig == null || testList[0] > threeBig) return recThirdBig(testList.drop(1), oneBig, twoBig, testList[0])
return recThirdBig(testList.drop(1), oneBig, twoBig, threeBig)
}
This is similar to a prior problem.
Here’s a solution in C.
#include <stdbool.h> #include <stdio.h> #include <stdlib.h> #define N_ENTERED 10 #define N_RETAINED 3 int main(void) { int array[N_RETAINED]; for (size_t i = 0; i < N_ENTERED; ++i) { int x; while (true) { printf("Enter Number %zu: ", i+1); int n_assigned = scanf("%d", &x); size_t n_trailing = 0; while (getchar() != '\n') ++n_trailing; if (n_assigned == 1 && n_trailing == 0) break; } bool reorder = true; if (i < N_RETAINED) { array[i] = x; } else if (x > array[0]) { array[0] = x; } else { reorder = false; } if (reorder) { for (size_t j = 1; j < N_RETAINED; ++j) { if (j > i) break; if (array[j-1] > array[j]) { int tmp = array[j-1]; array[j-1] = array[j]; array[j] = tmp; } } } } printf("Biggest Number %d: %d\n", N_RETAINED, array[0]); return EXIT_SUCCESS; }Example: