Third Best Price

Time limit: 2 s

Memory limit: 64 MB

 

Description

Steve would like to buy a new car. He isn’t wealthy, so he would prefer a reasonably cheap car. The only problem is that the quality of the cheapest cars is… let’s say questionable.

Thus Steve decided to make a list of car prices and to buy a car with the third lowest price.

You will be given a list of integers prices. The same price may occur multiple times in prices, but it should count only once in the ordering of available prices. See Example 2 for further clarification.

Find the third lowest price in this list. If there are less than three different car prices in prices, you should output -1.

Input Format

The first line is an integer (1 <= <= 50), the number of prices.

The next line contains of N space-separated integers prices (1 <= prices[i] <= 1000).

Output Format

An integer, the third lowest price, or -1 if there are less than three different car prices.

 

Sample Input 1

9
10 40 50 20 70 80 30 90 60

Sample Output 1

30

 

Sample Input 2

10
10 10 10 10 20 20 30 30 40 40

Sample Output 2

30

Note

The lowest price is 10, the second lowest is 20 and the third lowest is 30.

 

Sample Input 3

1
10

Sample Output 3

-1

 

Sample Input 4

5
80 90 80 90 80

Sample Output 4

-1

Code Solution

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int N,x,i,j,a = 0, pos = 0, temp;
    int diff = 1, result = 0;
    scanf("%d",&N);
    if(N>=1 && N <= 50){
        long int price[N];
            while (a < N) {
                if (scanf("%d", &x) != 1) break;
                price[a] = x;
                 if (!(price[a] >=1 &&price[a] <=1000 )){
                        return 0;
                 }
                a++;
            }
        for(i=0;i<N-1;i++){
            pos=i;
            for(j=i+1;j<N;j++){
                if(price[j]<price[pos]){
                    pos=j;
                }
            }
            if (pos!=i){
                temp=price[pos];
                price[pos]=price[i];
                price[i]=temp;
            }
        }
        for(i=0;i<N-1;i++){
           if(price[i] != price[i+1]){
               diff++;
               if(diff == 3){
                    result = price[i+1];
               }
            }
        }
        if(diff<3){
            printf("-1\n");
        }
        else{

            printf("%d\n",result);
        }

    }
    return 0;
}

 

Programming Problem Set #Third Best Price
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