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Check if two items can be selected from two different categories without exceeding price

  • Last Updated : 15 Jun, 2021

Given two arrays prices[], type[] and an integer S, the task is to check if two items can be selected from two different categories without excedding the total price S. Each element in the type[] array denotes the category of the ith element and each element in the prices[] array denotes the price of the ith element.

Examples: 

Input: S = 10, type[] = {0, 1, 1, 0}, prices[] = {3, 8, 6, 5} 
Output: Yes 
Explanation: 
Elements prices[0] and prices[2] can be selected 
Total prices = prices[0] + prices[2] = 3 + 6 = 9

Input: S = 15, type[] = {0, 1, 1, 0}, prices[] = {5, 7, 6, 5} 
Output: No 
Explanation: 
There is no possible solution such that total price is less than 15. 
 

Approach: The idea is to iterate choose every possible pair using two nested loops. For each pair check that if their category is different and their total price is less than S, If yes then there is a way to pick two items otherwise there are no such pairs items.



Below is the implementation of the above approach: 

C++14




// C++ implementation to check if
// two items can be selected from
// two different categories without
// exceeding the total price
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if
// two items can be selected from
// two different categories without
// exceeding the total price
string check(int S, int prices[],
             int type[], int n)
{
 
    // Loop to choose two different
    // pairs using two nested loops
    for (int j = 0; j < n; j++) {
        for (int k = j + 1; k < n; k++) {
 
            // Condition to check if the price
            // of these two elements is less than S
            if ((type[j] == 0 && type[k] == 1)
                || (type[j] == 1 && type[k] == 0)) {
 
                if (prices[j] + prices[k] <= S) {
                    return "Yes";
                }
            }
        }
    }
    return "No";
}
 
int main()
{
    int prices[] = { 3, 8, 6, 5 };
    int type[] = { 0, 1, 1, 0 };
    int S = 10;
    int n = 4;
 
    // Function Call
    cout << check(S, prices, type, n);
    return 0;
}

Java




// Java implementation to check if
// two items can be selected from
// two different categories without
// exceeding the total price
import java.util.*;
class GFG{
 
// Function to check if
// two items can be selected from
// two different categories without
// exceeding the total price
static String check(int S, int prices[],
                    int type[], int n)
{
 
    // Loop to choose two different
    // pairs using two nested loops
    for (int j = 0; j < n; j++)
    {
        for (int k = j + 1; k < n; k++)
        {
 
            // Condition to check if the price
            // of these two elements is less than S
            if ((type[j] == 0 && type[k] == 1) ||
                (type[j] == 1 && type[k] == 0))
            {
                if (prices[j] + prices[k] <= S)
                {
                    return "Yes";
                }
            }
        }
    }
    return "No";
}
 
// Driver Code
public static void main(String[] args)
{
    int prices[] = { 3, 8, 6, 5 };
    int type[] = { 0, 1, 1, 0 };
    int S = 10;
    int n = 4;
 
    // Function Call
    System.out.print(check(S, prices, type, n));
}
}
 
// This code is contributed by sapnasingh4991

Python3




# Python3 implementation to check if
# two items can be selected from
# two different categories without
# exceeding the total price
 
# Function to check if
# two items can be selected from
# two different categories without
# exceeding the total price
def check(S, prices, type1, n):
 
    # Loop to choose two different
    # pairs using two nested loops
    for j in range(0, n):
        for k in range(j + 1, n):
 
            # Condition to check if the price
            # of these two elements is less than S
            if ((type1[j] == 0 and type1[k] == 1) or
                (type1[j] == 1 and type1[k] == 0)):
 
                if (prices[j] + prices[k] <= S):
                    return "Yes";
         
    return "No";
 
# Driver Code
prices = [ 3, 8, 6, 5 ];
type1 = [ 0, 1, 1, 0 ];
S = 10;
n = 4;
 
# Function Call
print(check(S, prices, type1, n));
 
# This code is contributed by Code_Mech

C#




// C# implementation to check if
// two items can be selected from
// two different categories without
// exceeding the total price
using System;
 
class GFG{
 
// Function to check if two items 
// can be selected from two
// different categories without
// exceeding the total price
static String check(int S, int []prices,
                    int []type, int n)
{
 
    // Loop to choose two different
    // pairs using two nested loops
    for(int j = 0; j < n; j++)
    {
       for(int k = j + 1; k < n; k++)
       
        
          // Condition to check if the price
          // of these two elements is less than S
          if ((type[j] == 0 && type[k] == 1) ||
              (type[j] == 1 && type[k] == 0))
          {
              if (prices[j] + prices[k] <= S)
              {
                  return "Yes";
              }
          }
       }
    }
    return "No";
}
 
// Driver Code
public static void Main(String[] args)
{
    int []prices = { 3, 8, 6, 5 };
    int []type = { 0, 1, 1, 0 };
    int S = 10;
    int n = 4;
 
    // Function call
    Console.Write(check(S, prices, type, n));
}
}
 
// This code is contributed by sapnasingh4991

Javascript




<script>
 
// Javascript implementation to check if
// two items can be selected from
// two different categories without
// exceeding the total price
 
// Function to check if two items 
// can be selected from two
// different categories without
// exceeding the total price
function check(S, prices, type, n)
{
     
    // Loop to choose two different
    // pairs using two nested loops
    for(let j = 0; j < n; j++)
    {
       for(let k = j + 1; k < n; k++)
       
            
          // Condition to check if the price
          // of these two elements is less than S
          if ((type[j] == 0 && type[k] == 1) ||
              (type[j] == 1 && type[k] == 0))
          {
              if (prices[j] + prices[k] <= S)
              {
                  return "Yes";
              }
          }
       }
    }
    return "No";
}
 
// Driver code
let prices = [ 3, 8, 6, 5 ];
let type = [ 0, 1, 1, 0 ];
let S = 10;
let n = 4;
 
// Function call
document.write(check(S, prices, type, n));
 
// This code is contributed by mukesh07
 
</script>
Output: 
Yes

 




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