Check whether the given integers a, b, c and d are in proportion

0 0 3 minutes read

Given four integers a, b, c and d. The task is to check whether it is possible to pair them up such that they are in proportion. We are allowed to shuffle the order of the numbers.
Examples:

Input: arr[] = {1, 2, 4, 2}
Output: Yes
1 / 2 = 2 / 4
Input: arr[] = {1, 2, 5, 2}
Output: No

Approach: If four numbers a, b, c and d are in proportion then a:b = c:d. The solution is to sort the four numbers and pair up the first 2 together and the last 2 together and check their ratios this is because, in order for them to be in proportion, the product of means has to be equal to the product of extremes. So, a * d has to be equal to c * b.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if the
// given four integers are in proportion
bool inProportion(int arr[])
{
 
    // Array will consist of
    // only four integers
    int n = 4;
 
    // Sort the array
    sort(arr, arr + n);
 
    // Find the product of extremes and means
    long extremes = (long)arr[0] * (long)arr[3];
    long means = (long)arr[1] * (long)arr[2];
 
    // If the products are equal
    if (extremes == means)
        return true;
    return false;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 4, 2 };
 
    if (inProportion(arr))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
     
// Function that returns true if the
// given four integers are in proportion
static boolean inProportion(int []arr)
{
 
    // Array will consist of
    // only four integers
    int n = 4;
 
    // Sort the array
    Arrays.sort(arr);
 
    // Find the product of extremes and means
    long extremes = (long)arr[0] * (long)arr[3];
    long means = (long)arr[1] * (long)arr[2];
 
    // If the products are equal
    if (extremes == means)
        return true;
    return false;
}
 
// Driver code
public static void main(String args[]) 
{
    int arr[] = { 1, 2, 4, 2 };
 
    if (inProportion(arr))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach 
 
# Function that returns true if the 
# given four integers are in proportion 
def inProportion(arr) :
 
    # Array will consist of 
    # only four integers 
    n = 4; 
 
    # Sort the array 
    arr.sort()
 
    # Find the product of extremes and means 
    extremes = arr[0] * arr[3]; 
    means = arr[1] * arr[2]; 
 
    # If the products are equal 
    if (extremes == means) :
        return True; 
         
    return False; 
 
# Driver code 
if __name__ == "__main__" : 
 
    arr = [ 1, 2, 4, 2 ]; 
 
    if (inProportion(arr)) :
        print("Yes"); 
    else :
        print("No"); 
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
                     
class GFG
{
     
// Function that returns true if the
// given four integers are in proportion
static bool inProportion(int []arr)
{
 
    // Array will consist of
    // only four integers
    int n = 4;
 
    // Sort the array
    Array.Sort(arr);
 
    // Find the product of extremes and means
    long extremes = (long)arr[0] * (long)arr[3];
    long means = (long)arr[1] * (long)arr[2];
 
    // If the products are equal
    if (extremes == means)
        return true;
    return false;
}
 
// Driver code
public static void Main(String []args) 
{
    int []arr = { 1, 2, 4, 2 };
 
    if (inProportion(arr))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function that returns true if the
// given four integers are in proportion
function inProportion(arr)
{
 
    // Array will consist of
    // only four integers
    var n = 4;
 
    // Sort the array
    arr.sort();
 
    // Find the product of extremes and means
    var extremes = arr[0] * arr[3];
    var means = arr[1] * arr[2];
 
    // If the products are equal
    if (extremes == means)
        return true;
    return false;
}
 
// Driver code
var arr = [ 1, 2, 4, 2 ]
if (inProportion(arr))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by rutvik_56.
</script>

Output:

Yes

Time Complexity: O(n * log n)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!