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++ 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 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 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# 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 |
<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> |
Yes
Time Complexity: O(n * log n)