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Divide array into two sub-arrays such that their averages are equal

  • Difficulty Level : Easy
  • Last Updated : 22 Apr, 2021

Given an integer array, the task is to divide an integer array into two sub-arrays to make their averages equal if possible.
Examples : 
 

Input : arr[] = {1, 5, 7, 2, 0};    
Output : (0  1) and (2  4) 
Subarrays arr[0..1] and arr[2..4] have
same average.

Input : arr[] = {4, 3, 5, 9, 11};
Output : Not possible

Asked in Microsoft 
 

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A Naive Approach is to run two loops and find subarrays whose averages are equal. 
 



C++




// Simple C++ program to find subarrays
// whose averages are equal
#include<bits/stdc++.h>
using namespace std;
 
// Finding two subarrays
// with equal average.
void findSubarrays(int arr[], int n)
{
    bool found = false;
    int lsum = 0;
    for (int i = 0; i < n - 1; i++)
    {
        lsum += arr[i];
        int rsum = 0;
        for (int j = i + 1; j < n; j++)
            rsum += arr[j];
 
        // If averages of arr[0...i] and
        // arr[i+1..n-1] are same. To avoid
        // floating point problems we compare
        // "lsum*(n-i+1)" and "rsum*(i+1)"
        // instead of "lsum/(i+1)" and
        // "rsum/(n-i+1)"
        if (lsum * (n - i - 1) ==
               rsum * (i + 1))
        {
            printf("From (%d %d) to (%d %d)\n",
                           0, i, i + 1, n - 1);
            found = true;
        }
    }
 
    // If no subarrays found
    if (found == false)
        cout << "Subarrays not found"
             << endl;
}
 
// Driver code
int main()
{
    int arr[] = {1, 5, 7, 2, 0};
    int n = sizeof(arr) / sizeof(arr[0]);
    findSubarrays(arr, n);
    return 0;
}

Java




// Simple Java program to find subarrays
// whose averages are equal
 
public class GFG {
     
    // Finding two subarrays
    // with equal average.
    static void findSubarrays(int[] arr, int n)
    {
        boolean found = false;
        int lsum = 0;
         
        for (int i = 0; i < n - 1; i++)
        {
            lsum += arr[i];
            int rsum = 0;
             
            for (int j = i + 1; j < n; j++)
                rsum += arr[j];
     
            // If averages of arr[0...i] and
            // arr[i+1..n-1] are same. To avoid
            // floating point problems we compare
            // "lsum*(n-i+1)" and "rsum*(i+1)"
            // instead of "lsum/(i+1)" and
            // "rsum/(n-i+1)"
            if (lsum * (n - i - 1) ==
                                rsum * (i + 1))
            {
                System.out.println("From (0 " + i
                        + ") to (" +(i + 1) + " "
                                + (n - 1)+ ")");
                             
                found = true;
            }
        }
     
        // If no subarrays found
        if (found == false)
            System.out.println( "Subarrays not "
                                    + "found");
    }
     
    // Driver code
    static public void main (String[] args)
    {
        int[] arr = {1, 5, 7, 2, 0};
        int n = arr.length;
        findSubarrays(arr, n);
    }
}
 
// This code is contributed by Mukul Singh.

Python 3




# Simple Python 3 program to find subarrays
# whose averages are equal
 
# Finding two subarrays with equal average.
def findSubarrays(arr, n):
 
    found = False
    lsum = 0
    for i in range(n - 1):
     
        lsum += arr[i]
        rsum = 0
        for j in range(i + 1, n):
            rsum += arr[j]
 
        # If averages of arr[0...i] and
        # arr[i+1..n-1] are same. To avoid
        # floating point problems we compare
        # "lsum*(n-i+1)" and "rsum*(i+1)"
        # instead of "lsum/(i+1)" and
        # "rsum/(n-i+1)"
        if (lsum * (n - i - 1) == rsum * (i + 1)):
            print("From", "(", 0, i, ")",
                  "to", "(", i + 1, n - 1, ")")
            found = True
 
    # If no subarrays found
    if (found == False):
        print("Subarrays not found")
 
# Driver code
if __name__ == "__main__":
     
    arr = [1, 5, 7, 2, 0]
    n = len(arr)
    findSubarrays(arr, n)
 
# This code is contributed by ita_c

C#




// Simple C# program to find subarrays
// whose averages are equal
using System;
 
public class GFG {
     
    // Finding two subarrays
    // with equal average.
    static void findSubarrays(int []arr, int n)
    {
        bool found = false;
        int lsum = 0;
         
        for (int i = 0; i < n - 1; i++)
        {
            lsum += arr[i];
            int rsum = 0;
             
            for (int j = i + 1; j < n; j++)
                rsum += arr[j];
     
            // If averages of arr[0...i] and
            // arr[i+1..n-1] are same. To avoid
            // floating point problems we compare
            // "lsum*(n-i+1)" and "rsum*(i+1)"
            // instead of "lsum/(i+1)" and
            // "rsum/(n-i+1)"
            if (lsum * (n - i - 1) ==
                                  rsum * (i + 1))
            {
                Console.WriteLine("From ( 0 " + i
                        + ") to(" + (i + 1) + " "
                                + (n - 1) + ")");
                             
                found = true;
            }
        }
     
        // If no subarrays found
        if (found == false)
            Console.WriteLine( "Subarrays not "
                                    + "found");
    }
     
    // Driver code
    static public void Main ()
    {
        int []arr = {1, 5, 7, 2, 0};
        int n = arr.Length;
        findSubarrays(arr, n);
    }
}
 
// This code is contributed by anuj_67.

PHP




<?php
// Simple PHP program to find subarrays
// whose averages are equal
 
// Finding two subarrays
// with equal average.
function findSubarrays( $arr, $n)
{
    $found = false;
    $lsum = 0;
    for ( $i = 0; $i < $n - 1; $i++)
    {
        $lsum += $arr[$i];
        $rsum = 0;
        for ( $j = $i + 1; $j < $n; $j++)
            $rsum += $arr[$j];
 
        // If averages of arr[0...i] and
        // arr[i+1..n-1] are same. To avoid
        // floating point problems we compare
        // "lsum*(n-i+1)" and "rsum*(i+1)"
        // instead of "lsum/(i+1)" and "rsum/(n-i+1)"
        if ($lsum * ($n - $i - 1) ==
                $rsum * ($i + 1))
        {
        echo "From ( 0 ", $i," )".
             " to (", $i + 1," ", $n - 1,")\n";
                                             
            $found = true;
        }
    }
 
    // If no subarrays found
    if ($found == false)
        echo "Subarrays not found" ;
}
 
// Driver code
$arr = array(1, 5, 7, 2, 0);
$n = count($arr);
findSubarrays($arr, $n);
 
// This code is contributed by vt_m
?>

Javascript




<script>
 
// Simple Javascript program to find subarrays
// whose averages are equal
     
    // Finding two subarrays
    // with equal average.
    function findSubarrays(arr,n)
    {
        let found = false;
        let lsum = 0;
           
        for (let i = 0; i < n - 1; i++)
        {
            lsum += arr[i];
            let rsum = 0;
               
            for (let j = i + 1; j < n; j++)
                rsum += arr[j];
       
            // If averages of arr[0...i] and
            // arr[i+1..n-1] are same. To avoid
            // floating point problems we compare
            // "lsum*(n-i+1)" and "rsum*(i+1)"
            // instead of "lsum/(i+1)" and
            // "rsum/(n-i+1)"
            if (lsum * (n - i - 1) ==
                                rsum * (i + 1))
            {
                document.write("From (0 " + i
                        + ") to (" +(i + 1) + " "
                                + (n - 1)+ ")");
                               
                found = true;
            }
        }
       
        // If no subarrays found
        if (found == false)
            document.write( "Subarrays not "
                                    + "found");
    }
     
    // Driver code
    let arr=[1, 5, 7, 2, 0];
    let n = arr.length;
    findSubarrays(arr, n);
     
    // This code is contributed by avanitrachhadiya2155
     
</script>

Output : 

From (0  1) to (2  4)

Time complexity : O(n2
Auxiliary Space : O(1)
An Efficient solution is to find sum of array elements. Initialize leftsum as zero. Run a loop and find leftsum by adding elements of array. For rightsum, we subtract leftsum from total sum then we find rightsum and find average of leftsum and rightsum as according to their index.
 

1) Compute sum of all array elements. Let this 
   sum be "sum"
2) Initialize leftsum = 0.
3) Run a loop for i=0 to n-1.
    a) leftsum  = leftsum + arr[i]
    b) rightsum = sum - leftsum
    c) If average of left and right are same, 
       print current index as output.

Below is the implementation for above approach:
 

C++




// Efficient C++ program for
// dividing array to make
// average equal
#include<bits/stdc++.h>
using namespace std;
 
void findSubarrays(int arr[], int n)
{
    // Find array sum
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
 
    bool found = false;
    int lsum = 0;
    for (int i = 0; i < n - 1; i++)
    {
        lsum += arr[i];
        int rsum = sum - lsum;
 
        // If averages of arr[0...i]
        // and arr[i+1..n-1] are same.
        // To avoid floating point problems
        // we compare "lsum*(n-i+1)"
        // and "rsum*(i+1)" instead of
        // "lsum/(i+1)" and "rsum/(n-i+1)"
        if (lsum * (n - i - 1) == rsum * (i + 1))
        {
            printf("From (%d %d) to (%d %d)\n",
                               0, i, i+1, n-1);
            found = true;
        }
    }
 
    // If no subarrays found
    if (found == false)
        cout << "Subarrays not found"
             << endl;
}
 
// Driver code
int main()
{
    int arr[] = {1, 5, 7, 2, 0};
    int n = sizeof(arr) / sizeof(arr[0]);
    findSubarrays(arr, n);
    return 0;
}

Java




// Efficient Java program for
// dividing array to make
// average equal
import java.util.*;
     
class GFG
{
static void findSubarrays(int arr[], int n)
{
    // Find array sum
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
 
    boolean found = false;
    int lsum = 0;
    for (int i = 0; i < n - 1; i++)
    {
        lsum += arr[i];
        int rsum = sum - lsum;
 
        // If averages of arr[0...i]
        // and arr[i+1..n-1] are same.
        // To avoid floating point problems
        // we compare "lsum*(n-i+1)"
        // and "rsum*(i+1)" instead of
        // "lsum/(i+1)" and "rsum/(n-i+1)"
        if (lsum * (n - i - 1) == rsum * (i + 1))
        {
            System.out.printf("From (%d %d) to (%d %d)\n",
                                      0, i, i + 1, n - 1);
            found = true;
        }
    }
 
    // If no subarrays found
    if (found == false)
        System.out.println("Subarrays not found");
}
 
// Driver code
static public void main ( String []arg)
{
    int arr[] = {1, 5, 7, 2, 0};
    int n = arr.length;
    findSubarrays(arr, n);
}
}
 
// This code is contributed by Princi Singh

Python3




# Efficient Python program for
# dividing array to make
# average equal
 
def findSubarrays(arr, n):
 
    # Find array sum
    sum = 0;
    for i in range(n):
        sum += arr[i];
 
    found = False;
    lsum = 0;
    for i in range(n - 1):
        lsum += arr[i];
        rsum = sum - lsum;
 
        # If averages of arr[0...i]
        # and arr[i + 1..n - 1] are same.
        # To avoid floating poproblems
        # we compare "lsum*(n - i + 1)"
        # and "rsum*(i + 1)" instead of
        # "lsum / (i + 1)" and "rsum/(n - i + 1)"
        if (lsum * (n - i - 1) == rsum * (i + 1)):
            print("From (%d %d) to (%d %d)\n"%
                   (0, i, i + 1, n - 1));
            found = True;
         
    # If no subarrays found
    if (found == False):
        print("Subarrays not found");
 
# Driver code
if __name__ == '__main__':
    arr = [ 1, 5, 7, 2, 0 ];
    n = len(arr);
    findSubarrays(arr, n);
 
# This code is contributed by Rajput-Ji

C#




// Efficient C# program for
// dividing array to make
// average equal
using System;
     
class GFG
{
static void findSubarrays(int []arr, int n)
{
    // Find array sum
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
 
    bool found = false;
    int lsum = 0;
    for (int i = 0; i < n - 1; i++)
    {
        lsum += arr[i];
        int rsum = sum - lsum;
 
        // If averages of arr[0...i]
        // and arr[i+1..n-1] are same.
        // To avoid floating point problems
        // we compare "lsum*(n-i+1)"
        // and "rsum*(i+1)" instead of
        // "lsum/(i+1)" and "rsum/(n-i+1)"
        if (lsum * (n - i - 1) == rsum * (i + 1))
        {
            Console.Write("From ({0} {1}) to ({2} {3})\n",
                                      0, i, i + 1, n - 1);
            found = true;
        }
    }
 
    // If no subarrays found
    if (found == false)
        Console.WriteLine("Subarrays not found");
}
 
// Driver code
static public void Main ( String []arg)
{
    int []arr = {1, 5, 7, 2, 0};
    int n = arr.Length;
    findSubarrays(arr, n);
}
}
     
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Efficient Javascript program for
// dividing array to make
// average equal
     
    function findSubarrays(arr,n)
    {
    // Find array sum
    let sum = 0;
    for (let i = 0; i < n; i++)
        sum += arr[i];
 
    let found = false;
    let lsum = 0;
    for (let i = 0; i < n - 1; i++)
    {
        lsum += arr[i];
        let rsum = sum - lsum;
 
        // If averages of arr[0...i]
        // and arr[i+1..n-1] are same.
        // To avoid floating point problems
        // we compare "lsum*(n-i+1)"
        // and "rsum*(i+1)" instead of
        // "lsum/(i+1)" and "rsum/(n-i+1)"
        if (lsum * (n - i - 1) == rsum * (i + 1))
        {
            document.write(
       "From (0 "+i+") to ("+(i+1)+" "+(n-1)+")\n"
      );
                                       
            found = true;
        }
    }
 
    // If no subarrays found
    if (found == false)
        document.write("Subarrays not found");
    }
    // Driver code
    let arr=[1, 5, 7, 2, 0];
    let n = arr.length;
    findSubarrays(arr, n);
     
    // This code is contributed by rag2127
 
</script>

Output : 

From (0  1) to (2  4)

Time complexity : O(n) 
Auxiliary Space : O(1)
This article is contributed by DANISH_RAZA . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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