Given an array of integers arr[], the task is to count the number of sub-arrays such that the average of elements present in the sub-array is greater than the average of elements that are not present in the sub-array.
Examples:
Input: arr[] = {6, 3, 5}
Output: 3
The sub-arrays are {6}, {5} and {6, 3, 5} because their averages
are greater than {3, 5}, {6, 3} and {} respectively.
Input: arr[] = {2, 1, 4, 1}
Output: 5
Approach: The problem can be solved easily by calculating the prefix sum array of the given array. The ith element of the prefix sum array will contain sum of elements up to i. So, the sum of elements between any two indexes i and j can be found using the prefix sum array. Using a nested loop, find all the possible sub-arrays such that its average sum is greater than average of elements not present in the array.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int countSubarrays( int a[], int n)
{
int count = 0;
int pre[n + 1] = { 0 };
for ( int i = 1; i < n + 1; i++) {
pre[i] = pre[i - 1] + a[i - 1];
}
for ( int i = 1; i < n + 1; i++) {
for ( int j = i; j < n + 1; j++) {
int sum1 = pre[j] - pre[i - 1], count1 = j - i + 1;
int sum2 = pre[n] - sum1, count2 = ((n - count1) == 0) ? 1 : (n - count1);
int includ = sum1 / count1;
int exclud = sum2 / count2;
if (includ > exclud)
count++;
}
}
return count;
}
int main()
{
int arr[] = { 6, 3, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << countSubarrays(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int countSubarrays( int a[], int n)
{
int count = 0 ;
int []pre = new int [n + 1 ];
Arrays.fill(pre, 0 );
for ( int i = 1 ; i < n + 1 ; i++)
{
pre[i] = pre[i - 1 ] + a[i - 1 ];
}
for ( int i = 1 ; i < n + 1 ; i++)
{
for ( int j = i; j < n + 1 ; j++)
{
int sum1 = pre[j] - pre[i - 1 ], count1 = j - i + 1 ;
int sum2 = pre[n] - sum1, count2 =
((n - count1) == 0 ) ? 1 : (n - count1);
int includ = sum1 / count1;
int exclud = sum2 / count2;
if (includ > exclud)
count++;
}
}
return count;
}
public static void main(String args[])
{
int arr[] = { 6 , 3 , 5 };
int n = arr.length;
System.out.println(countSubarrays(arr, n));
}
}
|
Python3
def countSubarrays(a, n):
count = 0
pre = [ 0 for i in range (n + 1 )]
for i in range ( 1 , n + 1 ):
pre[i] = pre[i - 1 ] + a[i - 1 ]
for i in range ( 1 , n + 1 ):
for j in range (i, n + 1 ):
sum1 = pre[j] - pre[i - 1 ]
count1 = j - i + 1
sum2 = pre[n] - sum1
if n - count1 = = 0 :
count2 = 1
else :
count2 = n - count1
includ = sum1 / / count1
exclud = sum2 / / count2
if (includ > exclud):
count + = 1
return count
arr = [ 6 , 3 , 5 ]
n = len (arr)
print (countSubarrays(arr, n))
|
C#
using System;
class GFG
{
static int countSubarrays( int []a, int n)
{
int count = 0;
int []pre = new int [n + 1];
Array.Fill(pre, 0);
for ( int i = 1; i < n + 1; i++)
{
pre[i] = pre[i - 1] + a[i - 1];
}
for ( int i = 1; i < n + 1; i++)
{
for ( int j = i; j < n + 1; j++)
{
int sum1 = pre[j] - pre[i - 1], count1 = j - i + 1;
int sum2 = pre[n] - sum1, count2 =
((n - count1) == 0) ? 1 : (n - count1);
int includ = sum1 / count1;
int exclud = sum2 / count2;
if (includ > exclud)
count++;
}
}
return count;
}
public static void Main()
{
int []arr = { 6, 3, 5 };
int n = arr.Length;
Console.WriteLine(countSubarrays(arr, n));
}
}
|
PHP
<?php
function countSubarrays( $a , $n )
{
$count = 0;
$pre = array_fill (0, $n + 1, 0);
for ( $i = 1; $i < $n + 1; $i ++)
{
$pre [ $i ] = $pre [ $i - 1] + $a [ $i - 1];
}
for ( $i = 1; $i < $n + 1; $i ++)
{
for ( $j = $i ; $j < $n + 1; $j ++)
{
$sum1 = $pre [ $j ] - $pre [ $i - 1] ;
$count1 = $j - $i + 1;
$sum2 = $pre [ $n ] - $sum1 ;
$count2 = (( $n - $count1 ) == 0) ?
1 : ( $n - $count1 );
$includ = floor ( $sum1 / $count1 );
$exclud = floor ( $sum2 / $count2 );
if ( $includ > $exclud )
$count ++;
}
}
return $count ;
}
$arr = array ( 6, 3, 5 );
$n = count ( $arr ) ;
echo countSubarrays( $arr , $n );
?>
|
Javascript
<script>
function countSubarrays(a, n)
{
let count = 0;
let pre = new Uint8Array(n + 1);
for (let i = 1; i < n + 1; i++) {
pre[i] = pre[i - 1] + a[i - 1];
}
for (let i = 1; i < n + 1; i++) {
for (let j = i; j < n + 1; j++) {
let sum1 = pre[j] - pre[i - 1], count1 = j - i + 1;
let sum2 = pre[n] - sum1, count2 = ((n - count1) == 0) ? 1 : (n - count1);
let includ = Math.floor(sum1 / count1);
let exclud = Math.floor(sum2 / count2);
if (includ > exclud)
count++;
}
}
return count;
}
let arr = [ 6, 3, 5 ];
let n = arr.length;
document.write(countSubarrays(arr, n));
</script>
|
Time Complexity: O(N^2) where N is the length of the array, as are using nested loops to traverse N*N times.
Auxiliary Space: O(N), as we are using pre array of size N while is extra space.