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Count the number of sub-arrays such that the average of elements present in the subarray is greater than that not present in the sub-array in C++
Given an array arr[ ] of positive integers. The goal is to find the count of subarrays of arr[ ] that have an average of its elements greater than the average of the rest of the elements of arr[ ] that are not present in it.
For Example
Input
arr[ ] = { 3, 2, 4 }Output
Count of number of sub-arrays such that the average of elements present in the sub−array is greater than that do not present in the sub−array are: 2
Explanation
The subarrays are − [ 3 ], [ 2 ], [ 4 ], [ 3,2 ], [ 2,4 ], [ 3,2,4 ]. Average of [ 4 ] is 4 which is more than the average of [ 2,3 ]. Average of [ 3,2,4 ] is 3 which is more than the average of [ ]
Input
arr[ ] = { 3, 3, 3 }Output
Count of number of sub−arrays such that the average of elements present in the sub−array is greater than that do not present in the sub−array are: 1
Explanation
The subarrays are − [ 3 ], [ 3 ], [ 3 ], [ 3,3 ], [ 3,3 ], [ 3,3,3 ]. Average of [ 3,3,3 ] is 3 which is more than the average of [ ]
Approach used in the below program is as follows −
In this approach create a prefix sum array which will store the sum of elements upto index i in new_arr[i] . Now we have sums upto previous element then calculate sum upto arr[i] and count of elements as j−i+1 and calculate averages.
Take an array arr[] as input.
Function count(int arr[], int size) takes arr[ ] and returns the count of sub−arrays such that the average of elements present in the sub−array is greater than that not present in the sub−array.
Take an array new_arr[size] to store sum upto previous index elements.
Traverse from i=0 to i<size and set new_arr[i] with new_arr[i−1]+arr[i−1].
Now traverse new_arr[ ] using two for loops.
Now calculate total_1 as the sum of previous subarrays. And elements in it as count_1.
Calculate total_2 as sum of next subarrays and elements in it as count_2.
Calculate averages as check_1 = total_1 / count_1; and check_2 = total_2 / count_2;
If average check_1 > check_2 then increment count.
At the end of for loops return count as result.
Example
#include <bits/stdc++.h>
using namespace std;
int count(int arr[], int size){
int count = 0;
int new_size = size + 1;
int new_arr[new_size] = { 0 };
for (int i = 1; i < new_size; i++){
new_arr[i] = new_arr[i − 1] + arr[i − 1];
}
for (int i = 1; i < new_size; i++){
for (int j = i; j < new_size; j++){
int total_1 = new_arr[j] − new_arr[i − 1];
int count_1 = j − i + 1;
int total_2 = new_arr[size] − total_1;
int count_2 = 0;
if((size − count_1) == 0){
count_2 = 1;
} else {
count_2 = size − count_1;
}
int check_1 = total_1 / count_1;
int check_2 = total_2 / count_2;
if (check_1 > check_2){
count++;
}
}
}
return count;
}
int main(){
int arr[] = { 2, 6, 2, 4 };
int size = sizeof(arr) / sizeof(arr[0]);
cout<<"Count of number of sub−arrays such that the average of elements present in
the sub−array "<< "is greater than that not present in the sub−array are: "<<count(arr, size);
return 0;
}Output
If we run the above code it will generate the following output −
Count the number of sub−arrays such that the average of elements present in the subarrayis greater than that not present in the sub-array are: 6
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