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++

C++Server Side ProgrammingProgramming

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

 Live Demo

#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
raja
Published on 05-Jan-2021 11:00:25
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