Count number of pairs (i, j) such that arr[i] * arr[j] > arr[i] + arr[j]

Given an array arr[] of non-negative integers, the task is to count pairs of indices (i, j such that arr[i] * arr[j] > arr[i] + arr[j] where i < j.
Examples: 
 

Input: arr[] = { 5, 0, 3, 1, 2 } 
Output: 3
Input: arr[] = { 1, 1, 1 } 
Output: 0 
 

Naive Approach: Run two nested loops and check for every pair whether the condition is satisfied. If the condition is satisfied for any pair then update count = count + 1 and print the count in the end.
Below is the implementation of the above approach: 
 

3

Efficient Approach: Consider the following cases: 
 

1) arr[i] = 0 or arr[i] = 1 and arr[j] = any element 
In this case, arr[j] * arr[i] will always be less than arr[i] + arr[j]. 
Hence we can discard all pairs which have one element either 0 or 1.
2) arr[i] = 2 and arr[j] <= 2 
In this case, arr[j] * arr[i] will always be less than or equal to arr[i] + arr[j]. 
Hence again we can discard all such pairs.
3) arr[i] = 2 and arr[j] > 2 
This case will produce valid pairs. If count_2 is the count of ‘2’s and count_others 
is the count of elements greater than 2, 
then number of pairs will be count_2 * count_others.
4) arr[i] > 2 and arr[j] > 2 
This case will also produce valid pairs. Let count_others be the number of elements 
greater than 2, then every two elements among these count_others elements 
will form a valid pair. Hence the number of pairs will be 
 

Therefore, total count = (count_2 * count_others) + (count_others * (count_others – 1)) / 2. 
 

Below is the implementation of the above approach: 
 

3

Time Complexity: O(N)
Auxiliary Space: O(1)