Given an array of n integers, find the sum of f(a[i], a[j]) of all pairs (i, j) such that (1 <= i < j <= n).
If |a[j]-a[i]| > 1 f(a[i], a[j]) = a[j] - a[i] Else // if |a[j]-a[i]| <= 1 f(a[i], a[j]) = 0
Input : 6 6 4 4 Output : -8 Explanation: All pairs are: (6 - 6) + (6 - 6) + (6 - 6) + (4 - 6) + (4 - 6) + (4 - 6) + (4 - 6) + (4 - 4) + (4 - 4) = -8 Input: 1 2 3 1 3 Output: 4 Explanation: the pairs that add up are: (3, 1), (3, 1) to give 4, rest all pairs according to condition gives 0.
A naive approach is to iterate through all pairs and calculate f(a[i], a[j]) and summing it while traversing in two nested loops will give us our answer.
Time Complexity: O(n^2)
A efficient approach will be to use a map/hash function to keep a count of every occurring numbers and then traverse through the list. While traversing through the list, we multiply the count of numbers that are before it and the number itself. Then subtract this result with the pre-sum of the number before that number to get the sum of difference of all pairs possible with that number. To remove all pairs whose absolute difference is <=1, simply subtract the count of occurrence of (number-1) and (number+1) from the previously computed sum. Here we subtract count of (number-1) from the computed sum as it had been previously added to the sum, and we add (number+1) count since the negative has been added to the pre-computed sum of all pairs.
Time Complexity : O(n)
Below is the implementation of the above approach :
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