Find total number of (i,j) pairs in an array such that i<j and a[i]+a[j]=i+j. [closed]

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Closed last year.

I have a naive solution which uses two loops, but I want to improve the time complexity as O(nlogn). Any better approach available?

The array is unsorted and can have negative values too.

Sample Test Case: Array: 1 0 3 2

Output: 4

Explanation: Indices - (0,1), (0,3), (1,2), (2,3) are the pairs which satisfy the given constraints.

回答1:

Problem: find (i,j), i<j, s.t. a[i] + a[j] = i + j

I propose the following approach:

1. Set b[i] = a[i] - i
   The problem becomes: find (i,j), i<j, s.t. b[i] + b[j] = 0
   Complexity: O(n)

2. Create objects c[i] = (b[i],i), a struct for example, in a std::vector (or std::array)
   Complexity: O(n)

3. Sort the c[i] pairs according to b values -> get d[j] = (v[j], i[j]) , sorted array
   Complexity: O(nlogn)

4. Find all pairs j,k such that v[j] = -v[k]
   Complexity: the array being sorted, should be O(n)

5. keep the cases i[j] < i[k]
   Complexity: O(n)

来源：https://stackoverflow.com/questions/53343473/find-total-number-of-i-j-pairs-in-an-array-such-that-ij-and-aiaj-ij