Author and Editorialist: sumeet-varma
Pre-requisites: - Segment tree
Problem: Given N intervals(L[i], R[i]), you have to minimize the magic for Q queries where each query has two parameters A and B and magic is calculated as follows.
magic = 0
for i in range(0, N)
x = choose any point which lies in i<sup>th</sup> interval (x ≥ L[i] and x ≤ R[i]) y = choose A or B magic += |x – y|
For a given query (A <= B), intervals having non-zero magic can be split in 3 different categories.
Case 1: L[i] <= R[i] <= A
Case 2: A < L[i] <= R[i] < B
Case 3: B < L[i] <= R[i]
For case 1, magic += A - R[i]. It can be solved in O(logn) per query after sorting the intervals on end point and building range sum segment tree/BIT over it. Case 3 can be solved in a similar way.
For case 2,
Lemma: If (L[i] + R[i]) <= (A + B) , then magic += L[i] - A else magic += B - R[i]
Proof: Let magic += L[i] - A
\therefore min(L[i] - A, B - R[i]) = L[i] - A
\implies L[i] - A \le B - R[i]
\implies L[i] + R[i] \le A + B
Intuitive Proof: If midpoint([L[i], R[i]]) \le midpoint([A, B]), then L[i] should be closer to A than R[i] is from B.
Thus, we again split the intervals of case 3 in two types.
Case 3A: A < L[i] <= R[i] < B and L[i] + R[i] <= A + B
Case 3B: A < L[i] <= R[i] < B and L[i] + R[i] > A + B
For case 3A, we sort the intervals based on L[i] + R[i] and build a segment tree over it in which in each node, we again sort the intervals based on L[i] and also build a prefix sum over sorted L[i] in each node.
For each query, we find \sum L[i] and count of all intervals such that L[i] > A and L[i] + R[i] <= A + B. We don’t need to check for R[i] < B. Think why?
This can be done with the help of binary search on array of sorted intervals for determining prefix of intervals having L[i] + R[i] <= A + B and then a range query in segment tree for that prefix.
And finally, magic += \sum L[i] - count * A
Case 3B can be solved in a similar way.
Time Complexity: O((N + Q) * log^2N)
Space Complexity: O(N * log^2N)