MINMAGIC - editorial

editorial
ipc15flb
medium
minmagic
segment-tree

#1

Problem Link:
[Contest][1]
[Practice][2]

Author and Editorialist: [sumeet-varma][3]

Difficulty: Medium

Pre-requisites: - Segment tree

Problem: Given N intervals(L*, R*), 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 ith interval (x ≥ L* and x ≤ R*)

  y = choose A or B

  magic += |x – y|

Explanation:
For a given query (A <= B), intervals having non-zero magic can be split in 3 different categories.

Case 1: L* <= R* <= A

Case 2: A < L* <= R* < B

Case 3: B < L* <= R*

For case 1, magic += A - R*. 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 + R) <= (A + B) , then magic += L* - A else magic += B - R***

Proof: Let magic += L* - A

herefore min(L* - A, B - R*) = L* - A

\implies L* - A \le B - R*

\implies L* + R* \le A + B

Intuitive Proof: If midpoint([L*, R*]) \le midpoint([A, B]), then L* should be closer to A than R* is from B.

Thus, we again split the intervals of case 3 in two types.

Case 3A: A < L* <= R* < B and L* + R* <= A + B

Case 3B: A < L* <= R* < B and L* + R* > A + B

For case 3A, we sort the intervals based on L + R** and build a segment tree over it in which in each node, we again sort the intervals based on L* and also build a prefix sum over sorted L* in each node.

For each query, we find \sum L* and count of all intervals such that L* > A and L* + R* <= A + B. We don’t need to check for R* < 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* + R* <= A + B and then a range query in segment tree for that prefix.

And finally, magic += \sum L* - count * A

Case 3B can be solved in a similar way.

Time Complexity: O((N + Q) * log^2N)

Space Complexity: O(N * log^2N)

[1]: https://www.codechef.com/IPC15FLB/problems/MINMAGIC
[2]: https://www.codechef.com/problems/MINMAGIC
[3]: https://www.codechef.com/users/sumeet_varma