The Upside Down Complete Solution

This article is part of the Scaler Topics Fortnightly Contest - 24

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Solution Approach

Effectively, only 2 portals would be used at max, once while entering and once while exiting.
Say, we use the portals more than 2 times. We will be using them in pairs.
Consider that the first portal we used was at (i,j) and the last exiting portal we used was (k,l).
The minimum time to reach from (i,j) to (k,l) would be when we don’t switch portals and stay in the Upside-Down. Using more portals won’t improve the time spent.
So, as we are only using 2 portals, we can brute force all the combinations of 2 portals and check the minimum. We will also consider the case where we don’t use any portal.

Time Complexity: $O$ ( $R^2$ * $C^2$ )