I don't get the logic of this question

#1

I don’t get the logic of this question

Rain water trapping
#2

hi @aakshatmalhotra100
An element of the array can store water if there are higher bars on left and right. We can find the amount of water to be stored in every element by finding the heights of bars on left and right sides. The idea is to compute the amount of water that can be stored in every element of array. For example, consider the array {3, 0, 0, 2, 0, 4}, we can store three units of water at indexes 1 and 2, and one unit of water at index 3, and three units of water at index 4.

A Simple Solution is to traverse every array element and find the highest bars on left and right sides. Take the smaller of two heights. The difference between the smaller height and height of the current element is the amount of water that can be stored in this array element. Time complexity of this solution is O(n2). But this soln may give TLE on some test cases…

Better Approach

An element of an array can store water if there are higher bars on left and right. We can find the amount of water to be stored in every element by finding the heights of bars on the left and right sides. The idea is to compute the amount of water that can be stored in every element of the array. For example, consider the array {3, 0, 0, 2, 0, 4}, we can store two units of water at indexes 1 and 2, and one unit of water at index 2.

Pre-compute highest bar on left and right of every bar in O(n) time. Then use these pre-computed values to find the amount of water in every array element.

#3

hi @aakshatmalhotra100
i hope it’s clear now??

#4

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