Showing wrong answer for 5 tc
hello @akshitmehta
i m not getting ur approach.
also ur solution time complexity is quadratic which will tle
for most of the cases.
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).
optimisation
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.
