Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

Examples:

Input: arr[] = {2, 0, 2} Output: 2 Structure is like below | | |_| We can trap 2 units of water in the middle gap. Input: arr[] = {3, 0, 0, 2, 0, 4} Output: 10 Structure is like below | | | | | | |__|_| We can trap "3*2 units" of water between 3 an 2, "1 unit" on top of bar 2 and "3 units" between 2 and 4. See below diagram also. Input: arr[] = [0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1] Output: 6 | | || | _|_||_|||||| Trap "1 unit" between first 1 and 2, "4 units" between first 2 and 3 and "1 unit" between second last 1 and last 2

Source: http://qa.geeksforgeeks.org/1875/trapping-rain-water

## We strongly recommend that you click here and practice it, before moving on to the solution.

An element of array can store water if there are higher bars on left and right. We can find 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 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 two units of water at indexes 1 and 2, and one unit of water at index 2.

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 smaller height and height of current element is the amount of water that can be stored in this array element. Time complexity of this solution is O(n^{2}).

An **Efficient Solution** is to prre-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. Below is C++ implementation of this solution.

## C++

// C++ program to find maximum amount of water that can // be trapped within given set of bars. #include<bits/stdc++.h> using namespace std; int findWater(int arr[], int n) { // left[i] contains height of tallest bar to the // left of i'th bar including itself int left[n]; // Right [i] contains height of tallest bar to // the right of ith bar including itself int right[n]; // Initialize result int water = 0; // Fill left array left[0] = arr[0]; for (int i = 1; i < n; i++) left[i] = max(left[i-1], arr[i]); // Fill right array right[n-1] = arr[n-1]; for (int i = n-2; i >= 0; i--) right[i] = max(right[i+1], arr[i]); // Calculate the accumulated water element by element // consider the amount of water on i'th bar, the // amount of water accumulated on this particular // bar will be equal to min(left[i], right[i]) - arr[i] . for (int i = 0; i < n; i++) water += min(left[i],right[i]) - arr[i]; return water; } // Driver program int main() { int arr[] = {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Maximum water that can be accumulated is " << findWater(arr, n); return 0; }

## Java

// Java program to find maximum amount of water that can // be trapped within given set of bars. class Test { static int arr[] = new int[]{0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1}; // Method for maximum amount of water static int findWater(int n) { // left[i] contains height of tallest bar to the // left of i'th bar including itself int left[] = new int[n]; // Right [i] contains height of tallest bar to // the right of ith bar including itself int right[] = new int[n]; // Initialize result int water = 0; // Fill left array left[0] = arr[0]; for (int i = 1; i < n; i++) left[i] = Math.max(left[i-1], arr[i]); // Fill right array right[n-1] = arr[n-1]; for (int i = n-2; i >= 0; i--) right[i] = Math.max(right[i+1], arr[i]); // Calculate the accumulated water element by element // consider the amount of water on i'th bar, the // amount of water accumulated on this particular // bar will be equal to min(left[i], right[i]) - arr[i] . for (int i = 0; i < n; i++) water += Math.min(left[i],right[i]) - arr[i]; return water; } // Driver method to test the above function public static void main(String[] args) { System.out.println("Maximum water that can be accumulated is " + findWater(arr.length)); } }

Output:

Maximum water that can be accumulated is 6

Time Complexity: O(n)

Auxiliary Space: O(n)

**Space Optimization in above solution :**

Instead of maintaing two arrays of size n for storing left and right max of each element, we will just maintain two variables to store the maximum till that point. Since water trapped at any element = min( max_left, max_right) – arr[i] we will calculate water trapped on smaller element out of A[lo] and A[hi] first and move the pointers till *lo* doesn’t cross *hi*.

## C++

// C++ program to find maximum amount of water that can // be trapped within given set of bars. // Space Complexity : O(1) #include<iostream> using namespace std; int findWater(int arr[], int n) { // initialize output int result = 0; // maximum element on left and right int left_max = 0, right_max = 0; // indices to traverse the array int lo = 0, hi = n-1; while(lo <= hi) { if(arr[lo] < arr[hi]) { if(arr[lo] > left_max) // update max in left left_max = arr[lo]; else // water on curr element = max - curr result += left_max - arr[lo]; lo++; } else { if(arr[hi] > right_max) // update right maximum right_max = arr[hi]; else result += right_max - arr[hi]; hi--; } } return result; } int main() { int arr[] = {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Maximum water that can be accumulated is " << findWater(arr, n); } // This code is contributed by Aditi Sharma

## Java

// JAVA Code For Trapping Rain Water import java.util.*; class GFG { static int findWater(int arr[], int n) { // initialize output int result = 0; // maximum element on left and right int left_max = 0, right_max = 0; // indices to traverse the array int lo = 0, hi = n-1; while(lo <= hi) { if(arr[lo] < arr[hi]) { if(arr[lo] > left_max) // update max in left left_max = arr[lo]; else // water on curr element = // max - curr result += left_max - arr[lo]; lo++; } else { if(arr[hi] > right_max) // update right maximum right_max = arr[hi]; else result += right_max - arr[hi]; hi--; } } return result; } /* Driver program to test above function */ public static void main(String[] args) { int arr[] = {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1}; int n = arr.length; System.out.println("Maximum water that " + "can be accumulated is " + findWater(arr, n)); } } // This code is contributed by Arnav Kr. Mandal.

Output:

Maximum water that can be accumulated is 6

Thanks to Gaurav Ahirwar and **Aditi Sharma **for above solution.

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