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# Maximum water that can be stored between two buildings

• Difficulty Level : Hard
• Last Updated : 15 Feb, 2023

Given an integer array that represents the heights of N buildings. The task is to delete N – 2 buildings such that the water that can be trapped between the remaining two buildings is maximum. The total water trapped between two buildings is a gap between them (the number of buildings removed) multiplied by the height of the smaller building.

Examples:

Input: arr[] = {1, 3, 4}
Output:
Explanation: We have to calculate the maximum water that can be stored between any 2 buildings.
Water between the buildings of height 1 and height 3 = 0.
Water between the buildings of height 1 and height 4 = 1.
Water between the buildings of height 3 and height 4 = 0.
Hence maximum of all the cases is 1.

Input: arr[] = {2, 1, 3, 4, 6, 5}
Output:
We remove the middle 4 buildings and get the total water stored as 2 * 4 = 8

Recommended Practice

Naive approach:

Check for all possible pairs and the pair which can hold maximum water will be the answer. Water stored between two buildings of heights h1 and h2 would be equal to minimum(h1, h2)*(distance between the buildings – 1), maximize this value to get the answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach` `#include ` `using` `namespace` `std;`   `// Return the maximum water that can be stored` `int` `maxWater(``int` `height[], ``int` `n)` `{` `    ``int` `maximum = 0;`   `    ``// Check all possible pairs of buildings` `    ``for` `(``int` `i = 0; i < n - 1; i++) {` `        ``for` `(``int` `j = i + 1; j < n; j++) {` `            ``int` `current` `                ``= (min(height[i], height[j]) * (j - i - 1));`   `            ``// Maximum so far` `            ``maximum = max(maximum, current);` `        ``}` `    ``}` `    ``return` `maximum;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `height[] = { 2, 1, 3, 4, 6, 5 };` `    ``int` `n = ``sizeof``(height) / ``sizeof``(height);` `    ``cout << maxWater(height, n);` `    ``return` `0;` `}`

## Java

 `// Java implementation of the above approach` `class` `GFG {`   `    ``// Return the maximum water that can be stored` `    ``static` `int` `maxWater(``int` `height[], ``int` `n)` `    ``{` `        ``int` `max = ``0``;`   `        ``// Check all possible pairs of buildings` `        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {` `            ``for` `(``int` `j = i + ``1``; j < n; j++) {` `                ``int` `current` `                    ``= (Math.min(height[i], height[j])` `                       ``* (j - i - ``1``));`   `                ``// Maximum so far` `                ``max = Math.max(max, current);` `            ``}` `        ``}` `        ``return` `max;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `height[] = { ``2``, ``1``, ``3``, ``4``, ``6``, ``5` `};` `        ``int` `n = height.length;` `        ``System.out.print(maxWater(height, n));` `    ``}` `}`

## Python3

 `# Python3 implementation of the above approach`   `# Return the maximum water` `# that can be stored`     `def` `maxWater(height, n):` `    ``maximum ``=` `0`   `    ``# Check all possible pairs of buildings` `    ``for` `i ``in` `range``(n ``-` `1``):` `        ``for` `j ``in` `range``(i ``+` `1``, n):` `            ``current ``=` `min``(height[i],` `                          ``height[j]) ``*` `(j ``-` `i ``-` `1``)`   `            ``# Maximum so far` `            ``maximum ``=` `max``(maximum, current)`   `    ``return` `maximum`     `# Driver code` `if` `__name__ ``=``=` `"__main__"``:` `    ``height ``=` `[``2``, ``1``, ``3``, ``4``, ``6``, ``5``]`   `    ``n ``=` `len``(height)` `    ``print``(maxWater(height, n))`

## C#

 `// C# implementation of the above approach` `using` `System;`   `class` `GFG {`   `    ``// Return the maximum water that can be stored` `    ``static` `int` `maxWater(``int``[] height, ``int` `n)` `    ``{` `        ``int` `max = 0;`   `        ``// Check all possible pairs of buildings` `        ``for` `(``int` `i = 0; i < n - 1; i++) {` `            ``for` `(``int` `j = i + 1; j < n; j++) {` `                ``int` `current` `                    ``= (Math.Min(height[i], height[j])` `                       ``* (j - i - 1));`   `                ``// Maximum so far` `                ``max = Math.Max(max, current);` `            ``}` `        ``}` `        ``return` `max;` `    ``}`   `    ``// Driver code` `    ``static` `public` `void` `Main()` `    ``{` `        ``int``[] height = { 2, 1, 3, 4, 6, 5 };` `        ``int` `n = height.Length;` `        ``Console.WriteLine(maxWater(height, n));` `    ``}` `}`

## Javascript

 ``

Output

`8`

Time Complexity: O(N*N)
Auxiliary Space: O(1)

## Maximum water that can be stored between two buildings using sorting

Efficient approach:

Sort the array according to increasing height without affecting the original indices i.e. make pairs of (element, index). Then for every element, assume it is the building with the minimum height among the two buildings required then the height of the required water will be equal to the height of the chosen building and the width will be equal to the index difference between the chosen building and the building to be found.

In order to choose the other building which maximizes the water, the other building has to be as far as possible and must be greater in height as compared to the currently chosen building.

Now, the problem gets reduced to finding the minimum and maximum indices on the right for every building in the sorted array.

Follow the steps below to implement the idea:

• Create an array of pairs pairs[] of size N with each pair of the type (i, arr[i]) and sort pairs[] in increasing order of second element of pair.
• Initialize minIndSoFar = pairs[n – 1].first and maxIndSoFar = pairs[n – 1].first as this would be the index of largest buildings on either sides of i and a variable maxi that will store the value of maximum water that can be stored.
• Run a for loop with counter i from N – 2 to
• Calculate the water that can be filled between the building on index i and minIndSoFar as ith building will be of same of less height than the building on minIndSoFar, so left = (pairs[i].second * (pairs[i].first – minIndSoFar – 1)).
• Calculate the water that can be filled between the building on index i and maxIndSoFar as ith building will be of same of less height than the building on maxIndSoFar, so right = (pairs[i].second *(maxIndSoFar – pairs[i].first – 1)).
• Now maximize maxi with max of left, right and maxi, update maxIndSoFar with max of maxIndSoFar and i, and minIndSoFar with min of maxIndSoFar and i.
• Return maxi.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach` `#include ` `using` `namespace` `std;`   `bool` `compareTo(pair<``int``, ``int``> p1, pair<``int``, ``int``> p2)` `{` `    ``return` `p1.second < p2.second;` `}`   `// Return the maximum water that` `// can be stored` `int` `maxWater(``int` `height[], ``int` `n)` `{`   `    ``// Make pairs with indices` `    ``pair<``int``, ``int``> pairs[n];` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``pairs[i] = make_pair(i, height[i]);`   `    ``// Sort array based on heights` `    ``sort(pairs, pairs + n, compareTo);`   `    ``// To store the min and max index so far` `    ``// from the right` `    ``int` `minIndSoFar = pairs[n - 1].first;` `    ``int` `maxIndSoFar = pairs[n - 1].first;` `    ``int` `maxi = 0;`   `    ``for` `(``int` `i = n - 2; i >= 0; i--) {`   `        ``// Current building paired with` `        ``// the building greater in height` `        ``// and on the extreme left` `        ``int` `left = 0;` `        ``if` `(minIndSoFar < pairs[i].first) {` `            ``left = (pairs[i].second` `                    ``* (pairs[i].first - minIndSoFar - 1));` `        ``}`   `        ``// Current building paired with` `        ``// the building greater in height` `        ``// and on the extreme right` `        ``int` `right = 0;` `        ``if` `(maxIndSoFar > pairs[i].first) {` `            ``right = (pairs[i].second` `                     ``* (maxIndSoFar - pairs[i].first - 1));` `        ``}`   `        ``// Maximum so far` `        ``maxi = max(left, max(right, maxi));`   `        ``// Update the maximum and minimum so far` `        ``minIndSoFar = min(minIndSoFar, pairs[i].first);` `        ``maxIndSoFar = max(maxIndSoFar, pairs[i].first);` `    ``}` `    ``return` `maxi;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `height[] = { 2, 1, 3, 4, 6, 5 };` `    ``int` `n = ``sizeof``(height) / ``sizeof``(height);`   `    ``cout << maxWater(height, n);` `}`

## Java

 `// Java implementation of the above approach` `import` `java.util.Arrays;`   `// Class to store the pairs` `class` `Pair ``implements` `Comparable {` `    ``int` `ind, val;`   `    ``Pair(``int` `ind, ``int` `val)` `    ``{` `        ``this``.ind = ind;` `        ``this``.val = val;` `    ``}`   `    ``@Override` `public` `int` `compareTo(Pair o)` `    ``{` `        ``if` `(``this``.val > o.val)` `            ``return` `1``;` `        ``return` `-``1``;` `    ``}` `}`   `class` `GFG {`   `    ``// Return the maximum water that can be stored` `    ``static` `int` `maxWater(``int` `height[], ``int` `n)` `    ``{`   `        ``// Make pairs with indices` `        ``Pair pairs[] = ``new` `Pair[n];` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``pairs[i] = ``new` `Pair(i, height[i]);`   `        ``// Sort array based on heights` `        ``Arrays.sort(pairs);`   `        ``// To store the min and max index so far` `        ``// from the right` `        ``int` `minIndSoFar = pairs[n - ``1``].ind;` `        ``int` `maxIndSoFar = pairs[n - ``1``].ind;` `        ``int` `max = ``0``;` `        ``for` `(``int` `i = n - ``2``; i >= ``0``; i--) {`   `            ``// Current building paired with the building` `            ``// greater in height and on the extreme left` `            ``int` `left = ``0``;` `            ``if` `(minIndSoFar < pairs[i].ind) {` `                ``left = (pairs[i].val` `                        ``* (pairs[i].ind - minIndSoFar - ``1``));` `            ``}`   `            ``// Current building paired with the building` `            ``// greater in height and on the extreme right` `            ``int` `right = ``0``;` `            ``if` `(maxIndSoFar > pairs[i].ind) {` `                ``right` `                    ``= (pairs[i].val` `                       ``* (maxIndSoFar - pairs[i].ind - ``1``));` `            ``}`   `            ``// Maximum so far` `            ``max = Math.max(left, Math.max(right, max));`   `            ``// Update the maximum and minimum so far` `            ``minIndSoFar` `                ``= Math.min(minIndSoFar, pairs[i].ind);` `            ``maxIndSoFar` `                ``= Math.max(maxIndSoFar, pairs[i].ind);` `        ``}`   `        ``return` `max;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `height[] = { ``2``, ``1``, ``3``, ``4``, ``6``, ``5` `};` `        ``int` `n = height.length;`   `        ``System.out.print(maxWater(height, n));` `    ``}` `}`

## Python3

 `# Python3 implementation of the above approach` `from` `functools ``import` `cmp_to_key`     `def` `compareTo(p1, p2):`   `    ``return` `p1[``1``] ``-` `p2[``1``]`   `# Return the maximum water that` `# can be stored`     `def` `maxWater(height, n):`   `    ``# Make pairs with indices` `    ``pairs ``=` `[``0` `for` `i ``in` `range``(n)]` `    ``for` `i ``in` `range``(n):` `        ``pairs[i] ``=` `[i, height[i]]`   `    ``# Sort array based on heights` `    ``pairs.sort(key``=``cmp_to_key(compareTo))`   `    ``# To store the min and max index so far` `    ``# from the right` `    ``minIndSoFar ``=` `pairs[n ``-` `1``][``0``]` `    ``maxIndSoFar ``=` `pairs[n ``-` `1``][``0``]` `    ``maxi ``=` `0`   `    ``for` `i ``in` `range``(n``-``2``, ``-``1``, ``-``1``):`   `        ``# Current building paired with` `        ``# the building greater in height` `        ``# and on the extreme left` `        ``left ``=` `0` `        ``if` `(minIndSoFar < pairs[i][``0``]):` `            ``left ``=` `(pairs[i][``1``] ``*` `                    ``(pairs[i][``0``] ``-` `                     ``minIndSoFar ``-` `1``))`   `        ``# Current building paired with` `        ``# the building greater in height` `        ``# and on the extreme right` `        ``right ``=` `0` `        ``if` `(maxIndSoFar > pairs[i][``0``]):` `            ``right ``=` `(pairs[i][``1``] ``*` `                     ``(maxIndSoFar ``-` `                         ``pairs[i][``0``] ``-` `1``))`   `        ``# Maximum so far` `        ``maxi ``=` `max``(left, ``max``(right, maxi))`   `        ``# Update the maximum and minimum so far` `        ``minIndSoFar ``=` `min``(minIndSoFar,` `                          ``pairs[i][``0``])` `        ``maxIndSoFar ``=` `max``(maxIndSoFar,` `                          ``pairs[i][``0``])` `    ``return` `maxi`     `# Driver code` `height ``=` `[``2``, ``1``, ``3``, ``4``, ``6``, ``5``]` `n ``=` `len``(height)`   `print``(maxWater(height, n))`

## C#

 `// C# implementation of the approach` `using` `System;` `using` `System.Linq;`   `class` `Program` `{`   `    ``// Method to compare two pairs based on the second element` `    ``static` `int` `CompareTo(Tuple<``int``, ``int``> p1, Tuple<``int``, ``int``> p2)` `    ``{` `        ``return` `p1.Item2.CompareTo(p2.Item2);` `    ``}`   `    ``// Return the maximum water that can be stored` `    ``static` `int` `MaxWater(``int``[] height, ``int` `n)` `    ``{` `        ``// Make pairs with indices` `        ``Tuple<``int``, ``int``>[] pairs = ``new` `Tuple<``int``, ``int``>[n];` `        ``for` `(``int` `i = 0; i < n; i++)` `            ``pairs[i] = Tuple.Create(i, height[i]);`   `        ``// Sort array based on heights` `        ``Array.Sort(pairs, CompareTo);`   `        ``// To store the min and max index so far from the right` `        ``int` `minIndSoFar = pairs[n - 1].Item1;` `        ``int` `maxIndSoFar = pairs[n - 1].Item1;` `        ``int` `maxi = 0;`   `        ``for` `(``int` `i = n - 2; i >= 0; i--)` `        ``{` `            ``// Current building paired with the building ` `           ``// greater in height and on the extreme left` `            ``int` `left = 0;` `            ``if` `(minIndSoFar < pairs[i].Item1)` `            ``{` `                ``left = (pairs[i].Item2 * (pairs[i].Item1 - minIndSoFar - 1));` `            ``}`   `            ``// Current building paired with the building greater ` `          ``// in height and on the extreme right` `            ``int` `right = 0;` `            ``if` `(maxIndSoFar > pairs[i].Item1)` `            ``{` `                ``right = (pairs[i].Item2 * (maxIndSoFar - pairs[i].Item1 - 1));` `            ``}`   `            ``// Maximum so far` `            ``maxi = Math.Max(left, Math.Max(right, maxi));`   `            ``// Update the maximum and minimum so far` `            ``minIndSoFar = Math.Min(minIndSoFar, pairs[i].Item1);` `            ``maxIndSoFar = Math.Max(maxIndSoFar, pairs[i].Item1);` `        ``}` `        ``return` `maxi;` `    ``}` `  ``static` `void` `Main(``string``[] args)` `    ``{` `        ``int``[] height = { 2, 1, 3, 4, 6, 5 };` `        ``int` `n = height.Length;`   `        ``Console.WriteLine(MaxWater(height, n));` `    ``}` `}`

## Javascript

 ``

Output

`8`

Time Complexity : O(N*log(N))
Auxiliary Space: O(N)

## Maximum water that can be stored between two buildings using Two pointer approach:

Below is the idea to solve the problem

Take two pointers i and j pointing to the first and the last building respectively and calculate the water that can be stored between these two buildings. Now increment i if height[i] < height[j] else decrement j. This is because the water that can be trapped is dependent on the height of the small building and moving from the greater height building will just reduce the amount of water instead of maximizing it. In the end, print the maximum amount of water calculated so far.

Follow the below steps to implement the idea:

• Initialize variable maximum to store maximum water that can be stored, i and j pointing to the first and the last.
• Run a while loop till i < j
• If height[i] < height[j] update maximum = max(maximum, (j – i – 1) * height[i]) and increment i by 1.
• Else maximum will be updated according to right height i.e. building at j maximum = max(maximum, (j – i – 1) * height[j]) and decrement j by 1.
• return maximum.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Return the maximum water that can be stored` `int` `maxWater(``int` `height[], ``int` `n)` `{`   `    ``// To store the maximum water so far` `    ``int` `maximum = 0;`   `    ``// Both the pointers are pointing at the first` `    ``// and the last buildings respectively` `    ``int` `i = 0, j = n - 1;`   `    ``// While the water can be stored between` `    ``// the currently chosen buildings` `    ``while` `(i < j) {`   `        ``// Update maximum water so far and increment i` `        ``if` `(height[i] < height[j]) {` `            ``maximum = max(maximum, (j - i - 1) * height[i]);` `            ``i++;` `        ``}`   `        ``// Update maximum water so far and decrement j` `        ``else` `{` `            ``maximum = max(maximum, (j - i - 1) * height[j]);` `            ``j--;` `        ``}` `    ``}`   `    ``return` `maximum;` `}`   `// Driver code` `int` `main()` `{`   `    ``int` `height[] = { 2, 1, 3, 4, 6, 5 };`   `    ``int` `n = ``sizeof``(height) / ``sizeof``(height);`   `    ``cout << (maxWater(height, n));` `}`   `// This code is contributed by CrazyPro`

## Java

 `// Java implementation of the approach` `import` `java.util.Arrays;`   `class` `GFG {`   `    ``// Return the maximum water that can be stored` `    ``static` `int` `maxWater(``int` `height[], ``int` `n)` `    ``{`   `        ``// To store the maximum water so far` `        ``int` `max = ``0``;`   `        ``// Both the pointers are pointing at the first` `        ``// and the last buildings respectively` `        ``int` `i = ``0``, j = n - ``1``;`   `        ``// While the water can be stored between` `        ``// the currently chosen buildings` `        ``while` `(i < j) {`   `            ``// Update maximum water so far and increment i` `            ``if` `(height[i] < height[j]) {` `                ``max = Math.max(max,` `                               ``(j - i - ``1``) * height[i]);` `                ``i++;` `            ``}`   `            ``// Update maximum water so far and decrement j` `            ``else` `if` `(height[j] < height[i]) {` `                ``max = Math.max(max,` `                               ``(j - i - ``1``) * height[j]);` `                ``j--;` `            ``}`   `            ``// Any of the pointers can be updated (or both)` `            ``else` `{` `                ``max = Math.max(max,` `                               ``(j - i - ``1``) * height[i]);` `                ``i++;` `                ``j--;` `            ``}` `        ``}`   `        ``return` `max;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `height[] = { ``2``, ``1``, ``3``, ``4``, ``6``, ``5` `};` `        ``int` `n = height.length;`   `        ``System.out.print(maxWater(height, n));` `    ``}` `}`

## Python3

 `# Python3 implementation of the approach`   `# Return the maximum water that can be stored`     `def` `maxWater(height, n):`   `    ``# To store the maximum water so far` `    ``maximum ``=` `0`   `    ``# Both the pointers are pointing at the first` `    ``# and the last buildings respectively` `    ``i ``=` `0` `    ``j ``=` `n ``-` `1`   `    ``# While the water can be stored between` `    ``# the currently chosen buildings` `    ``while` `(i < j):`   `        ``# Update maximum water so far and increment i` `        ``if` `(height[i] < height[j]):` `            ``maximum ``=` `max``(maximum, (j ``-` `i ``-` `1``) ``*` `height[i])` `            ``i ``+``=` `1`   `        ``# Update maximum water so far and decrement j` `        ``elif` `(height[j] < height[i]):` `            ``maximum ``=` `max``(maximum, (j ``-` `i ``-` `1``) ``*` `height[j])` `            ``j ``-``=` `1`   `        ``# Any of the pointers can be updated (or both)` `        ``else``:` `            ``maximum ``=` `max``(maximum, (j ``-` `i ``-` `1``) ``*` `height[i])` `            ``i ``+``=` `1` `            ``j ``-``=` `1`   `    ``return` `maximum`     `# Driver code` `height ``=` `[``2``, ``1``, ``3``, ``4``, ``6``, ``5``]`   `n ``=` `len``(height)`   `print``(maxWater(height, n))`   `# This code is contributed by CrazyPro`

## C#

 `// C# implementation of the approach` `using` `System;`   `class` `GFG {`   `    ``// Return the maximum water that can be stored` `    ``static` `int` `maxWater(``int``[] height, ``int` `n)` `    ``{`   `        ``// To store the maximum water so far` `        ``int` `max = 0;`   `        ``// Both the pointers are pointing at the first` `        ``// and the last buildings respectively` `        ``int` `i = 0, j = n - 1;`   `        ``// While the water can be stored between` `        ``// the currently chosen buildings` `        ``while` `(i < j) {`   `            ``// Update maximum water so far and increment i` `            ``if` `(height[i] < height[j]) {` `                ``max = Math.Max(max,` `                               ``(j - i - 1) * height[i]);` `                ``i++;` `            ``}`   `            ``// Update maximum water so far and decrement j` `            ``else` `if` `(height[j] < height[i]) {` `                ``max = Math.Max(max,` `                               ``(j - i - 1) * height[j]);` `                ``j--;` `            ``}`   `            ``// Any of the pointers can be updated (or both)` `            ``else` `{` `                ``max = Math.Max(max,` `                               ``(j - i - 1) * height[i]);` `                ``i++;` `                ``j--;` `            ``}` `        ``}`   `        ``return` `max;` `    ``}`   `    ``// Driver code` `    ``static` `public` `void` `Main()` `    ``{`   `        ``int``[] height = { 2, 1, 3, 4, 6, 5 };` `        ``int` `n = height.Length;`   `        ``Console.Write(maxWater(height, n));` `    ``}` `}`   `// This code is contributed by jit_t`

## Javascript

 ``

Output

`8`

Time Complexity: O(N)
Auxiliary Space: O(1)

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