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# Number of elements that can be seen from right side

• Difficulty Level : Easy
• Last Updated : 11 Oct, 2022

Given an array of integers, consider the elements as the height of the building, find the number of buildings that can be seen from the right side.

Examples:

```Input : height[] = {2, 6, 2, 4, 0, 1}
Output : 3
we can see only 3 building i.e with height 1, 4 and 6.

Input : height[] = {4, 8, 2, 0, 0, 5}
Output : 2```

This problem seems to be finding longest increasing sub sequence from right but actually it is not.We have to just increase the count if we encounter any building with greater height found so far.

Below is the implementation of the above approach.

## C++

 `// CPP program to find number of elements` `// that can be seen from right side.` `#include ` `using` `namespace` `std;`   `int` `numberOfElements(``int` `height[], ``int` `n)` `{` `    ``int` `max_so_far = 0;` `    ``int` `count = 0;`   `    ``for` `(``int` `i = n - 1; i >= 0; i--) {` `        ``if` `(height[i] > max_so_far) {` `            ``max_so_far = height[i];` `            ``count++;` `        ``}` `    ``}` `    ``return` `count;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 6;` `    ``int` `height[] = { 4, 8, 2, 0, 0, 5 };` `    ``cout << numberOfElements(height, n);` `    ``return` `0;` `}`

## Java

 `// Java program to find number of elements` `// that can be seen from right side.` `import` `java.util.*;` `class` `Solution` `{` ` `  `static` `int` `numberOfElements(``int` `height[], ``int` `n)` `{` `    ``int` `max_so_far = ``0``;` `    ``int` `coun = ``0``;` ` `  `    ``for` `(``int` `i = n - ``1``; i >= ``0``; i--) {` `        ``if` `(height[i] > max_so_far) {` `            ``max_so_far = height[i];` `            ``coun++;` `        ``}` `    ``}` `    ``return` `coun;` `}`   `// Driver code ` `public` `static` `void` `main(String args[])` `{` `    ``int` `n = ``6``;` `    ``int` `height[] = { ``4``, ``8``, ``2``, ``0``, ``0``, ``5` `};` `    ``System.out.println( numberOfElements(height, n));` `  `  `}`   `}` `//contributed by Arnab Kundu`

## Python3

 `# Python3 program to find ` `# number of elements` `# that can be seen from right side`   `def` `numberOfElements(height, n):` `    `  `    ``max_so_far ``=` `0` `    ``coun ``=` `0` `    `  `    ``for` `i ``in` `range``(n``-``1``,``-``1``,``-``1``):` `        ``if` `height[i] > max_so_far:` `            ``max_so_far ``=` `height[i]` `            ``coun ``=` `coun ``+` `1` `    ``return` `coun`   `#Driver code` `if` `__name__``=``=``'__main__'``:` `    ``n ``=` `6` `    ``height ``=` `[``4``, ``8``, ``2``, ``0``, ``0``, ``5``]` `    ``print``(numberOfElements(height, n))` `    `  `# This code is contributed by ` `# Shashank_Sharma     `

## C#

 `// C# program to find number of elements` `// that can be seen from right side.` `using` `System;`   `class` `GFG` `{` `public` `static` `int` `numberOfElements(``int` `[]height,` `                                   ``int` `n)` `{` `    ``int` `max_so_far = 0;` `    ``int` `coun = 0;`   `    ``for` `(``int` `i = n - 1; i >= 0; i--)` `    ``{` `        ``if` `(height[i] > max_so_far)` `        ``{` `            ``max_so_far = height[i];` `            ``coun++;` `        ``}` `    ``}` `    ``return` `coun;` `}`   `// Driver code ` `public` `static` `void` `Main()` `{` `    ``int` `n = 6;` `    ``int` `[]height = { 4, 8, 2, 0, 0, 5 };` `    ``Console.WriteLine(numberOfElements(height, n));` `}` `}`   `// This code is contributed by Soumik `

## PHP

 `= 0; ``\$i``--) ` `    ``{` `        ``if` `(``\$height``[``\$i``] > ``\$max_so_far``) ` `        ``{` `            ``\$max_so_far` `= ``\$height``[``\$i``];` `            ``\$coun``++;` `        ``}` `    ``}` `    ``return` `\$coun``;` `}`   `// Driver code` `\$n` `= 6;` `\$height` `= ``array``(4, 8, 2, 0, 0, 5 );` `echo` `numberOfElements(``\$height``, ``\$n``);`   `// This code is contributed` `// by Akanksha Rai`

## Javascript

 ``

Output

`2`

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

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