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# Number of strictly increasing Buildings from right with distinct Colors

• Difficulty Level : Medium
• Last Updated : 31 May, 2022

Given an integer and two integer arrays H[] and C[] of size where H[] stores the height of consecutive buildings and C[] stores the color codes for those building in which they are painted.
The task is to determine how many colors are visible at once from the view on the right i.e. right of the rightmost building.
Examples:

Input: K = 5, H[] = {5, 4, 3, 2, 3}, C[] = {1, 2, 3, 4, 5}
Output: Input: K = 5, H[] = {1, 2, 3, 4, 5}, C[] = {3, 3, 3, 3, 3}
Output:

Approach: On observing carefully, the above problem can be simplified to find the number of strictly increasing buildings from right with distinct colors.

1. Store the Last element of Height array in max variable.
2. Now in an array Arr, at position corresponding to the element at the last of the colour array store 1.
3. Now start traversing the Height array from n-2 to 0.
4. If we get element greater than max then store that variable in max and again in array Arr, at position correspond to the ith element in the colour array store 1

5. At last Count the number of 1’s present in the array Arr. It gives the total number of colour visible from the end.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach` `#include ``using` `namespace` `std;` `// Function to return the number of``// colors visible``int` `colourVisible(``int` `height[], ``int` `colour[], ``int` `K)``{``    ``int` `arr[K + 1] = { 0 }, visible = 0;` `    ``int` `max = height[K - 1];``    ``arr[colour[K - 1]] = 1;` `    ``for` `(``int` `i = K - 2; i >= 0; i--) {``        ``if` `(height[i] > max) {``            ``max = height[i];``            ``arr[colour[i]] = 1;``        ``}``    ``}` `    ``// Count the Number of 1's``    ``for` `(``int` `i = 1; i <= K; i++) {``        ``if` `(arr[i] == 1)``            ``visible++;``    ``}` `    ``return` `visible;``}` `// Driver code``int` `main()``{``    ``int` `height[] = { 3, 5, 1, 2, 3 };``    ``int` `colour[] = { 1, 2, 3, 4, 3 };``    ``int` `K = ``sizeof``(colour) / ``sizeof``(colour);` `    ``cout << colourVisible(height, colour, K);` `    ``return` `0;``}`

## Java

 `//Java  implementation of above approach` `import` `java.io.*;` `class` `GFG {``    ``// Function to return the number of``// colors visible``static` `int` `colourVisible(``int` `height[], ``int` `colour[], ``int` `K)``{``    ``int` `arr[]=``new` `int``[K + ``1``] ;``    ``int` `visible = ``0``;` `    ``int` `max = height[K - ``1``];``    ``arr[colour[K - ``1``]] = ``1``;` `    ``for` `(``int` `i = K - ``2``; i >= ``0``; i--) {``        ``if` `(height[i] > max) {``            ``max = height[i];``            ``arr[colour[i]] = ``1``;``        ``}``    ``}` `    ``// Count the Number of 1's``    ``for` `(``int` `i = ``1``; i <= K; i++) {``        ``if` `(arr[i] == ``1``)``            ``visible++;``    ``}` `    ``return` `visible;``}` `// Driver code``    ` `    ``public` `static` `void` `main (String[] args) {``    ` `    ``int` `height[] = { ``3``, ``5``, ``1``, ``2``, ``3` `};``    ``int` `colour[] = { ``1``, ``2``, ``3``, ``4``, ``3` `};``    ``int` `K = colour.length;` `    ``System.out.println (colourVisible(height, colour, K));``    ``}``}`

## Python3

 `# Python3 implementation of above approach` `# Function to return the number of``# colors visible``def` `colourVisible(height, colour, K):``    ``arr ``=` `[``0` `for` `i ``in` `range``(K ``+` `1``)]``    ``visible ``=` `0` `    ``max` `=` `height[K ``-` `1``]``    ``arr[colour[K ``-` `1``]] ``=` `1``    ` `    ``i ``=` `K ``-` `2``    ``while``(i >``=` `0``):``        ``if` `(height[i] > ``max``):``            ``max` `=` `height[i]``            ``arr[colour[i]] ``=` `1``        ``i ``-``=` `1``    ` `    ``# Count the Number of 1 complement``    ``for` `i ``in` `range``(``1``, K ``+` `1``, ``1``):``            ``if` `(arr[i] ``=``=` `1``):``                ``visible ``+``=` `1``    ` `    ``return` `visible` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``height ``=` `[``3``, ``5``, ``1``, ``2``, ``3``]``    ``colour ``=` `[``1``, ``2``, ``3``, ``4``, ``3``]``    ``K ``=` `len``(colour)` `    ``print``(colourVisible(height, colour, K))` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# implementation of above approach``using` `System;` `class` `GFG``{``// Function to return the number of``// colors visible``static` `int` `colourVisible(``int` `[]height,``                         ``int` `[]colour, ``int` `K)``{``    ``int` `[]arr=``new` `int``[K + 1] ;``    ``int` `visible = 0;` `    ``int` `max = height[K - 1];``    ``arr[colour[K - 1]] = 1;` `    ``for` `(``int` `i = K - 2; i >= 0; i--)``    ``{``        ``if` `(height[i] > max)``        ``{``            ``max = height[i];``            ``arr[colour[i]] = 1;``        ``}``    ``}` `    ``// Count the Number of 1's``    ``for` `(``int` `i = 1; i <= K; i++)``    ``{``        ``if` `(arr[i] == 1)``            ``visible++;``    ``}` `    ``return` `visible;``}` `// Driver code``static` `public` `void` `Main ()``{``    ``int` `[]height = { 3, 5, 1, 2, 3 };``    ``int` `[]colour = { 1, 2, 3, 4, 3 };``    ``int` `K = colour.Length;``    ` `    ``Console.WriteLine(colourVisible(height, colour, K));``}``}` `// This code is contributed by Sach_Code`

## PHP

 `= 0; ``\$i``--)``    ``{``        ``if` `(``\$height``[``\$i``] > ``\$max``)``        ``{``            ``\$max` `= ``\$height``[``\$i``];``            ``\$arr``[``\$colour``[``\$i``]] = 1;``        ``}``    ``}` `    ``// Count the Number of 1's``    ``for` `(``\$i` `= 1; ``\$i` `<= ``\$K``; ``\$i``++)``    ``{``        ``if` `(``\$arr``[``\$i``] == 1)``            ``\$visible``++;``    ``}` `    ``return` `\$visible``;``}` `// Driver code``\$height` `= ``array``( 3, 5, 1, 2, 3 );``\$colour` `= ``array``( 1, 2, 3, 4, 3 );``\$K` `= ``count``(``\$colour``);` `echo` `colourVisible(``\$height``, ``\$colour``, ``\$K``);` `// This code is contributed by mits``?>`

## Javascript

 ``

Output:

`2`

Time Complexity: O(K), where K is the size of colour array
Auxiliary Space: O(K), as an extra size of K is used

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