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: 3
Input: K = 5, H[] = {1, 2, 3, 4, 5}, C[] = {3, 3, 3, 3, 3}
Output: 1
Approach: On observing carefully, the above problem can be simplified to find the number of strictly increasing buildings from right with distinct colors.
- Store the Last element of Height array in max variable.
- Now in an array Arr, at position corresponding to the element at the last of the colour array store 1.
- Now start traversing the Height array from n-2 to 0.
- 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.
- 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 <bits/stdc++.h>using namespace std;// Function to return the number of// colors visibleint 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 codeint main(){ int height[] = { 3, 5, 1, 2, 3 }; int colour[] = { 1, 2, 3, 4, 3 }; int K = sizeof(colour) / sizeof(colour[0]); 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 visibledef 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 codeif __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 approachusing 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
<?php// PHP implementation of above approach// Function to return the number of// colors visiblefunction colourVisible($height, $colour, $K){ $arr = array_fill(0, $K + 1, 0); $visible = 0; $max = $height[$K - 1]; $arr[$colour[$K - 1]] = 1; for ($i = $K - 2; $i >= 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
<script>// Javascript implementation of above approach// Function to return the number of// colors visiblefunction colourVisible(height, colour, K){ var arr = Array(K+1).fill(0), visible = 0; var max = height[K - 1]; arr[colour[K - 1]] = 1; for (var i = K - 2; i >= 0; i--) { if (height[i] > max) { max = height[i]; arr[colour[i]] = 1; } } // Count the Number of 1's for (var i = 1; i <= K; i++) { if (arr[i] == 1) visible++; } return visible;}// Driver codevar height = [ 3, 5, 1, 2, 3 ];var colour = [ 1, 2, 3, 4, 3 ];var K = colour.length;document.write( colourVisible(height, colour, K));</script> |
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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