There is a street of length n and as we know it has two sides. Therefore a total of 2 * n spots are available. In each of these spots either a house or an office can be built with following 2 restrictions:
1. No two offices on the same side of the street can be adjacent.
2. No two offices on different sides of the street can be exactly opposite to each other i.e. they can’t overlook each other.
There are no restrictions on building houses and each spot must either have a house or office.
Given length of the street n, find total number of ways to build the street.
Examples:
Input : 2 Output : 7 Please see below diagram for explanation. Input : 3 Output : 17
Following image depicts the 7 possible ways for building the street with N = 2
Ways for building street with length 1
with 2 houses: (H H) = {1}
with 1 office and 1 house: (O H) (H O) = {2}
(O O) is not allowed according to the problem statement.
Total = 1 + 2 = 3
For length = 2,
with 2 houses: (H H) can be added to all
the cases of length 1:
(H H) (H H) (H H)
(H H) (O H) (H O) = {3}
with 1 office and 1 house:
(O H) and (H O) can both be added to all
the cases of length 1 with last row (H H)
(O H) (H O)
(H H) (H H) = {2}
when last row of a case of length 1
contains 1 office, it can only be
extended in one way to build an office
in row 2.
(H O) (O H)
(O H) (H O) = {2}
(O H) (H O) (O O)
(O H) (H O) (O H) etc are not allowed.
Total = 3 + 2 + 2 = 7
Since the problem can be solved by finding solution for smaller subproblems and then extending the same logic, it can be solved using dynamic programming. We move in steps of one unit length. For each row we have two options:
Build houses in both the spots
Build one house and one office
The first one can be done without any constraints. There is one way of building houses in both the spots at length i. So total ways using this choice = total ways for length i – 1.
For the second choice, if row (i-1) had houses in both spots we have two ways of building a office i.e. (H O) and (O H)
if row(i-1) had an office in one of its two spots we only have one way to build an office in row i.If prev row had (O H) curr row would have (H O) and similarly for prev row = (H O) curr row = (O H).
From the above logic, total ways with this choice = 2 * (choice1(i-1)) + choice2(i-1)
We will build a 2D dp for this.
dp[0][i] indicates choice1 and dp[1][i] indicates choice2 for row i.
Below is the implementation of above idea :
C++
// C++ program to count ways to build street // under given constraints#include <bits/stdc++.h>using namespace std;// function to count ways of building// a street of n rowslong countWays(int n){ long dp[2][n + 1]; // base case dp[0][1] = 1; dp[1][1] = 2; for (int i = 2; i <= n; i++) { // ways of building houses in both // the spots of ith row dp[0][i] = dp[0][i - 1] + dp[1][i - 1]; // ways of building an office in one of // the two spots of ith row dp[1][i] = dp[0][i - 1] * 2 + dp[1][i - 1]; } // total ways for n rows return dp[0][n] + dp[1][n];}// driver program for checking above functionint main(){ int n = 5; cout << "Total no of ways with n = " << n << " are: " << countWays(n) << endl;} |
Java
// Java program to count ways to build street // under given constraintspublic class GFG {// function to count ways of building// a street of n rows static long countWays(int n) { long dp[][] = new long[2][n + 1]; // base case dp[0][1] = 1; dp[1][1] = 2; for (int i = 2; i <= n; i++) { // ways of building houses in both // the spots of ith row dp[0][i] = dp[0][i - 1] + dp[1][i - 1]; // ways of building an office in one of // the two spots of ith row dp[1][i] = dp[0][i - 1] * 2 + dp[1][i - 1]; } // total ways for n rows return dp[0][n] + dp[1][n]; }// driver program for checking above function public static void main(String[] args) { int n = 5; System.out.print("Total no of ways with n = " + n + " are: " + countWays(n)); }}/*This code is contributed by PrinciRaj1992*/ |
Python3
# Python3 program to count ways to build # street under given constraints # function to count ways of building # a street of n rows def countWays(n) : dp = [[0] * (n + 1) for i in range(2)] # base case dp[0][1] = 1 dp[1][1] = 2 for i in range(2, n + 1) : # ways of building houses in both # the spots of ith row dp[0][i] = dp[0][i - 1] + dp[1][i - 1] # ways of building an office in one of # the two spots of ith row dp[1][i] = (dp[0][i - 1] * 2 + dp[1][i - 1]) # total ways for n rows return dp[0][n] + dp[1][n] # Driver Code if __name__ == "__main__" : n = 5 print("Total no of ways with n =", n, "are:", countWays(n)) # This code is contributed by Ryuga |
C#
// C# program to count ways to build street // under given constraints using System;public class GFG{// function to count ways of building // a street of n rows static long countWays(int n) { long [,]dp = new long[2 , n + 1]; // base case dp[0,1] = 1; dp[1,1] = 2; for (int i = 2; i <= n; i++) { // ways of building houses in both // the spots of ith row dp[0 , i] = dp[0 , i - 1] + dp[1 , i - 1]; // ways of building an office in one of // the two spots of ith row dp[1 , i] = dp[0 , i - 1] * 2 + dp[1 , i - 1]; } // total ways for n rows return dp[0 , n] + dp[1 , n]; } // driver program for checking above function public static void Main() { int n = 5; Console.Write("Total no of ways with n = " + n + " are: " + countWays(n)); } } /*This code is contributed by PrinciRaj1992*/ |
PHP
<?php// PHP program to count ways to build street // under given constraints// function to count ways of building// a street of n rowsfunction countWays($n){ // base case $dp[0][1] = 1; $dp[1][1] = 2; for ($i = 2; $i <= $n; $i++) { // ways of building houses in both // the spots of ith row $dp[0][$i] = $dp[0][$i - 1] + $dp[1][$i - 1]; // ways of building an office in one of // the two spots of ith row $dp[1][$i] = $dp[0][$i - 1] * 2 + $dp[1][$i - 1]; } // total ways for n rows return $dp[0][$n] + $dp[1][$n];} // Driver Code $n = 5; echo "Total no of ways with n = ",$n, " are: " ,countWays($n),"\n";// This code is contributed by jit_t?> |
Javascript
<script>// Javascript program to count ways to // build street under given constraints // Function to count ways of building// a street of n rowsfunction countWays(n){ let dp = new Array(2); for(let i = 0; i < 2; i++) { dp[i] = new Array(n + 1); } // Base case dp[0][1] = 1; dp[1][1] = 2; for(let i = 2; i <= n; i++) { // Ways of building houses in both // the spots of ith row dp[0][i] = dp[0][i - 1] + dp[1][i - 1]; // Ways of building an office in one of // the two spots of ith row dp[1][i] = dp[0][i - 1] * 2 + dp[1][i - 1]; } // Total ways for n rows return dp[0][n] + dp[1][n];}// Driver codelet n = 5;document.write("Total no of ways with n = " + n + " are: " + countWays(n)); // This code is contributed by suresh07</script> |
Total no of ways with n = 5 are: 99
Time Complexity: O(N)
Auxiliary Space: O(N)
Another Method
Suppose you have A(n-1) and A(n-2) ,where A(i) depicts ways for 2*i buildings, then :
A(n) = (Number of ways you have (H H) in A(n-1) )*3 + (Number of ways you have (O H) in A(n-1))*2
How these cases arise ? Simply you can think of making 2 states as examined in above explanation.
Number of ways you have (H H) in A(n-1) = Fix H H in (n-1)th position, this is equal to number of ways of n-2 position (fixing n-1 to H H).
Number of ways you have (O H) in A(n-1) = (Number of ways in n-1 ) – (Number of ways of H H in A(n-1) position).
So A(n) = 3*A(n-2) + 2*(A(n-1)-A(n-2)) = 2*A(n-1)+A(n-2)
C++
// C++ program of above approach#include <iostream>using namespace std;// Program to count waysint countways(long long n){ long long A[n+1]; A[0] = 1; A[1] = 3; A[2] = 7; // Iterate from 2 to n for(int i=2;i<=n;i++) { A[i] = 2*A[i-1]+A[i-2]; } return A[n];}// Driver codeint main() { int n = 5; // Count Ways cout << countways(5) << endl; return 0;} |
Java
// Java program of above approachimport java.io.*;class GFG{ // Program to count waysstatic int countways(int n){ int [] A = new int[n+1]; A[0] = 1; A[1] = 3; A[2] = 7; // Iterate from 2 to n for(int i = 2; i <= n; i++) { A[i] = 2 * A[i - 1] + A[i - 2]; } return A[n];}// Driver codepublic static void main(String[] args) { int n = 5; // Count Ways System.out.println(countways(5));}}// This code is contributed by Ankita saini |
Python3
# Python3 program of above approach# Program to count waysdef countways(n): A = [0 for i in range(n + 2)] A[0] = 1 A[1] = 3 A[2] = 7 # Iterate from 2 to n for i in range(2, n + 1): A[i] = 2 * A[i - 1] + A[i - 2] return A[n]# Driver coden = 5# Count Waysprint(countways(5))# This code is contributed by Sanjit_Prasad |
C#
// C# program of above approachusing System;class GFG{ // Program to count waysstatic int countways(int n){ int [] A = new int[n+1]; A[0] = 1; A[1] = 3; A[2] = 7; // Iterate from 2 to n for(int i = 2; i <= n; i++) { A[i] = 2 * A[i - 1] + A[i - 2]; } return A[n];}// Driver codepublic static void Main(){ int n = 5; // Count Ways Console.Write(countways(n));}}// This code is contributed by bunnyram19. |
Javascript
<script>// Javascript program of above approach// Program to count waysfunction countways(n){ let A = new Array(n+1).fill(0); A[0] = 1; A[1] = 3; A[2] = 7; // Iterate from 2 to n for(let i=2;i<=n;i++) { A[i] = 2*A[i-1]+A[i-2]; } return A[n];}// driver program let n = 5; // Count Ways document.write(countways(5)) </script> |
99
Time Complexity: O(N)
Auxiliary Space: O(N)
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