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# Minimize cost of painting N houses such that adjacent houses have different colors

#### Given an integer N and a 2D arraycost[][3], where cost[i][0], cost[i][1], and cost[i][2] is the cost of painting ith house with colors red, blue, and green respectively, the task is to find the minimum cost to paint all the houses such that no two adjacent houses have the same color.

Examples:

Input: N = 3, cost[][3] = {{14, 2, 11}, {11, 14, 5}, {14, 3, 10}}
Output: 10
Explanation:
Paint house 0 as blue. Cost = 2. Paint house 1 as green. Cost = 5. Paint house 2 as blue. Cost = 3.
Therefore, the total cost = 2 + 5 + 3 = 10.

Input: N = 2, cost[][3] = {{1, 2, 3}, {1, 4, 6}}
Output: 3

Naive Approach: The simplest approach to solve the given problem is to generate all possible ways of coloring all the houses with the colors red, blue, and green and find the minimum cost among all the possible combinations such that no two adjacent houses have the same colors.
Time Complexity: (3N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by using Dynamic Programming as there are overlapping subproblems that can be stored to minimize the number of recursive calls. The idea is to find the minimum cost of painting the current house by any color on the basis of the minimum cost of the other two colors of previously colored houses. Follow the steps below to solve the given problem:

Follow the steps below to solve the problem:

• Create an auxiliary 2D dp[][3] array to store the minimum cost of previously colored houses.
• Initialize dp[0][0], dp[0][1], and dp[0][2] as the cost of cost[i][0], cost[i][1], and cost[i][2] respectively.
• Traverse the given array cost[][3] over the range [1, N] and update the cost of painting the current house with colors red, blue, and green with the minimum of the cost other two colors in dp[i][0], dp[i][1], and dp[i][2] respectively.
• After completing the above steps, print the minimum of dp[N – 1][0], dp[N – 1][1], and dp[N – 1][2] as the minimum cost of painting all the houses with different adjacent colors.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find the minimum cost of` `// coloring the houses such that no two` `// adjacent houses has the same color` `int` `minCost(vector >& costs,` `            ``int` `N)` `{` `    ``// Corner Case` `    ``if` `(N == 0)` `        ``return` `0;`   `    ``// Auxiliary 2D dp array` `    ``vector > dp(` `        ``N, vector<``int``>(3, 0));`   `    ``// Base Case` `    ``dp[0][0] = costs[0][0];` `    ``dp[0][1] = costs[0][1];` `    ``dp[0][2] = costs[0][2];`   `    ``for` `(``int` `i = 1; i < N; i++) {`   `        ``// If current house is colored` `        ``// with red the take min cost of` `        ``// previous houses colored with` `        ``// (blue and green)` `        ``dp[i][0] = min(dp[i - 1][1],` `                       ``dp[i - 1][2])` `                   ``+ costs[i][0];`   `        ``// If current house is colored` `        ``// with blue the take min cost of` `        ``// previous houses colored with` `        ``// (red and green)` `        ``dp[i][1] = min(dp[i - 1][0],` `                       ``dp[i - 1][2])` `                   ``+ costs[i][1];`   `        ``// If current house is colored` `        ``// with green the take min cost of` `        ``// previous houses colored with` `        ``// (red and blue)` `        ``dp[i][2] = min(dp[i - 1][0],` `                       ``dp[i - 1][1])` `                   ``+ costs[i][2];` `    ``}`   `    ``// Print the min cost of the` `    ``// last painted house` `    ``cout << min(dp[N - 1][0],` `                ``min(dp[N - 1][1],` `                    ``dp[N - 1][2]));` `}`   `// Driver Code` `int` `main()` `{` `    ``vector > costs{ { 14, 2, 11 },` `                                ``{ 11, 14, 5 },` `                                ``{ 14, 3, 10 } };` `    ``int` `N = costs.size();`   `    ``// Function Call` `    ``minCost(costs, N);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.io.*;` `import` `java.lang.*;` `import` `java.util.*;`   `class` `GFG {`   `  ``// Function to find the minimum cost of` `  ``// coloring the houses such that no two` `  ``// adjacent houses has the same color` `  ``static` `void` `minCost(``int` `costs[][], ``int` `N)` `  ``{`   `    ``// Corner Case` `    ``if` `(N == ``0``)` `      ``return``;`   `    ``// Auxiliary 2D dp array` `    ``int` `dp[][] = ``new` `int``[N][``3``];`   `    ``// Base Case` `    ``dp[``0``][``0``] = costs[``0``][``0``];` `    ``dp[``0``][``1``] = costs[``0``][``1``];` `    ``dp[``0``][``2``] = costs[``0``][``2``];`   `    ``for` `(``int` `i = ``1``; i < N; i++) {`   `      ``// If current house is colored` `      ``// with red the take min cost of` `      ``// previous houses colored with` `      ``// (blue and green)` `      ``dp[i][``0``] = Math.min(dp[i - ``1``][``1``], dp[i - ``1``][``2``])` `        ``+ costs[i][``0``];`   `      ``// If current house is colored` `      ``// with blue the take min cost of` `      ``// previous houses colored with` `      ``// (red and green)` `      ``dp[i][``1``] = Math.min(dp[i - ``1``][``0``], dp[i - ``1``][``2``])` `        ``+ costs[i][``1``];`   `      ``// If current house is colored` `      ``// with green the take min cost of` `      ``// previous houses colored with` `      ``// (red and blue)` `      ``dp[i][``2``] = Math.min(dp[i - ``1``][``0``], dp[i - ``1``][``1``])` `        ``+ costs[i][``2``];` `    ``}`   `    ``// Print the min cost of the` `    ``// last painted house` `    ``System.out.println(` `      ``Math.min(dp[N - ``1``][``0``],` `               ``Math.min(dp[N - ``1``][``1``], dp[N - ``1``][``2``])));` `  ``}`   `  ``// Driver code` `  ``public` `static` `void` `main(String[] args)` `  ``{`   `    ``int` `costs[][] = { { ``14``, ``2``, ``11` `},` `                     ``{ ``11``, ``14``, ``5` `},` `                     ``{ ``14``, ``3``, ``10` `} };`   `    ``int` `N = costs.length;`   `    ``// Function Call` `    ``minCost(costs, N);` `  ``}` `}`   `// This code is contributed by Kingash.`

## Python3

 `# Python 3 program for the above approach`   `# Function to find the minimum cost of` `# coloring the houses such that no two` `# adjacent houses has the same color` `def` `minCost(costs, N):` `  `  `    ``# Corner Case` `    ``if` `(N ``=``=` `0``):` `        ``return` `0`   `    ``# Auxiliary 2D dp array` `    ``dp ``=` `[[``0` `for` `i ``in` `range``(``3``)] ``for` `j ``in` `range``(``3``)]`   `    ``# Base Case` `    ``dp[``0``][``0``] ``=` `costs[``0``][``0``]` `    ``dp[``0``][``1``] ``=` `costs[``0``][``1``]` `    ``dp[``0``][``2``] ``=` `costs[``0``][``2``]`   `    ``for` `i ``in` `range``(``1``, N, ``1``):` `      `  `        ``# If current house is colored` `        ``# with red the take min cost of` `        ``# previous houses colored with` `        ``# (blue and green)` `        ``dp[i][``0``] ``=` `min``(dp[i ``-` `1``][``1``], dp[i ``-` `1``][``2``]) ``+` `costs[i][``0``]`   `        ``# If current house is colored` `        ``# with blue the take min cost of` `        ``# previous houses colored with` `        ``# (red and green)` `        ``dp[i][``1``] ``=` `min``(dp[i ``-` `1``][``0``], dp[i ``-` `1``][``2``]) ``+` `costs[i][``1``]`   `        ``# If current house is colored` `        ``# with green the take min cost of` `        ``# previous houses colored with` `        ``# (red and blue)` `        ``dp[i][``2``] ``=` `min``(dp[i ``-` `1``][``0``], dp[i ``-` `1``][``1``]) ``+` `costs[i][``2``]`   `    ``# Print the min cost of the` `    ``# last painted house` `    ``print``(``min``(dp[N ``-` `1``][``0``], ``min``(dp[N ``-` `1``][``1``],dp[N ``-` `1``][``2``])))`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``costs ``=` `[[``14``, ``2``, ``11``],` `             ``[``11``, ``14``, ``5``],` `             ``[``14``, ``3``, ``10``]]` `    ``N ``=` `len``(costs)` `    `  `    ``# Function Call` `    ``minCost(costs, N)` `    `  `    ``# This code is contributed by ipg2016107.`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{`   `  ``// Function to find the minimum cost of` `  ``// coloring the houses such that no two` `  ``// adjacent houses has the same color` `  ``static` `int` `minCost(List>costs,` `                     ``int` `N)` `  ``{` `    ``// Corner Case` `    ``if` `(N == 0)` `      ``return` `0;`   `    ``// Auxiliary 2D dp array` `    ``List<``int``> temp = ``new` `List<``int``>();` `    ``for``(``int` `i=0;i<3;i++)` `      ``temp.Add(0);` `    ``List> dp = ``new` `List>();` `    ``for``(``int` `i=0;i>costs = ``new` `List>();` `    ``costs.Add(``new` `List<``int``>(){14, 2, 11});` `    ``costs.Add(``new` `List<``int``>(){11, 14, 5 });` `    ``costs.Add(``new` `List<``int``>(){14, 3, 10 });` `    ``int` `N = 3;`   `    ``// Function Call` `    ``Console.WriteLine((``int``)(minCost(costs, N)));` `  ``}` `}`   `// This code is contributed by bgangwar59.`

## Javascript

 ``

Output

`10`

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

Efficient Approach using Dp with Constant Space: If we see the efficient approach that’s discussed above we can observe one thing for painting current house you only required information of paint house just before that is for painting ith house  you only required information for i-1th house so we rather than making a dp of size 3*N we can just make 6 variable that is 3 for current and 3 for last and space complexity will reduce to O(1)

Below is the implementation of the above approach:

## C++

 `#include ` `using` `namespace` `std;`   `/*package whatever //do not write package name here */`   `// c++ program for the above approach`   `// Function to find the minimum cost of` `// coloring the houses such that no two` `// adjacent houses has the same color` `void` `minCost(vector> costs, ``int` `N)` `{`   `    ``// Corner Case` `    ``if` `(N == 0)` `        ``return``;`   `    ``// Base Case` `    ``int` `previous_red = costs[0][0];` `    ``int` `previous_blue = costs[0][1];` `    ``int` `previous_green = costs[0][2];` `    ``int` `current_red;` `    ``int` `current_blue;` `    ``int` `current_green;` `    ``for` `(``int` `i = 1; i < N; i++) {`   `        ``// for coloring current house to red we will` `        ``// take previous blue and green` `        ``current_red = min(previous_blue, previous_green) + costs[i][0];`   `        ``// for coloring current house to blue we will` `        ``// take previous red and green` `        ``current_blue = min(previous_red, previous_green) + costs[i][1];`   `        ``// for coloring current house to green we will` `        ``// take previous red and blue` `        ``current_green = min(previous_red, previous_blue) + costs[i][2];`   `        ``// setting previous value to current for next` `        ``// iteration` `        ``previous_red = current_red;` `        ``previous_blue = current_blue;` `        ``previous_green = current_green;` `    ``}`   `    ``// Print the min cost of the` `    ``// last painted house` `    ``cout << (min(previous_red, min(previous_blue,previous_green)));` `}`   `int` `main(){` `    ``vector> costs = { { 14, 2, 11 },` `                      ``{ 11, 14, 5 },` `                      ``{ 14, 3, 10 } };`   `    ``int` `N = costs.size();`   `    ``// Function Call` `    ``minCost(costs, N);` `    ``return` `0;` `}`   `// This code is contributed by Nidhi goel.`

## Java

 `/*package whatever //do not write package name here */`   `// Java program for the above approach` `import` `java.io.*;` `import` `java.lang.*;` `import` `java.util.*;`   `class` `GFG {`   `    ``// Function to find the minimum cost of` `    ``// coloring the houses such that no two` `    ``// adjacent houses has the same color` `    ``static` `void` `minCost(``int` `costs[][], ``int` `N)` `    ``{`   `        ``// Corner Case` `        ``if` `(N == ``0``)` `            ``return``;`   `        ``// Base Case` `        ``int` `previous_red = costs[``0``][``0``];` `        ``int` `previous_blue = costs[``0``][``1``];` `        ``int` `previous_green = costs[``0``][``2``];` `        ``int` `current_red;` `        ``int` `current_blue;` `        ``int` `current_green;` `        ``for` `(``int` `i = ``1``; i < N; i++) {`   `            ``// for coloring current house to red we will` `            ``// take previous blue and green` `            ``current_red` `                ``= Math.min(previous_blue, previous_green)` `                  ``+ costs[i][``0``];`   `            ``// for coloring current house to blue we will` `            ``// take previous red and green` `            ``current_blue` `                ``= Math.min(previous_red, previous_green)` `                  ``+ costs[i][``1``];`   `            ``// for coloring current house to green we will` `            ``// take previous red and blue` `            ``current_green` `                ``= Math.min(previous_red, previous_blue)` `                  ``+ costs[i][``2``];`   `            ``// setting previous value to current for next` `            ``// iteration` `            ``previous_red = current_red;` `            ``previous_blue = current_blue;` `            ``previous_green = current_green;` `        ``}`   `        ``// Print the min cost of the` `        ``// last painted house` `        ``System.out.println(` `            ``Math.min(previous_red,` `                     ``Math.min(previous_blue,previous_green)));` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``int` `costs[][] = { { ``14``, ``2``, ``11` `},` `                          ``{ ``11``, ``14``, ``5` `},` `                          ``{ ``14``, ``3``, ``10` `} };`   `        ``int` `N = costs.length;`   `        ``// Function Call` `        ``minCost(costs, N);` `    ``}` `}`   `// This code is contributed by Vikas Bishnoi.`

## Python3

 `#Python program for the above approach` `def` `minCost(costs, N):` ` ``# Corner Case` `  ``if` `N ``=``=` `0``:` `      ``return`   `  ``# Base Case` `  ``previous_red ``=` `costs[``0``][``0``]` `  ``previous_blue ``=` `costs[``0``][``1``]` `  ``previous_green ``=` `costs[``0``][``2``]` `  ``current_red ``=` `0` `  ``current_blue ``=` `0` `  ``current_green ``=` `0` `  ``for` `i ``in` `range``(``1``, N):`   `      ``# for coloring current house to red we will` `      ``# take previous blue and green` `      ``current_red ``=` `min``(previous_blue, previous_green) ``+` `costs[i][``0``]`   `      ``# for coloring current house to blue we will` `      ``# take previous red and green` `      ``current_blue ``=` `min``(previous_red, previous_green) ``+` `costs[i][``1``]`   `      ``# for coloring current house to green we will` `      ``# take previous red and blue` `      ``current_green ``=` `min``(previous_red, previous_blue) ``+` `costs[i][``2``]`   `      ``# setting previous value to current for next iteration` `      ``previous_red ``=` `current_red` `      ``previous_blue ``=` `current_blue` `      ``previous_green ``=` `current_green`   `  ``# Print the min cost of the last painted house` `  ``print``(``min``(previous_red, ``min``(previous_blue, previous_green)))` `costs ``=` `[[``14``, ``2``, ``11``], [``11``, ``14``, ``5``], [``14``, ``3``, ``10``]]` `N ``=` `len``(costs)` `# Function Call` `minCost(costs, N)`

## Javascript

 `// Javascript program for the above approach` `function` `minCost(costs, N) ` `{`   `// Corner Case` `if` `(N == 0) ` `{` `return``;` `}`   `// Base Case` `let previous_red = costs[0][0];` `let previous_blue = costs[0][1];` `let previous_green = costs[0][2];` `let current_red = 0;` `let current_blue = 0;` `let current_green = 0;` `for` `(let i = 1; i < N; i++) {`   `    ``// for coloring current house to red we will` `    ``// take previous blue and green` `    ``current_red = Math.min(previous_blue, previous_green) + costs[i][0];`   `    ``// for coloring current house to blue we will` `    ``// take previous red and green` `    ``current_blue = Math.min(previous_red, previous_green) + costs[i][1];`   `    ``// for coloring current house to green we will` `    ``// take previous red and blue` `    ``current_green = Math.min(previous_red, previous_blue) + costs[i][2];`   `    ``// setting previous value to current for next iteration` `    ``previous_red = current_red;` `    ``previous_blue = current_blue;` `    ``previous_green = current_green;` `}`   `// Print the min cost of the last painted house` `console.log(Math.min(previous_red, Math.min(previous_blue, previous_green)));` `}`   `let costs = [[14, 2, 11], [11, 14, 5], [14, 3, 10]];` `let N = costs.length;`   `// Function Call` `minCost(costs, N);`

## C#

 `using` `System;`   `public` `class` `GFG` `{` `    ``// Function to find the minimum cost of` `    ``// coloring the houses such that no two` `    ``// adjacent houses has the same color` `    ``static` `void` `minCost(``int``[][] costs, ``int` `N)` `    ``{`   `        ``// Corner Case` `        ``if` `(N == 0)` `            ``return``;`   `        ``// Base Case` `        ``int` `previous_red = costs[0][0];` `        ``int` `previous_blue = costs[0][1];` `        ``int` `previous_green = costs[0][2];` `        ``int` `current_red;` `        ``int` `current_blue;` `        ``int` `current_green;` `        ``for` `(``int` `i = 1; i < N; i++)` `        ``{`   `            ``// for coloring current house to red we will` `            ``// take previous blue and green` `            ``current_red = Math.Min(previous_blue, previous_green) + costs[i][0];`   `            ``// for coloring current house to blue we will` `            ``// take previous red and green` `            ``current_blue = Math.Min(previous_red, previous_green) + costs[i][1];`   `            ``// for coloring current house to green we will` `            ``// take previous red and blue` `            ``current_green = Math.Min(previous_red, previous_blue) + costs[i][2];`   `            ``// setting previous value to current for next` `            ``// iteration` `            ``previous_red = current_red;` `            ``previous_blue = current_blue;` `            ``previous_green = current_green;` `        ``}`   `        ``// Print the min cost of the` `        ``// last painted house` `        ``Console.WriteLine(Math.Min(previous_red, Math.Min(previous_blue, previous_green)));` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{`   `        ``int``[][] costs = { ``new` `int``[] { 14, 2, 11 },` `                        ``new` `int``[] { 11, 14, 5 },` `                        ``new` `int``[] { 14, 3, 10 } };`   `        ``int` `N = costs.Length;`   `        ``// Function Call` `        ``minCost(costs, N);` `    ``}` `}`

Output

`10`

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

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