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Return the number of unique ways you can climb the staircase

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$begingroup$

There exists a staircase with N steps, and you can climb up either 1 or 2 steps at a time. Given N, write a function that returns the number of unique ways you can climb the staircase. The order of the steps matters.

For example, if N is 4, then there are 5 unique ways:

1, 1, 1, 1
2, 1, 1
1, 2, 1
1, 1, 2
2, 2

What if, instead of being able to climb 1 or 2 steps at a time, you could climb any number from a set of positive integers X? For example, if X = 1, 3, 5, you could climb 1, 3, or 5 steps at a time.

public class DailyCodingProblem12 
public static void main(String args[]) 
 int[] X = 1, 2 ;
 int N = 4;
 int result = solution(N, X);
 System.out.println(result);

 int[] X1 = 1, 2, 3 ;
 N = 4;
 result = solution(N, X1);
 System.out.println(result);

 int[] X2 = 1, 2, 3 ;
 N = 3;
 result = solution(N, X2);
 System.out.println(result);


static int solution(int N, int[] X) 
 int[] memory = new int[N + 1];
 memory[0] = 1;
 memory[1] = 1;
 return noOfWays(N, X, memory);


static int noOfWays(int N, int[] X, int[] memory) 
 if (memory[N] != 0) 
 return memory[N];
 
 int noOfWays = 0;
 for (int i = 0; i < X.length && (N - X[i] >= 0); i++) 
 memory[N - X[i]] = noOfWays(N - X[i], X, memory);
 noOfWays += memory[N - X[i]];
 
 return noOfWays;

How do I improve my solution? There is a way to save more space?

java programming-challenge

share|improve this question

asked 11 mins ago

Maclean PintoMaclean Pinto

2887

$endgroup$

add a comment | 

0

$begingroup$

There exists a staircase with N steps, and you can climb up either 1 or 2 steps at a time. Given N, write a function that returns the number of unique ways you can climb the staircase. The order of the steps matters.

For example, if N is 4, then there are 5 unique ways:

1, 1, 1, 1
2, 1, 1
1, 2, 1
1, 1, 2
2, 2

What if, instead of being able to climb 1 or 2 steps at a time, you could climb any number from a set of positive integers X? For example, if X = 1, 3, 5, you could climb 1, 3, or 5 steps at a time.

public class DailyCodingProblem12 
public static void main(String args[]) 
 int[] X = 1, 2 ;
 int N = 4;
 int result = solution(N, X);
 System.out.println(result);

 int[] X1 = 1, 2, 3 ;
 N = 4;
 result = solution(N, X1);
 System.out.println(result);

 int[] X2 = 1, 2, 3 ;
 N = 3;
 result = solution(N, X2);
 System.out.println(result);


static int solution(int N, int[] X) 
 int[] memory = new int[N + 1];
 memory[0] = 1;
 memory[1] = 1;
 return noOfWays(N, X, memory);


static int noOfWays(int N, int[] X, int[] memory) 
 if (memory[N] != 0) 
 return memory[N];
 
 int noOfWays = 0;
 for (int i = 0; i < X.length && (N - X[i] >= 0); i++) 
 memory[N - X[i]] = noOfWays(N - X[i], X, memory);
 noOfWays += memory[N - X[i]];
 
 return noOfWays;

How do I improve my solution? There is a way to save more space?

java programming-challenge

share|improve this question

asked 11 mins ago

Maclean PintoMaclean Pinto

2887

$endgroup$

add a comment | 

$begingroup$

There exists a staircase with N steps, and you can climb up either 1 or 2 steps at a time. Given N, write a function that returns the number of unique ways you can climb the staircase. The order of the steps matters.

For example, if N is 4, then there are 5 unique ways:

1, 1, 1, 1
2, 1, 1
1, 2, 1
1, 1, 2
2, 2

What if, instead of being able to climb 1 or 2 steps at a time, you could climb any number from a set of positive integers X? For example, if X = 1, 3, 5, you could climb 1, 3, or 5 steps at a time.

public class DailyCodingProblem12 
public static void main(String args[]) 
 int[] X = 1, 2 ;
 int N = 4;
 int result = solution(N, X);
 System.out.println(result);

 int[] X1 = 1, 2, 3 ;
 N = 4;
 result = solution(N, X1);
 System.out.println(result);

 int[] X2 = 1, 2, 3 ;
 N = 3;
 result = solution(N, X2);
 System.out.println(result);


static int solution(int N, int[] X) 
 int[] memory = new int[N + 1];
 memory[0] = 1;
 memory[1] = 1;
 return noOfWays(N, X, memory);


static int noOfWays(int N, int[] X, int[] memory) 
 if (memory[N] != 0) 
 return memory[N];
 
 int noOfWays = 0;
 for (int i = 0; i < X.length && (N - X[i] >= 0); i++) 
 memory[N - X[i]] = noOfWays(N - X[i], X, memory);
 noOfWays += memory[N - X[i]];
 
 return noOfWays;

How do I improve my solution? There is a way to save more space?

java programming-challenge

share|improve this question

asked 11 mins ago

Maclean PintoMaclean Pinto

2887

$endgroup$

1, 1, 1, 1
2, 1, 1
1, 2, 1
1, 1, 2
2, 2

public class DailyCodingProblem12 
public static void main(String args[]) 
 int[] X = 1, 2 ;
 int N = 4;
 int result = solution(N, X);
 System.out.println(result);

 int[] X1 = 1, 2, 3 ;
 N = 4;
 result = solution(N, X1);
 System.out.println(result);

 int[] X2 = 1, 2, 3 ;
 N = 3;
 result = solution(N, X2);
 System.out.println(result);


static int solution(int N, int[] X) 
 int[] memory = new int[N + 1];
 memory[0] = 1;
 memory[1] = 1;
 return noOfWays(N, X, memory);


static int noOfWays(int N, int[] X, int[] memory) 
 if (memory[N] != 0) 
 return memory[N];
 
 int noOfWays = 0;
 for (int i = 0; i < X.length && (N - X[i] >= 0); i++) 
 memory[N - X[i]] = noOfWays(N - X[i], X, memory);
 noOfWays += memory[N - X[i]];
 
 return noOfWays;

share|improve this question

asked 11 mins ago

Maclean PintoMaclean Pinto

2887

share|improve this question

asked 11 mins ago

Maclean PintoMaclean Pinto

2887

asked 11 mins ago

Maclean PintoMaclean Pinto

2887

asked 11 mins ago

Maclean PintoMaclean Pinto

2887