Return the number of unique ways you can climb the staircase
$begingroup$
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
asked 5 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
asked 5 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
asked 5 mins ago
Maclean PintoMaclean Pinto
2887
$endgroup$
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
java programming-challenge
asked 5 mins ago
Maclean PintoMaclean Pinto
2887
asked 5 mins ago
Maclean PintoMaclean Pinto
2887
asked 5 mins ago
Maclean PintoMaclean Pinto
2887
asked 5 mins ago
Maclean PintoMaclean Pinto
2887
asked 5 mins ago
Maclean PintoMaclean Pinto
2887
2887
add a comment |
add a comment |
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