Objective: A child is climbing up a staircase with n steps, and can hop either 1 step, 2 steps, or 3 steps at a time. Implement a method to count how many possible ways the child can jump up the stairs.
Example:
Number of stairs : 3
Number of ways = 4 ( {1,1,1}, {1,2}, {2,1}, {3} )
Approach:
- Say child has to take n steps.
- At every step the child has 3 options, to take 1 step, 2 step or 3 steps.
- So if child take 1 step then find the number of ways to complete n-1 steps +1.
- Similarly if child take 2 steps then find the number of ways to complete n-2 steps +1.
- If child take 3 step then find the number of ways to complete n-3 steps +1.
- So total number of ways to complete n steps = No of ways to complete (n-1)steps + No of ways to complete (n-2)steps + No of ways to complete (n-3)steps +1.
Using Recursion:
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| public class StepsPossiblePathsRecur { | |
| public int possibleWays(int n) { | |
| if (n < 1) { | |
| return 0; | |
| } | |
| return 1 + possibleWays(n – 1) + possibleWays(n – 2) | |
| + possibleWays(n – 3); | |
| } | |
| public static void main(String[] args) { | |
| // TODO Auto-generated method stub | |
| int n = 3; | |
| StepsPossiblePathsRecur s = new StepsPossiblePathsRecur(); | |
| System.out.println(s.possibleWays(n)); | |
| } | |
| } |
If we solve this problem using recursion then all the sub-problems will be calculated repeatedly.
Using Dynamic Programming:
- We will solve it Top-Down approach.
- We need to store the solutions for the sub problems in an array.
DP Code:
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| public class StepsPossiblePathsRecur { | |
| public int possibleWaysDyna(int n, int[] dyn) { | |
| if (n == 0) | |
| return 1; | |
| if (n < 0) | |
| return 0; | |
| if (dyn[n] > 0) { | |
| return dyn[n]; | |
| } | |
| dyn[n] = possibleWaysDyna(n – 1, dyn) + possibleWaysDyna(n – 2, dyn) | |
| + possibleWaysDyna(n – 3, dyn); | |
| return dyn[n]; | |
| } | |
| public static void main(String[] args) { | |
| // TODO Auto-generated method stub | |
| int n = 3; | |
| StepsPossiblePathsRecur s = new StepsPossiblePathsRecur(); | |
| int[] dyn = new int[n + 1]; | |
| System.out.println(s.possibleWaysDyna(n, dyn)); | |
| } | |
| } |
Output:
4