Twenty Percent Knowledge: October 2016

Thursday, October 20, 2016

Recursion: Davis' Staircase

Recursion, if computation is repeated, memorization can be used to make it linear. Some of them, if asking summary of something, DP can usually be used to solve it.

In this post, I am providing DP solution instead of recursion, which should be easily be written with memorization as an optimization.

Davis has

staircases in his house and he likes to climb each staircase

, or

steps at a time. Being a very precocious child, he wonders how many ways there are to reach the top of the staircase.
Given the respective heights for each of the

staircases in his house, find and print the number of ways he can climb each staircase on a new line.

Input Format

The first line contains a single integer,

, denoting the number of staircases in his house.
Each line

of the

subsequent lines contains a single integer,

, denoting the height of staircase

Constraints

Subtasks

for of the maximum score.

Output Format

For each staircase, print the number of ways Davis can climb it in a new line.

Sample Input

Sample Output

1
4
44

Explanation

Let's calculate the number of ways of climbing the first two of the Davis'

staircases:

The first staircase only has step, so there is only one way for him to climb it (i.e., by jumping step). Thus, we print on a new line.
The second staircase has steps and he can climb it in any of the four following ways:

Thus, we print

on a new line.

public class Solution {
    static void numWays(int n){
        //let dp[i] is num ways to reach stair i. 
        //each dp[i] is an accumulation of ways to reach to there
        int dp[] = new int[n+1];
        //assuming you are at nth stair, to reach there
        //you can step 1 from n-1th stair
        //or step 2 from n-2th stair or
        //step 3 from n-3th stair
        //so you have that many ways to reach to nth stair. 
        //adding them up is the ways to reach to nth stair
        //dp[3]=dp[2]+dp[1]+dp[0];
        
        /*
        to reach to stair 1, you have one way:dp[1]=1=dp[0]
        to reach to stair 2, you have: 1 from dp[0], that is 1 way,2. from dp[1], that's 1 way.
        to reach to stair 3, you can : 1. dp[0]->dp[3],2.dp[2]->dp[3],3. dp[1]->dp[3]
        whatever you use to reach dp[2] now contributes to dp[3]
        */
        if(n==1){
            System.out.println(1);
        }else if(n==2){
            System.out.println(2);
        }else{
            dp[0]=1;//0 is base 
            dp[1]=1;dp[2]=dp[1]+dp[0];

            for(int i=3;i<=n;i++){
                dp[i]=dp[i-1]+dp[i-2]+dp[i-3];
            }
            System.out.println(dp[n]);
        }
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int s = in.nextInt();
        for(int a0 = 0; a0 < s; a0++){
            int n = in.nextInt();
            numWays(n);
        }
        
    }
}

Coin Change-- number of ways

Given a number of dollars,

, and a list of dollar values for

distinct coins,

, find and print the number of different ways you can make change for

dollars if each coin is available in an infinite quantity.

Hints:

You can solve this problem recursively, but you must optimize your solution to eliminate overlapping subproblems using Dynamic Programming if you wish to pass all test cases. More specifically, think of ways to store the checked solutions and use the stored values to avoid repeatedly calculating the same values.
Think about the degenerate cases:
- How many ways can you make change for dollars?
- How many ways can you make change for less than dollars if you have no coins?
If you are having trouble defining the storage for your precomputed values, then think about it in terms of the base case .

Input Format

The first line contain two space-separated integers describing the respective values of

and

.
The second line contains

space-separated integers describing the respective values of

, where each integer denotes the dollar value of a distinct coin available in an infinite quantity.

Constraints

The list of coins contains distinct integers where each integer denotes the dollar value of a coin available in an infinite quantity.

Output Format

Print a single integer denoting the number of ways we can make change for

dollars using an infinite supply of our

types of coins.

Sample Input 0

4 3
1 2 3

Sample Output 0

Explanation 0
For

and

there are four solutions:

Thus, we print

on a new line.

Sample Input 1

10 4
2 5 3 6

Sample Output 1

Explanation 1
For

and

there are five solutions:

Thus, we print

on a new line.

package dp;

import java.util.Scanner;

/**
 * first successful story by analyzing it and coming up an algorithm by myself
 * --even though I saw this question before, I basically forgot.
 * Them came up a acceptable solution within 15 minutes.
 * 
 * Cheers to myself.
 * @author Andrew Ma
 *
 */
public class CoinChangeNumOfWays {
 
    public static long makeChange(int[] coins, int money) {
     //I assumes 0 denom having 0 solutions to any amount of money
     //also assumes any denom having 0 solution to 0 amount of money
     //it turns out to be right! to be brave to make assumptions!
     //making room for 0 denom and 0 amount of money
        long dp[][] = new long[coins.length+1][money+1];
        //initialization the known solutions
        //0 denom
        for(int i=0;i<money+1;i++){
            dp[0][i]=0;//no solution for each money value
        }
        //0 money value
        for(int i=0;i<coins.length+1;i++){
            dp[i][0]=0;//no coin can do 0 change
        }
        //then build up values towards to final solution
        //c and m here are for dp array's dimensions
        /*
         * define dp[c][m] as accumulated solutions at c and m 
         * dp[c][m]=
         *    1. dp[c-1][m]. when money value is less than denom, then get solution from last denom
         *        --here it needs to have a 0 denom
         *    2. dp[c-1][m]+1. when money value equals denom value, we get one more solution
         *    3. dp[c-1][m] + dp[c][m-coins[c-1]]. when money valus is greater than denom value,
         *        it adds up accumulation from previous denom and same denom for meney value
         *        not including this denom.
         * damn, when did I become so able to analyze? I guess drawing it out and practice on paper or white board
         * really helped to see one's thoughts and then you just need to write the code to reflect the thoughts. 
         * and you are confident that you are able to write code to reflect your thoughts.
         */
        for(int c=1;c<coins.length+1;c++){
            for(int m=1;m<money+1;m++){
             //when it comes to refer values in coins, c need to be converted back to 0 based
                if(m<coins[c-1]){
                    //no change at this, copy the last denom's accumulated solutions
                    dp[c][m] = dp[c-1][m];
                }
                else if(m==coins[c-1]){
                    //we get one more solution
                    dp[c][m] = dp[c-1][m] +1;
                }
                else //if (m>coins[c-1])
                    {
                    //then it is last denom's accumulated solution + lesser value's accumulated solution
                    dp[c][m] = dp[c-1][m] + dp[c][m-coins[c-1]];
                }
            }
        }
        return dp[coins.length][money];
    }
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int m = in.nextInt();
        int coins[] = new int[m];
        for(int coins_i=0; coins_i < m; coins_i++){
            coins[coins_i] = in.nextInt();
        }
        System.out.println(makeChange(coins, n));
    }
}