Dynamic Programming Series

Dynamic Programming Series

Day 8

Decode Ways..!!

Problem Description

A message containing letters from a to z is being encoded to numbers using the following mapping :

Problem Description

A message containing letters from a to z is being encoded to numbers using the following mapping :

'a' -> 1

'b' -> 2

'c' -> 3

'z' -> 26

Problem Description

Task : Find out the numbers of ways of creating different codes.

(given string is non-empty)

Problem Description

Task : Find out the numbers of ways of creating different codes.

(given string is non-empty)

Example :

Problem Description

Task : Find out the numbers of ways of creating different codes.

(given string is non-empty)

Example :

"12"

Problem Description

Task : Find out the numbers of ways of creating different codes.

(given string is non-empty)

Example :

"12"

ab

l

Problem Description

Task : Find out the numbers of ways of creating different codes.

(given string is non-empty)

Example :

"12"

ab

l

Because a = 1 and b = 2 in case 1 and in case 2 we have l = 12

Problem Description

Task : Find out the numbers of ways of creating different codes.

(given string is non-empty)

Example 2:

"226"

Problem Description

Task : Find out the numbers of ways of creating different codes.

(given string is non-empty)

Example 2:

"226"

bz

vf

bff

Problem Description

Task : Find out the numbers of ways of creating different codes.

(given string is non-empty)

Example 2:

"226"

bz

vf

bff

Because bz(2,26) , vf(22,6) and bff(2,2,6)

So now we have understood the problem..!!

So now we have understood the problem..!!

Now the question arises How we can approach to this problem??

What we have learnt till now in this

series ?

Dividing a big problem into smaller subproblems.... that's the first step that

we should take..!!

What we have learnt till now in this

series ?

Dividing a big problem into smaller subproblems.... that's the first step that

we should take..!!

Now how we will do this in this particular problem??

Let us see with some basic cases

If we have a string of length 1

Suppose input is "5"

Then how many ways exist to form the code i.e., only a single way -> 'e'

Let us see with some basic cases

If we have a string of length 1

Suppose input is "5"

Then how many ways exist to form the code i.e., only a single way -> 'e'

This is what we can think at least..!!

If we have string of length 2

Suppose we have a string "15"

Then how many ways exist to form the code

If we have string of length 2

Suppose we have a string "15"

Then how many ways exist to form the code

That is 2 one is "ae" and the other is "o".

If we have string of length 2

Suppose we have a string "15"

Then how many ways exist to form the code

That is 2 one is "ae" and the other is "o".

Simple observations..!!

if we have input as "45"

if we have input as "45"

Then also the answer would be 2.

if we have input as "45"

Then also the answer would be 2.

Hey..!! But wait a minute "45" can exist as "de" but we do not have any character from a...z in 45 we only have 26 characters..!!

That's the game .. to make such important observations

Now following the traditions that we are following so far..!!

1. Developing a recursive approach.

2. Developing an iterative approach.

How recursive thing will work here?

Let us understand through an Example how we form a recursive approach for this question.

Given n = 226

How recursive thing will work here?

Let us understand through an Example how we form a recursive approach for this question.

Given n = 226

226

226

226

b

v

How recursive thing will work here?

Let us understand through an Example how we form a recursive approach for this question.

Given n = 226

226

226

226

b

v

226

bb

226

bz

226

bb

226

bz

How recursive thing will work here?

Let us understand through an Example how we form a recursive approach for this question.

Given n = 226

226

226

226

b

v

226

bb

226

bz

226

bb

226

bz

226

vf

How recursive thing will work here?

Let us understand through an Example how we form a recursive approach for this question.

Given n = 226

226

226

226

b

v

226

bb

226

bz

226

bb

226

bz

226

vf

1

2

3

Let's take one more example to make it real easy.

Take n = 2567

2567

b

y

2567

2567

Let's take one more example to make it real easy.

Take n = 2567

2567

b

y

2567

2567

bb

be

2567

2567

X

Let's take one more example to make it real easy.

Take n = 2567

2567

b

y

2567

2567

bb

be

2567

2567

X

yf

2567

X

2567

Let's take one more example to make it real easy.

Take n = 2567

2567

b

y

2567

2567

bb

be

2567

2567

X

yf

2567

X

2567

bb

bef

2567

X

2567

Let's take one more example to make it real easy.

Take n = 2567

2567

b

y

2567

2567

bb

be

2567

2567

X

yf

2567

X

2567

bb

bef

2567

X

2567

yfg

2567

X

2567

Let's take one more example to make it real easy.

Take n = 2567

2567

b

y

2567

2567

bb

be

2567

2567

X

yf

2567

X

2567

bb

bef

2567

X

2567

yfg

2567

X

2567

1

2

befg

2567

Now we are ready to go for a recursive code for this approach.

int decode(int i, string s)
{
    int n = s.size();
    if (i == n)
        return 1;
    if (s[i] == '0')
        return 0;
    int res = 0;
    res = decode(i+1, s);
    if (i <= n-2 && (s[i] == '1' || (s[i] == '2' && s[i+1] <= '6')))
        res += decode(i+2, s);
    return res;
}
int numDecodings(string s) {
    if (s.size() == 0)
        return 0;
    return decode(0, s);
}

Let's talk about the time complexity of this Approach.

On the worst case for every number we have two calls.

level 0 = 2^0 = 1

Total levels = n

level 1 = 2^1 = 2

level 2 = 2^2 = 4

Complexity\ =\ O(2^n)

Now To reduce the complexity we should be able to find the redundancy in the tree.

Take back our previous example n = 2567

2567

b

y

2567

2567

bb

be

2567

2567

X

yf

2567

X

2567

bb

bef

2567

X

2567

yfg

2567

X

2567

1

2

befg

2567

Now To reduce the complexity we should be able to find the redundancy in the tree.

Can we store the result of of these steps so we don't have to recompute it?

Yes Of course, That's where the DP comes.

Let's talk about the DP approach

Now what we will be doing that we will store the number ways possible till the ith index in our dp array.

DP[i]\ =\ number\ of\ ways\ possible\ till\ ith\ index

Now when we will compute the answer for (i+1)th index we will add the value of dp[i] and dp[i-1](if possible)

DP[i+1] = DP[i] + DP[i-1]

We will fill this DP array in Bottom Up Manner.

int n = s.length();
int* dp = new int[n+1]();
if(s[0]-'0'!=0)
dp[0] = 1;
if(s[0]-'0'!=0)
dp[1] = 1;
for(int  i = 2; i<=n; i++)
{
    if(s[i-1]-'0'!=0)
        dp[i] = dp[i-1];
    if(((s[i-2]-'0')*10 + (s[i-1] - '0') <= 26) && ((s[i-2]-'0')*10 + (s[i-1] - '0') 
                                                   >= 10))
        dp[i] += dp[i-2];
}
return dp[n];

So that's It for Day 8

Concepts that we have learnt

So that's It for Day 8

We have taken a very first step in Dynamic Programming.

Concepts that we have learnt

1. How to handle Edge cases.

So that's It for Day 8

We have taken a very first step in Dynamic Programming.

Concepts that we have learnt

1. How to handle Edge cases..

2. Use of optimal Substructure with memoization

So that's It for Day 8

We have taken a very first step in Dynamic Programming.

Concepts that we have learnt

1. How to handle Edge cases.

2. Use of optimal Substructure with memoization

Still, Have any Doubts???

Still, Have any Doubts???

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Send your doubt to email provided in description.

Copy of Copy of Day 6 vacation problem

By Chirayu Jain

Copy of Copy of Day 6 vacation problem

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Chirayu Jain