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Grokking Dynamic Programming Patterns for Coding Interviews

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Introduction

What is Dynamic Programming?

Pattern 1: 0/1 Knapsack

0/1 Knapsack

Equal Subset Sum Partition

Subset Sum

Minimum Subset Sum Difference

Count of Subset Sum

Target Sum

Pattern 2: Unbounded Knapsack

Unbounded Knapsack

Rod Cutting

Coin Change

Minimum Coin Change

Maximum Ribbon Cut

Pattern 3: Fibonacci Numbers

Fibonacci numbers

Staircase

Number factors

Minimum jumps to reach the end

Minimum jumps with fee

House thief

Pattern 4: Palindromic Subsequence

Longest Palindromic Subsequence

Longest Palindromic Substring

Count of Palindromic Substrings

Minimum Deletions in a String to make it a Palindrome

Palindromic Partitioning

Pattern 5: Longest Common Substring

Longest Common Substring

Longest Common Subsequence

Minimum Deletions & Insertions to Transform a String into another

Longest Increasing Subsequence

Maximum Sum Increasing Subsequence

Shortest Common Super-sequence

Minimum Deletions to Make a Sequence Sorted

Longest Repeating Subsequence

Subsequence Pattern Matching

Longest Bitonic Subsequence

Longest Alternating Subsequence

Edit Distance

Strings Interleaving

Appendix

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Shortest Common Super-sequence

We'll cover the following

Problem Statement

Basic Solution

Top-down Dynamic Programming with Memoization

Code

Bottom-up Dynamic Programming

Code

Problem Statement #

Given two sequences ‘s1’ and ‘s2’, write a method to find the length of the shortest sequence which has ‘s1’ and ‘s2’ as subsequences.

Example 2:

Input: s1: "abcf" s2:"bdcf"
Output: 5
Explanation: The shortest common super-sequence (SCS) is "abdcf".

Example 2:

Input: s1: "dynamic" s2:"programming"
Output: 15
Explanation: The SCS is "dynprogrammicng".

Basic Solution #

The problem is quite similar to the Longest Common Subsequence.

A basic brute-force solution could be to try all the super-sequences of the given sequences. We can process both of the sequences one character at a time, so at any step we must choose between:

If the sequences have a matching character, we can skip one character from both the sequences and make a recursive call for the remaining lengths to get SCS.

If the strings don’t match, we start two new recursive calls by skipping one character separately from each string. The minimum of these two recursive calls will have our answer.

Here is the code:

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class SCS {

public int findSCSLength(String s1, String s2) {
return findSCSLengthRecursive(s1, s2, 0, 0);
}

private int findSCSLengthRecursive(String s1, String s2, int i1, int i2) {
// if we have reached the end of a string, return the remaining length of the other string,
// as in this case we have to take all of the remaining other string
if(i1 == s1.length())
return s2.length()-i2;
if(i2 == s2.length())
return s1.length()-i1;

if(s1.charAt(i1) == s2.charAt(i2))
return 1 + findSCSLengthRecursive(s1, s2, i1+1, i2+1);

int length1 = 1 + findSCSLengthRecursive(s1, s2, i1, i2+1);
int length2 = 1 + findSCSLengthRecursive(s1, s2, i1+1, i2);

return Math.min(length1, length2);
}

public static void main(String[] args) {
SCS scs = new SCS();
System.out.println(scs.findSCSLength("abcf", "bdcf"));
System.out.println(scs.findSCSLength("dynamic", "programming"));
}
}

RUN

SAVERESET

The time complexity of the above algorithm is exponential O(2n+m)O(2^{n+m})O(2​n+m​​), where ‘n’ and ‘m’ are the lengths of the input sequences. The space complexity is O(n+m)O(n+m)O(n+m) which is used to store the recursion stack.

Top-down Dynamic Programming with Memoization #

Let’s use memoization to overcome the overlapping subproblems.

The two changing values to our recursive function are the two indexes, i1 and i2. Therefore, we can store the results of all the subproblems in a two-dimensional array. (Another alternative could be to use a hash-table whose key would be a string (i1 + “|” + i2)).

Code #

Here is the code:

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class SCS {

public int findSCSLength(String s1, String s2) {
Integer[][] dp = new Integer[s1.length()][s2.length()];
return findSCSLengthRecursive(dp, s1, s2, 0, 0);
}

private int findSCSLengthRecursive(Integer[][] dp, String s1, String s2, int i1, int i2) {
// if we have reached the end of a string, return the remaining length of the other string,
// as in this case we have to take all of the remaining other string
if(i1 == s1.length())
return s2.length()-i2;
if(i2 == s2.length())
return s1.length()-i1;

if(dp[i1][i2] == null) {
if(s1.charAt(i1) == s2.charAt(i2))
dp[i1][i2] = 1 + findSCSLengthRecursive(dp, s1, s2, i1+1, i2+1);
else {
int length1 = 1 + findSCSLengthRecursive(dp, s1, s2, i1, i2+1);
int length2 = 1 + findSCSLengthRecursive(dp, s1, s2, i1+1, i2);
dp[i1][i2] = Math.min(length1, length2);
}
}

return dp[i1][i2];
}

public static void main(String[] args) {
SCS scs = new SCS();
System.out.println(scs.findSCSLength("abcf", "bdcf"));
System.out.println(scs.findSCSLength("dynamic", "programming"));
}
}

RUN

SAVERESET

Bottom-up Dynamic Programming #

Since we want to match all the subsequences of the given sequences, we can use a two-dimensional array to store our results. The lengths of the two strings will define the size of the array’s dimensions. So for every index ‘i’ in sequence ‘s1’ and ‘j’ in sequence ‘s2’, we will choose one of the following two options:

If the character s1[i] matches s2[j], the length of the SCS would be the one plus the length of the SCS till i-1 and j-1 indexes in the two strings.

If the character s1[i] does not match s2[j], we will consider two SCS: one without s1[i] and one without s2[j]. Our required SCS length will be the shortest of these two super-sequences plus one.

So our recursive formula would be:

if s1[i] == s2[j]
dp[i][j] = 1 + dp[i-1][j-1]
else
dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])

Let’s draw this visually for “abcf” and “bdcf”. Starting with a subsequence of zero length. As we can see, if any strings have zero length, then the shortest super-sequence would be equal to the length of the other string:

if one of the strings is of zero length, SCS would be equal to the length of the other string

1 of 12

i=1, j=1 => since s1[i-1] != s2[j-1], therefore dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])

2 of 12

i=1, j=2 => since s1[i-1] == s2[j-1], therefore dp[i][j] = 1 + dp[i-1][j-1]

3 of 12

i=1, j=3-4 => since s1[i] != s2[j], therefore dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])

4 of 12

i=2, j=1 => since s1[i] != s2[j], therefore dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])

5 of 12

i=2, j=2 => since s1[i] != s2[j], therefore dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])

6 of 12

i=2, j=3-4 => since s1[i] != s2[j], therefore dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])

7 of 12

i=3, j=1-2 => since s1[i] != s2[j], therefore dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])

8 of 12

i=3, j=3 => since s1[i-1] == s2[j-1], therefore dp[i][j] = 1 + dp[i-1][j-1]

9 of 12

i=3, j=4 => since s1[i-1] == s2[j-1], therefore dp[i][j] = 1 + dp[i-1][j-1]

10 of 12

i=4, j=1-3 => since s1[i] != s2[j], therefore dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])

11 of 12

i=4, j=4 => since s1[i-1] == s2[j-1], therefore dp[i][j] = 1 + dp[i-1][j-1]

12 of 12

From the above visualization, we can clearly see that the longest increasing subsequence is of length ‘5’ – as shown by dp[4][4].

Code #

Here is the code for our bottom-up dynamic programming approach:

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class SCS {

public int findSCSLength(String s1, String s2) {
int[][] dp = new int[s1.length()+1][s2.length()+1];

// if one of the strings is of zero length, SCS would be equal to the length of the other string
for(int i=0; i<=s1.length(); i++)
dp[i][0] = i;
for(int j=0; j<=s2.length(); j++)
dp[0][j] = j;

for(int i=1; i <= s1.length(); i++) {
for(int j=1; j <= s2.length(); j++) {
if(s1.charAt(i-1) == s2.charAt(j-1))
dp[i][j] = 1 + dp[i-1][j-1];
else
dp[i][j] = 1 + Math.min(dp[i-1][j], dp[i][j-1]);
}
}
return dp[s1.length()][s2.length()];
}

public static void main(String[] args) {
SCS scs = new SCS();
System.out.println(scs.findSCSLength("abcf", "bdcf"));
System.out.println(scs.findSCSLength("dynamic", "programming"));
}
}

RUN

SAVERESET

The time and space complexity of the above algorithm is O(n∗m)O(n*m)O(n∗m).

Completed

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