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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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Edit Distance

We'll cover the following

Problem Statement

Basic Solution

Code

Top-down Dynamic Programming with Memoization

Code

Bottom-up Dynamic Programming

Code

Problem Statement #

Given strings s1 and s2, we need to transform s1 into s2 by deleting, inserting, or replacing characters. Write a function to calculate the count of the minimum number of edit operations.

Example 1:

Input: s1 = "bat"
s2 = "but"
Output: 1
Explanation: We just need to replace 'a' with 'u' to transform s1 to s2.

Example 2:

Input: s1 = "abdca"
s2 = "cbda"
Output: 2
Explanation: We can replace first 'a' with 'c' and delete second 'c'.

Example 3:

Input: s1 = "passpot"
s2 = "ppsspqrt"
Output: 3
Explanation: Replace 'a' with 'p', 'o' with 'q', and insert 'r'.

Basic Solution #

A basic brute-force solution could be to try all operations (one by one) on each character of s1. We can iterate through s1 and s2 together. Let’s assume index1 and index2 point to the current indexes of s1 and s2 respectively, so we have two options at every step:

If the strings have a matching character, we can recursively match for the remaining lengths.

If the strings don’t match, we start three new recursive calls representing the three edit operations. Whichever recursive call returns the minimum count of operations will be our answer.

Code #

Here is recursive implementation:

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class EditDistance {

public int findMinOperations(String s1, String s2) {
return findMinOperationsRecursive(s1, s2, 0, 0);
}

private int findMinOperationsRecursive(String s1, String s2, int i1, int i2) {

// if we have reached the end of s1, then we have to insert all the remaining characters of s2
if(i1 == s1.length())
return s2.length() - i2;

// if we have reached the end of s2, then we have to delete all the remaining characters of s1
if(i2 == s2.length())
return s1.length() - i1;

// If the strings have a matching character, we can recursively match for the remaining lengths.
if(s1.charAt(i1) == s2.charAt(i2))
return findMinOperationsRecursive(s1, s2, i1+1, i2+1);

int c1 = 1 + findMinOperationsRecursive(s1, s2, i1+1, i2); //perform deletion
int c2 = 1 + findMinOperationsRecursive(s1, s2, i1, i2+1); //perform insertion
int c3 = 1 + findMinOperationsRecursive(s1, s2, i1+1, i2+1); // perform replacement

return Math.min(c1, Math.min(c2, c3));
}

public static void main(String[] args) {
EditDistance editDisatnce = new EditDistance();
System.out.println(editDisatnce.findMinOperations("bat", "but"));
System.out.println(editDisatnce.findMinOperations("abdca", "cbda"));
System.out.println(editDisatnce.findMinOperations("passpot", "ppsspqrt"));
}
}

RUN

SAVERESET

Because of the three recursive calls, the time complexity of the above algorithm is exponential O(3m+n)O(3^{m+n})O(3​m+n​​), where ‘m’ and ‘n’ are the lengths of the two input strings. 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 #

We can use an array to store the already solved subproblems.

The two changing values in 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 for Top-down DP approach:

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class EditDistance {

public int findMinOperations(String s1, String s2) {
Integer[][] dp = new Integer[s1.length()+1][s2.length()+1];
return findMinOperationsRecursive(dp, s1, s2, 0, 0);
}

private int findMinOperationsRecursive(Integer[][] dp, String s1, String s2, int i1, int i2) {

if(dp[i1][i2] == null) {
// if we have reached the end of s1, then we have to insert all the remaining characters of s2
if(i1 == s1.length())
dp[i1][i2] = s2.length() - i2;

// if we have reached the end of s2, then we have to delete all the remaining characters of s1
else if(i2 == s2.length())
dp[i1][i2] = s1.length() - i1;

// If the strings have a matching character, we can recursively match for the remaining lengths
else if(s1.charAt(i1) == s2.charAt(i2))
dp[i1][i2] = findMinOperationsRecursive(dp, s1, s2, i1+1, i2+1);
else {
int c1 = findMinOperationsRecursive(dp, s1, s2, i1+1, i2); //delete
int c2 = findMinOperationsRecursive(dp, s1, s2, i1, i2+1); //insert
int c3 = findMinOperationsRecursive(dp, s1, s2, i1+1, i2+1); //replace
dp[i1][i2] = 1 + Math.min(c1, Math.min(c2, c3));
}
}

return dp[i1][i2];
}

public static void main(String[] args) {
EditDistance editDisatnce = new EditDistance();
System.out.println(editDisatnce.findMinOperations("bat", "but"));
System.out.println(editDisatnce.findMinOperations("abdca", "cbda"));
System.out.println(editDisatnce.findMinOperations("passpot", "ppsspqrt"));
}
}

RUN

SAVERESET

What is the time and space complexity of the above solution? Since our memoization array dp[s1.length()][s2.length()] stores the results for all the subproblems, we can conclude that we will not have more than m∗nm*nm∗n subproblems (where ‘m’ and ‘n’ are the lengths of the two input strings.). This means that our time complexity will be O(m∗n)O(m*n)O(m∗n).

The above algorithm will be using O(m∗n)O(m*n)O(m∗n) space for the memoization array. Other than that we will use O(m+n)O(m+n)O(m+n) space for the recursion call-stack. So the total space complexity will be O(m∗n+(m+n))O(m*n + (m+n))O(m∗n+(m+n)), which is asymptotically equivalent to O(m∗n)O(m*n)O(m∗n).

Bottom-up Dynamic Programming #

Since we want to match all the characters of the given two strings, we can use a two-dimensional array to store our results. The lengths of the two strings will define the size of the two dimensions of the array. So for every index ‘i1’ in string ‘s1’ and ‘i2’ in string ‘s2’, we will choose one of the following options:

If the character s1[i1] matches s2[i2], the count of the edit operations will be equal to the count of the edit operations for the remaining strings.

If the character s1[i1] does not match s2[i2], we will take the minimum count from the remaining strings after performing any of the three edit operations.

So our recursive formula would be:

if s1[i1] == s2[i2]
dp[i1][i2] = dp[i1-1][i2-1]
else
dp[i1][i2] = 1 + min(dp[i1-1][i2], // delete
dp[i1][i2-1], // insert
dp[i1-1][i2-1]) // replace

Code #

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

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class EditDistance {

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

// if s2 is empty, we can remove all the characters of s1 to make it empty too
for(int i1=0; i1 <= s1.length(); i1++)
dp[i1][0] = i1;

// if s1 is empty, we have to insert all the characters of s2
for(int i2=0; i2 <= s2.length(); i2++)
dp[0][i2] = i2;

for(int i1=1; i1 <= s1.length(); i1++) {
for(int i2=1; i2 <= s2.length(); i2++) {
// If the strings have a matching character, we can recursively match for the remaining lengths
if(s1.charAt(i1-1) == s2.charAt(i2-1))
dp[i1][i2] = dp[i1-1][i2-1];
else
dp[i1][i2] = 1 + Math.min(dp[i1-1][i2], //delete
Math.min(dp[i1][i2-1], //insert
dp[i1-1][i2-1])); //replace
}
}

return dp[s1.length()][s2.length()];
}

public static void main(String[] args) {
EditDistance editDisatnce = new EditDistance();
System.out.println(editDisatnce.findMinOperations("bat", "but"));
System.out.println(editDisatnce.findMinOperations("abdca", "cbda"));
System.out.println(editDisatnce.findMinOperations("passpot", "ppsspqrt"));
}
}

RUN

SAVERESET

The time and space complexity of the above algorithm is O(n∗m)O(n*m)O(n∗m), where ‘m’ and ‘n’ are the lengths of the two input strings.

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