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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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Subset Sum

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Problem Statement
*

Example 1:

Example 2:

Example 3:

Basic Solution

Bottom-up Dynamic Programming

Code

Challenge

Try it yourself

Problem Statement #

Given a set of positive numbers, determine if there exists a subset whose sum is equal to a given number ‘S’.

Example 1: #

Input: {1, 2, 3, 7}, S=6
Output: True
The given set has a subset whose sum is '6': {1, 2, 3}

Example 2: #

Input: {1, 2, 7, 1, 5}, S=10
Output: True
The given set has a subset whose sum is '10': {1, 2, 7}

Example 3: #

Input: {1, 3, 4, 8}, S=6
Output: False
The given set does not have any subset whose sum is equal to '6'.

Basic Solution #

This problem follows the 0/1 Knapsack pattern and is quite similar to Equal Subset Sum Partition.
A basic brute-force solution could be to try all subsets of the given numbers to see if any set has a sum equal to ‘S’.

So our brute-force algorithm will look like:

for each number 'i'
create a new set which INCLUDES number 'i' if it does not exceed 'S', and recursively
process the remaining numbers
create a new set WITHOUT number 'i', and recursively process the remaining numbers
return true if any of the above two sets has a sum equal to 'S', otherwise return false

Since this problem is quite similar to Equal Subset Sum Partition, let’s jump directly to the bottom-up dynamic programming solution.

Bottom-up Dynamic Programming #

We’ll try to find if we can make all possible sums with every subset to populate the array dp[TotalNumbers][S+1].

For every possible sum ‘s’ (where 0 <= s <= S), we have two options:

Exclude the number. In this case, we will see if we can get the sum ‘s’ from the subset excluding this number => dp[index-1][s]

Include the number if its value is not more than ‘s’. In this case, we will see if we can find a subset to get the remaining sum => dp[index-1][s-num[index]]

If either of the above two scenarios returns true, we can find a subset with a sum equal to ‘s’.

Let’s draw this visually, with the example input {1, 2, 3, 7}, and start with our base case of size zero:

'0' sum can always be found through an empty set

1 of 10

With only one number, we can form a subset only when the required sum is equal to that number

2 of 10

sum: 1, index:1=> (dp[index-1][sum] , as the 'sum' is less than the number at index '1' (i.e., 1 < 2)

3 of 10

sum: 2, index:1=> (dp[index-1][sum] || dp[index-1][sum-2])

4 of 10

sum: 3, index:1=> (dp[index-1][sum] || dp[index-1][sum-2])

5 of 10

sum: 4-6 index:1=> (dp[index-1][sum] || dp[index-1][sum-2])

6 of 10

sum: 1,2,3, index:2=> (dp[index-1][sum] || dp[index-1][sum-3])

7 of 10

sum: 4, index:2=> (dp[index-1][sum] || dp[index-1][sum-3])

8 of 10

sum: 5, 6, index:2=> (dp[index-1][sum] || dp[index-1][sum-3])

9 of 10

sum: 1-6, index:3=> (dp[index-1][sum] || dp[index-1][sum-7])

10 of 10

Code #

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

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class SubsetSum {

public boolean canPartition(int[] num, int sum) {
int n = num.length;
boolean[][] dp = new boolean[n][sum + 1];

// populate the sum=0 columns, as we can always form '0' sum with an empty set
for (int i = 0; i < n; i++)
dp[i][0] = true;

// with only one number, we can form a subset only when the required sum is
// equal to its value
for (int s = 1; s <= sum; s++) {
dp[0][s] = (num[0] == s ? true : false);
}

// process all subsets for all sums
for (int i = 1; i < num.length; i++) {
for (int s = 1; s <= sum; s++) {
// if we can get the sum 's' without the number at index 'i'
if (dp[i - 1][s]) {
dp[i][s] = dp[i - 1][s];
} else if (s >= num[i]) {
// else include the number and see if we can find a subset to get the remaining
// sum
dp[i][s] = dp[i - 1][s - num[i]];
}
}
}

// the bottom-right corner will have our answer.
return dp[num.length - 1][sum];
}

public static void main(String[] args) {
SubsetSum ss = new SubsetSum();
int[] num = { 1, 2, 3, 7 };
System.out.println(ss.canPartition(num, 6));
num = new int[] { 1, 2, 7, 1, 5 };
System.out.println(ss.canPartition(num, 10));
num = new int[] { 1, 3, 4, 8 };
System.out.println(ss.canPartition(num, 6));
}
}

RUN

SAVERESET

The above solution has time and space complexity of O(N∗S)O(N*S)O(N∗S), where ‘N’ represents total numbers and ‘S’ is the required sum.

Challenge #

Can we further improve our bottom-up DP solution? Can you find an algorithm that has O(S)O(S)O(S) space complexity?

Show Hint

Try it yourself #

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class SubsetSum {

static boolean canPartition(int[] num, int sum) {
//TODO: Write - Your - Code
return false;
}
}

TESTShow Solution

SAVERESET

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