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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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Minimum Subset Sum Difference

We'll cover the following

Problem Statement
*

Example 1:

Example 2:

Example 3:

Basic Solution

Code

Top-down Dynamic Programming with Memoization

Code

Bottom-up Dynamic Programming

Code

Problem Statement #

Given a set of positive numbers, partition the set into two subsets with a minimum difference between their subset sums.

Example 1: #

Input: {1, 2, 3, 9}
Output: 3
Explanation: We can partition the given set into two subsets where minimum absolute difference
between the sum of numbers is '3'. Following are the two subsets: {1, 2, 3} & {9}.

Example 2: #

Input: {1, 2, 7, 1, 5}
Output: 0
Explanation: We can partition the given set into two subsets where minimum absolute difference
between the sum of number is '0'. Following are the two subsets: {1, 2, 5} & {7, 1}.

Example 3: #

Input: {1, 3, 100, 4}
Output: 92
Explanation: We can partition the given set into two subsets where minimum absolute difference
between the sum of numbers is '92'. Here are the two subsets: {1, 3, 4} & {100}.

Basic Solution #

This problem follows the 0/1 Knapsack pattern and can be converted into a Subset Sum problem.

Let’s assume S1 and S2 are the two desired subsets. A basic brute-force solution could be to try adding each element either in S1 or S2, to find the combination that gives the minimum sum difference between the two sets.

So our brute-force algorithm will look like:

for each number 'i'
add number 'i' to S1 and recursively process the remaining numbers
add number 'i' to S2 and recursively process the remaining numbers
return the minimum absolute difference of the above two sets

Code #

Here is the code for the brute-force solution:

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class PartitionSet {

public int canPartition(int[] num) {
return this.canPartitionRecursive(num, 0, 0, 0);
}

private int canPartitionRecursive(int[] num, int currentIndex, int sum1, int sum2) {
// base check
if (currentIndex == num.length)
return Math.abs(sum1 - sum2);

// recursive call after including the number at the currentIndex in the first set
int diff1 = canPartitionRecursive(num, currentIndex+1, sum1+num[currentIndex], sum2);

// recursive call after including the number at the currentIndex in the second set
int diff2 = canPartitionRecursive(num, currentIndex+1, sum1, sum2+num[currentIndex]);

return Math.min(diff1, diff2);
}

public static void main(String[] args) {
PartitionSet ps = new PartitionSet();
int[] num = {1, 2, 3, 9};
System.out.println(ps.canPartition(num));
num = new int[]{1, 2, 7, 1, 5};
System.out.println(ps.canPartition(num));
num = new int[]{1, 3, 100, 4};
System.out.println(ps.canPartition(num));
}
}

RUN

SAVERESET

The time complexity of the above algorithm is exponential O(2n)O(2^n)O(2​n​​), where ‘n’ represents the total number. The space complexity is O(n)O(n)O(n) which is used to store the recursion stack.

Top-down Dynamic Programming with Memoization #

We can use memoization to overcome the overlapping sub-problems.

We will be using a two-dimensional array to store the results of the solved sub-problems. We can uniquely identify a sub-problem from ‘currentIndex’ and ‘Sum1’; as ‘Sum2’ will always be the sum of the remaining numbers.

Code #

Here is the code:

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class PartitionSet {

public int canPartition(int[] num) {
int sum = 0;
for (int i = 0; i < num.length; i++)
sum += num[i];

Integer[][] dp = new Integer[num.length][sum + 1];
return this.canPartitionRecursive(dp, num, 0, 0, 0);
}

private int canPartitionRecursive(Integer[][] dp, int[] num, int currentIndex, int sum1, int sum2) {
// base check
if(currentIndex == num.length)
return Math.abs(sum1 - sum2);

// check if we have not already processed similar problem
if(dp[currentIndex][sum1] == null) {
// recursive call after including the number at the currentIndex in the first set
int diff1 = canPartitionRecursive(dp, num, currentIndex + 1, sum1 + num[currentIndex], sum2);

// recursive call after including the number at the currentIndex in the second set
int diff2 = canPartitionRecursive(dp, num, currentIndex + 1, sum1, sum2 + num[currentIndex]);

dp[currentIndex][sum1] = Math.min(diff1, diff2);
}

return dp[currentIndex][sum1];
}

public static void main(String[] args) {
PartitionSet ps = new PartitionSet();
int[] num = {1, 2, 3, 9};
System.out.println(ps.canPartition(num));
num = new int[]{1, 2, 7, 1, 5};
System.out.println(ps.canPartition(num));
num = new int[]{1, 3, 100, 4};
System.out.println(ps.canPartition(num));
}
}

RUN

SAVERESET

Bottom-up Dynamic Programming #

Let’s assume ‘S’ represents the total sum of all the numbers. So what we are trying to achieve in this problem is to find a subset whose sum is as close to ‘S/2’ as possible, because if we can partition the given set into two subsets of an equal sum, we get the minimum difference i.e. zero. This transforms our problem to Subset Sum, where we try to find a subset whose sum is equal to a given number-- ‘S/2’ in our case. If we can’t find such a subset, then we will take the subset which has the sum closest to ‘S/2’. This is easily possible, as we will be calculating all possible sums with every subset.

Essentially, we need to calculate all the possible sums up to ‘S/2’ for all numbers. So how do we populate the array db[TotalNumbers][S/2+1] in the bottom-up fashion?

For every possible sum ‘s’ (where 0 <= s <= S/2), 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 two above scenarios is true, we can find a subset with a sum equal to ‘s’. We should dig into this before we can learn how to find the closest subset.

Let’s draw this visually, with the example input {1, 2, 3, 9}. Since the total sum is ‘15’, therefore, we will try to find a subset whose sum is equal to the half of it i.e. ‘7’.

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

1 of 11

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

2 of 11

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 11

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

4 of 11

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

5 of 11

sum: 4-7, index:1=> (dp[index-1][sum] || dp[index-1][sum-2])

6 of 11

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

7 of 11

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

8 of 11

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

9 of 11

sum: 7, index:2=> (dp[index-1][sum] || dp[index-1][sum-3])

10 of 11

sum: 1-7, index:1=> (dp[index-1][sum] , as the 'sum' is always less than the number (9)

11 of 11

The above visualization tells us that it is not possible to find a subset whose sum is equal to ‘7’. So what is the closest subset we can find? We can find such a subset if we start moving backward in the last row from the bottom right corner to find the first ‘T’. The first “T” in the above diagram is the sum ‘6’, which means we can find a subset whose sum is equal to ‘6’. This means the other set will have a sum of ‘9’, and the minimum difference will be ‘3’.

Code #

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

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class PartitionSet {

public int canPartition(int[] num) {
int sum = 0;
for (int i = 0; i < num.length; i++)
sum += num[i];

int n = num.length;
boolean[][] dp = new boolean[n][sum/2 + 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 that number
for(int s=1; s <= sum/2 ; 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/2; 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]];
}
}
}

int sum1 = 0;
// Find the largest index in the last row which is true
for (int i = sum / 2; i >= 0; i--) {
if (dp[n-1][i] == true) {
sum1 = i;
break;
}
}

int sum2 = sum - sum1;
return Math.abs(sum2-sum1);
}

public static void main(String[] args) {
PartitionSet ps = new PartitionSet();
int[] num = {1, 2, 3, 9};
System.out.println(ps.canPartition(num));
num = new int[]{1, 2, 7, 1, 5};
System.out.println(ps.canPartition(num));
num = new int[]{1, 3, 100, 4};
System.out.println(ps.canPartition(num));
}
}

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 total sum of all the numbers.

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