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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

Contact Us

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Coin Change

We'll cover the following

Introduction

Problem Statement

Basic Solution

Code

Top-down Dynamic Programming with Memoization

Bottom-up Dynamic Programming

Code

Introduction #

Given an infinite supply of ‘n’ coin denominations and a total money amount, we are asked to find the total number of distinct ways to make up that amount.

Example:

Denominations: {1,2,3}
Total amount: 5
Output: 5
Explanation: There are five ways to make the change for '5', here are those ways:
1. {1,1,1,1,1}
2. {1,1,1,2}
3. {1,2,2}
4. {1,1,3}
5. {2,3}

Problem Statement #

Given a number array to represent different coin denominations and a total amount ‘T’, we need to find all the different ways to make a change for ‘T’ with the given coin denominations. We can assume an infinite supply of coins, therefore, each coin can be chosen multiple times.

This problem follows the Unbounded Knapsack pattern.

Basic Solution #

A basic brute-force solution could be to try all combinations of the given coins to select the ones that give a total sum of ‘T’. This is what our algorithm will look like:

for each coin 'c'
create a new set which includes one quantity of coin 'c' if it does not exceed 'T', and
recursively call to process all coins
create a new set without coin 'c', and recursively call to process the remaining coins
return the count of sets who have a sum equal to 'T'

This problem is quite similar to Count of Subset Sum. The only difference here is that after including the item (i.e. coin), we recursively call to process all the items (including the current coin). In ‘Count of Subset Sum’, however, we were recursively calling to process only the remaining items.

Code #

Here is the code for the brute-force solution:

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class CoinChange {

public int countChange(int[] denominations, int total) {
return this.countChangeRecursive(denominations, total, 0);
}

private int countChangeRecursive(int[] denominations, int total, int currentIndex) {
// basic checks
if (total == 0)
return 1;

if(denominations.length == 0 || currentIndex >= denominations.length)
return 0;

// recursive call after selecting the coin at the currentIndex
// if the coin at currentIndex exceeds the total, we shouldn't process this
int sum1 = 0;
if( denominations[currentIndex] <= total )
sum1 = countChangeRecursive(denominations, total - denominations[currentIndex], currentIndex);

// recursive call after excluding the coin at the currentIndex
int sum2 = countChangeRecursive(denominations, total, currentIndex + 1);

return sum1 + sum2;
}

public static void main(String[] args) {
CoinChange cc = new CoinChange();
int[] denominations = {1, 2, 3};
System.out.println(cc.countChange(denominations, 5));
}
}

RUN

SAVERESET

The time complexity of the above algorithm is exponential O(2C+T)O(2^{C+T})O(2​C+T​​), where ‘C’ represents total coin denominations and ‘T’ is the total amount that we want to make change. The space complexity will be O(C+T)O(C+T)O(C+T).

Let’s try to find a better solution.

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 solved sub-problems. As mentioned above, we need to store results for every coin combination and for every possible sum:

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class CoinChange {

public int countChange(int[] denominations, int total)
{
Integer[][] dp = new Integer[denominations.length][total + 1];
return this.countChangeRecursive(dp, denominations, total, 0);
}

private int countChangeRecursive(Integer[][] dp, int[] denominations, int total, int currentIndex)
{
// base checks
if (total == 0)
return 1;

if(denominations.length == 0 || currentIndex >= denominations.length)
return 0;

// if we have already processed a similar sub-problem, return the result from memory
if(dp[currentIndex][total] != null)
return dp[currentIndex][total];

// recursive call after selecting the coin at the currentIndex
// if the number at currentIndex exceeds the total, we shouldn't process this
int sum1 = 0;
if( denominations[currentIndex] <= total )
sum1 = countChangeRecursive(dp, denominations, total - denominations[currentIndex], currentIndex);

// recursive call after excluding the number at the currentIndex
int sum2 = countChangeRecursive(dp, denominations, total, currentIndex + 1);

dp[currentIndex][total] = sum1 + sum2;
return dp[currentIndex][total];
}

public static void main(String[] args) {
CoinChange cc = new CoinChange();
int[] denominations = {1, 2, 3};
System.out.println(cc.countChange(denominations, 5));
}
}

RUN

SAVERESET

Bottom-up Dynamic Programming #

We will try to find if we can make all possible sums, with every combination of coins, to populate the array dp[TotalDenominations][Total+1].

So for every possible total ‘t’ (0<= t <= Total) and for every possible coin index (0 <= index < denominations.length), we have two options:

Exclude the coin. Count all the coin combinations without the given coin up to the total ‘t’ => dp[index-1][t]

Include the coin if its value is not more than ‘t’. In this case, we will count all the coin combinations to get the remaining total: dp[index][t-denominations[index]]

Finally, to find the total combinations, we will add both the above two values:

dp[index][t] = dp[index-1][t] + dp[index][t-denominations[index]]

Let’s draw this visually with the following example:

Denominations: [1, 2, 3]
Total: 5

Let’s start with our base case of zero total:

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

1 of 12

total: 1, index:0=> dp[index][total - denominations[index], since index =0, we didn't consider dp[index-1][total]

2 of 12

total: 2-5, index:0=> dp[index][total - denominations[index], since index =0, we didn't consider dp[index-1][total]

3 of 12

total: 1, index:1=> dp[index-1][total] , we didn't consider dp[index][total - denominations[index], since 'total < denominations[index]'

4 of 12

total: 2, index:1=> dp[index-1][total] + dp[index][total - denominations[index]

5 of 12

total: 3, index:1=> dp[index-1][total] + dp[index][total - denominations[index]

6 of 12

total: 4, index:1=> dp[index-1][total] + dp[index][total - denominations[index]

7 of 12

total: 5, index:1=> dp[index-1][total] + dp[index][total - denominations[index]

8 of 12

total: 1-2, index:2=> dp[index-1][total] , we didn't consider dp[index][total - denominations[index], since 'total < denominations[index]'

9 of 12

total: 3, index:2=> dp[index-1][total] + dp[index][total - denominations[index]

10 of 12

total: 4, index:2=> dp[index-1][total] + dp[index][total - denominations[index]

11 of 12

total: 5, index:2=> dp[index-1][total] + dp[index][total - denominations[index]

12 of 12

Code #

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

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class CoinChange {

public int countChange(int[] denominations, int total) {
int n = denominations.length;
int[][] dp = new int[n][total + 1];

// populate the total=0 columns, as we will always have an empty set for zero toal
for(int i=0; i < n; i++)
dp[i][0] = 1;

// process all sub-arrays for all capacities
for(int i=0; i < n; i++) {
for(int t=1; t <= total; t++) {
if(i > 0)
dp[i][t] = dp[i-1][t];
if(t >= denominations[i])
dp[i][t] += dp[i][t-denominations[i]];
}
}

// total combinations will be at the bottom-right corner.
return dp[n-1][total];
}

public static void main(String[] args) {
CoinChange cc = new CoinChange();
int[] denominations = {1, 2, 3};
System.out.println(cc.countChange(denominations, 5));
}
}

RUN

SAVERESET

The above solution has time and space complexity of O(C∗T)O(C*T)O(C∗T), where ‘C’ represents total coin denominations and ‘T’ is the total amount that we want to make change.

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