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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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Target Sum

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Problem Statement
*

Example 1:

Example 2:

Solution

Code

Space Optimized Solution

Problem Statement #

Given a set of positive numbers (non zero) and a target sum ‘S’. Each number should be assigned either a ‘+’ or ‘-’ sign. We need to find out total ways to assign symbols to make the sum of numbers equal to target ‘S’.

Example 1: #

Input: {1, 1, 2, 3}, S=1
Output: 3
Explanation: The given set has '3' ways to make a sum of '1': {+1-1-2+3} & {-1+1-2+3} & {+1+1+2-3}

Example 2: #

Input: {1, 2, 7, 1}, S=9
Output: 2
Explanation: The given set has '2' ways to make a sum of '9': {+1+2+7-1} & {-1+2+7+1}

Solution #

This problem follows the 0/1 Knapsack pattern and can be converted into Count of Subset Sum. Let’s dig into this.

We are asked to find two subsets of the given numbers whose difference is equal to the given target ‘S’. Take the first example above. As we saw, one solution is {+1-1-2+3}. So, the two subsets we are asked to find are {1, 3} & {1, 2} because,

(1 + 3) - (1 + 2 ) = 1

Now, let’s say ‘Sum(s1)’ denotes the total sum of set ‘s1’, and ‘Sum(s2)’ denotes the total sum of set ‘s2’. So the required equation is:

Sum(s1) - Sum(s2) = S

This equation can be reduced to the subset sum problem. Let’s assume that ‘Sum(num)’ denotes the total sum of all the numbers, therefore:

Sum(s1) + Sum(s2) = Sum(num)

Let’s add the above two equations:

=> Sum(s1) - Sum(s2) + Sum(s1) + Sum(s2) = S + Sum(num)
=> 2 * Sum(s1) =  S + Sum(num)
=> Sum(s1) = (S + Sum(num)) / 2

This essentially converts our problem to: “Find count of subsets of the given numbers whose sum is equal to”,

=> (S + Sum(num)) / 2

Code #

Let’s take the dynamic programming code of Count of Subset Sum and extend it to solve this problem:

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class TargetSum {

public int findTargetSubsets(int[] num, int s) {
int totalSum = 0;
for (int n : num)
totalSum += n;

// if 's + totalSum' is odd, we can't find a subset with sum equal to '(s + totalSum) / 2'
if(totalSum < s || (s + totalSum) % 2 == 1)
return 0;

return countSubsets(num, (s + totalSum) / 2);
}

// this function is exactly similar to what we have in 'Count of Subset Sum' problem.
private int countSubsets(int[] num, int sum) {
int n = num.length;
int[][] dp = new int[n][sum + 1];

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

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

// process all subsets for all sums
for(int i=1; i < num.length; i++) {
for(int s=1; s <= sum; s++) {
dp[i][s] = dp[i-1][s];
if(s >= num[i])
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) {
TargetSum ts = new TargetSum();
int[] num = {1, 1, 2, 3};
System.out.println(ts.findTargetSubsets(num, 1));
num = new int[]{1, 2, 7, 1};
System.out.println(ts.findTargetSubsets(num, 9));
}
}

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 desired sum.

We can further improve the solution to use only O(S)O(S)O(S) space.

Space Optimized Solution #

Here is the code for the space-optimized solution, using only a single array:

Java

Java

Java

JS

JS

Python3

Python3

C++

C++

class TargetSum {

public int findTargetSubsets(int[] num, int s) {
int totalSum = 0;
for (int n : num)
totalSum += n;

// if 's + totalSum' is odd, we can't find a subset with sum equal to '(s + totalSum) / 2'
if(totalSum < s || (s + totalSum) % 2 == 1)
return 0;

return countSubsets(num, (s + totalSum) / 2);
}

// this function is exactly similar to what we have in 'Count of Subset Sum' problem.
private int countSubsets(int[] num, int sum) {
int n = num.length;
int[] dp = new int[sum + 1];
dp[0] = 1;

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

// process all subsets for all sums
for(int i=1; i < num.length; i++) {
for(int s=sum; s >= 0; s--) {
if(s >= num[i])
dp[s] += dp[s-num[i]];
}
}

return dp[sum];
}

public static void main(String[] args) {
TargetSum ts = new TargetSum();
int[] num = {1, 1, 2, 3};
System.out.println(ts.findTargetSubsets(num, 1));
num = new int[]{1, 2, 7, 1};
System.out.println(ts.findTargetSubsets(num, 9));
}
}

RUN

SAVERESET

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