Introduction
Dynamics programming is a widely used concept and its often used for optimization. It refers to simplifying a complicated problem by breaking it down into simpler subproblems in a recursive manner usually Bottom up approach. There are two key attributes that a problem must have in order for dynamic programming to be applicable "Optimal substructure" and "Overlapping subproblems".To achieve its optimization, Dynamics programming uses a concept called Memorization
Remarks
Dynamic Programming is an improvement on Brute Force, see this example to understand how one can obtain a Dynamic Programming solution from Brute Force.
A Dynamic Programming Solution has 2 main requirements:

Overlapping Problems

Optimal Substructure
Overlapping Subproblems means that results of smaller versions of the problem are reused multiple times in order to arrive at the solution to the original problem
Optimal Substructure means that there is a method of calculating a problem from its subproblems.
A Dynamic Programming Solution has 2 main components, the State and the Transition
The State refers to a subproblem of the original problem.
The Transition is the method to solve a problem based on its subproblems
The time taken by a Dynamic Programming Solution can be calculated as No. of States * Transition Time
. Thus if a solution has N^2
states and the transition is O(N)
, then the solution would take roughly O(N^3)
time.
Edit Distance
The problem statement is like if we are given two string str1 and str2 then how many minimum number of operations can be performed on the str1 that it gets converted to str2.
Implementation in Java
public class EditDistance {
public static void main(String[] args) {
// TODO Autogenerated method stub
String str1 = "march";
String str2 = "cart";
EditDistance ed = new EditDistance();
System.out.println(ed.getMinConversions(str1, str2));
}
public int getMinConversions(String str1, String str2){
int dp[][] = new int[str1.length()+1][str2.length()+1];
for(int i=0;i<=str1.length();i++){
for(int j=0;j<=str2.length();j++){
if(i==0)
dp[i][j] = j;
else if(j==0)
dp[i][j] = i;
else if(str1.charAt(i1) == str2.charAt(j1))
dp[i][j] = dp[i1][j1];
else{
dp[i][j] = 1 + Math.min(dp[i1][j], Math.min(dp[i][j1], dp[i1][j1]));
}
}
}
return dp[str1.length()][str2.length()];
}
}
Output
3
Fibonacci Number
Bottom up approach for printing the nth Fibonacci number using Dynamic Programming.
Recursive Tree
fib(5)
/ \
fib(4) fib(3)
/ \ / \
fib(3) fib(2) fib(2) fib(1)
/ \ / \ / \
fib(2) fib(1) fib(1) fib(0) fib(1) fib(0)
/ \
fib(1) fib(0)
Overlapping Subproblems
Here fib(0),fib(1) and fib(3) are the overlapping subproblems.fib(0) is getting repeated 3 times, fib(1) is getting repeated 5 times and fib(3) is getting repeated 2 times.
Implementation
public int fib(int n){
int f[] = new int[n+1];
f[0]=0;f[1]=1;
for(int i=2;i<=n;i++){
f[i]=f[i1]+f[i2];
}
return f[n];
}
Time Complexity
O(n)
Knapsack Problem
01 Knapsack
The Knapsack Problem is a problem when given a set of items, each with a weight, a value and exactly 1 copy, determine the which item(s) to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible.
C++ Example:
Implementation:
int knapsack(vector<int> &value, vector<int> &weight, int N, int C){
int dp[C+1];
for (int i = 1; i <= C; ++i){
dp[i] = 100000000;
}
dp[0] = 0;
for (int i = 0; i < N; ++i){
for (int j = C; j >= weight[i]; j){
dp[j] = max(dp[j],dp[jweight[i]]+value[i]);
}
}
return dp[C];
}
Test:
3 5
5 2
2 1
3 2
Output:
3
That means the maximum value can be achieved is 3, which is achieved by choosing (2,1) and (3,2).
Unbounded Knapsack
The Unbounded Knapsack Problem is a problem which given a set of items, each with a weight, a value and infinite copies, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible.
Python(2.7.11) Example:
Implementation:
def unbounded_knapsack(w, v, c): # weight, value and capactiy
m = [0]
for r in range(1, c+1):
val = m[r1]
for i, wi in enumerate(w):
if wi > r:
continue
val = max(val, v[i] + m[rwi])
m.append(val)
return m[c] # return the maximum value can be achieved
The complexity of that implementation is O(nC)
, which n is number of items.
Test:
w = [2, 3, 4, 5, 6]
v = [2, 4, 6, 8, 9]
print unbounded_knapsack(w, v, 13)
Output:
20
That means the maximum value can be achieved is 20, which is achieved by choosing (5, 8), (5, 8) and (3, 4).
Longest Common Subsequence
If we are given with the two strings we have to find the longest common subsequence present in both of them.
Example
LCS for input Sequences “ABCDGH” and “AEDFHR” is “ADH” of length 3.
LCS for input Sequences “AGGTAB” and “GXTXAYB” is “GTAB” of length 4.
Implementation in Java
public class LCS {
public static void main(String[] args) {
// TODO Autogenerated method stub
String str1 = "AGGTAB";
String str2 = "GXTXAYB";
LCS obj = new LCS();
System.out.println(obj.lcs(str1, str2, str1.length(), str2.length()));
System.out.println(obj.lcs2(str1, str2));
}
//Recursive function
public int lcs(String str1, String str2, int m, int n){
if(m==0  n==0)
return 0;
if(str1.charAt(m1) == str2.charAt(n1))
return 1 + lcs(str1, str2, m1, n1);
else
return Math.max(lcs(str1, str2, m1, n), lcs(str1, str2, m, n1));
}
//Iterative function
public int lcs2(String str1, String str2){
int lcs[][] = new int[str1.length()+1][str2.length()+1];
for(int i=0;i<=str1.length();i++){
for(int j=0;j<=str2.length();j++){
if(i==0  j== 0){
lcs[i][j] = 0;
}
else if(str1.charAt(i1) == str2.charAt(j1)){
lcs[i][j] = 1 + lcs[i1][j1];
}else{
lcs[i][j] = Math.max(lcs[i1][j], lcs[i][j1]);
}
}
}
return lcs[str1.length()][str2.length()];
}
}
Output
4
Longest Common Substring
Given 2 string str1 and str2 we have to find the length of the longest common substring between them.
Examples
Input : X = "abcdxyz", y = "xyzabcd" Output : 4
The longest common substring is "abcd" and is of length 4.
Input : X = "zxabcdezy", y = "yzabcdezx" Output : 6
The longest common substring is "abcdez" and is of length 6.
Implementation in Java
public int getLongestCommonSubstring(String str1,String str2){
int arr[][] = new int[str2.length()+1][str1.length()+1];
int max = Integer.MIN_VALUE;
for(int i=1;i<=str2.length();i++){
for(int j=1;j<=str1.length();j++){
if(str1.charAt(j1) == str2.charAt(i1)){
arr[i][j] = arr[i1][j1]+1;
if(arr[i][j]>max)
max = arr[i][j];
}
else
arr[i][j] = 0;
}
}
return max;
}
Time Complexity
O(m*n)
Weighted Job Scheduling Algorithm
Weighted Job Scheduling Algorithm can also be denoted as Weighted Activity Selection Algorithm.
The problem is, given certain jobs with their start time and end time, and a profit you make when you finish the job, what is the maximum profit you can make given no two jobs can be executed in parallel?
This one looks like Activity Selection using Greedy Algorithm, but there's an added twist. That is, instead of maximizing the number of jobs finished, we focus on making the maximum profit. The number of jobs performed doesn't matter here.
Let's look at an example:
++++++++
 Name  A  B  C  D  E  F 
++++++++
(Start Time, Finish Time) (2,5)  (6,7)  (7,9)  (1,3)  (5,8)  (4,6) 
++++++++
 Profit  6  4  2  5  11  5 
++++++++
The jobs are denoted with a name, their start and finishing time and profit. After a few iterations, we can find out if we perform JobA and JobE, we can get the maximum profit of 17. Now how to find this out using an algorithm?
The first thing we do is sort the jobs by their finishing time in nondecreasing order. Why do we do this? It's because if we select a job that takes less time to finish, then we leave the most amount of time for choosing other jobs. We have:
++++++++
 Name  D  A  F  B  E  C 
++++++++
(Start Time, Finish Time) (1,3)  (2,5)  (4,6)  (6,7)  (5,8)  (7,9) 
++++++++
 Profit  5  6  5  4  11  2 
++++++++
We'll have an additional temporary array Acc_Prof of size n (Here, n denotes the total number of jobs). This will contain the maximum accumulated profit of performing the jobs. Don't get it? Wait and watch. We'll initialize the values of the array with the profit of each jobs. That means, Acc_Prof[i] will at first hold the profit of performing ith job.
++++++++
 Acc_Prof  5  6  5  4  11  2 
++++++++
Now let's denote position 2 with i, and position 1 will be denoted with j. Our strategy will be to iterate j from 1 to i1 and after each iteration, we will increment i by 1, until i becomes n+1.
j i
++++++++
 Name  D  A  F  B  E  C 
++++++++
(Start Time, Finish Time) (1,3)  (2,5)  (4,6)  (6,7)  (5,8)  (7,9) 
++++++++
 Profit  5  6  5  4  11  2 
++++++++
 Acc_Prof  5  6  5  4  11  2 
++++++++
We check if Job[i] and Job[j] overlap, that is, if the finish time of Job[j] is greater than Job[i]'s start time, then these two jobs can't be done together. However, if they don't overlap, we'll check if Acc_Prof[j] + Profit[i] > Acc_Prof[i]. If this is the case, we will update Acc_Prof[i] = Acc_Prof[j] + Profit[i]. That is:
if Job[j].finish_time <= Job[i].start_time
if Acc_Prof[j] + Profit[i] > Acc_Prof[i]
Acc_Prof[i] = Acc_Prof[j] + Profit[i]
endif
endif
Here Acc_Prof[j] + Profit[i] represents the accumulated profit of doing these two jobs toegther. Let's check it for our example:
Here Job[j] overlaps with Job[i]. So these to can't be done together. Since our j is equal to i1, we increment the value of i to i+1 that is 3. And we make j = 1.
j i
++++++++
 Name  D  A  F  B  E  C 
++++++++
(Start Time, Finish Time) (1,3)  (2,5)  (4,6)  (6,7)  (5,8)  (7,9) 
++++++++
 Profit  5  6  5  4  11  2 
++++++++
 Acc_Prof  5  6  5  4  11  2 
++++++++
Now Job[j] and Job[i] don't overlap. The total amount of profit we can make by picking these two jobs is: Acc_Prof[j] + Profit[i] = 5 + 5 = 10 which is greater than Acc_Prof[i]. So we update Acc_Prof[i] = 10. We also increment j by 1. We get,
j i
++++++++
 Name  D  A  F  B  E  C 
++++++++
(Start Time, Finish Time) (1,3)  (2,5)  (4,6)  (6,7)  (5,8)  (7,9) 
++++++++
 Profit  5  6  5  4  11  2 
++++++++
 Acc_Prof  5  6  10  4  11  2 
++++++++
Here, Job[j] overlaps with Job[i] and j is also equal to i1. So we increment i by 1, and make j = 1. We get,
j i
++++++++
 Name  D  A  F  B  E  C 
++++++++
(Start Time, Finish Time) (1,3)  (2,5)  (4,6)  (6,7)  (5,8)  (7,9) 
++++++++
 Profit  5  6  5  4  11  2 
++++++++
 Acc_Prof  5  6  10  4  11  2 
++++++++
Now, Job[j] and Job[i] don't overlap, we get the accumulated profit 5 + 4 = 9, which is greater than Acc_Prof[i]. We update Acc_Prof[i] = 9 and increment j by 1.
j i
++++++++
 Name  D  A  F  B  E  C 
++++++++
(Start Time, Finish Time) (1,3)  (2,5)  (4,6)  (6,7)  (5,8)  (7,9) 
++++++++
 Profit  5  6  5  4  11  2 
++++++++
 Acc_Prof  5  6  10  9  11  2 
++++++++
Again Job[j] and Job[i] don't overlap. The accumulated profit is: 6 + 4 = 10, which is greater than Acc_Prof[i]. We again update Acc_Prof[i] = 10. We increment j by 1. We get:
j i
++++++++
 Name  D  A  F  B  E  C 
++++++++
(Start Time, Finish Time) (1,3)  (2,5)  (4,6)  (6,7)  (5,8)  (7,9) 
++++++++
 Profit  5  6  5  4  11  2 
++++++++
 Acc_Prof  5  6  10  10  11  2 
++++++++
If we continue this process, after iterating through the whole table using i, our table will finally look like:
++++++++
 Name  D  A  F  B  E  C 
++++++++
(Start Time, Finish Time) (1,3)  (2,5)  (4,6)  (6,7)  (5,8)  (7,9) 
++++++++
 Profit  5  6  5  4  11  2 
++++++++
 Acc_Prof  5  6  10  14  17  8 
++++++++
_{* A few steps have been skipped to make the document shorter.}
If we iterate through the array Acc_Prof, we can find out the maximum profit to be 17! The pseudocode:
Procedure WeightedJobScheduling(Job)
sort Job according to finish time in nondecreasing order
for i > 2 to n
for j > 1 to i1
if Job[j].finish_time <= Job[i].start_time
if Acc_Prof[j] + Profit[i] > Acc_Prof[i]
Acc_Prof[i] = Acc_Prof[j] + Profit[i]
endif
endif
endfor
endfor
maxProfit = 0
for i > 1 to n
if maxProfit < Acc_Prof[i]
maxProfit = Acc_Prof[i]
return maxProfit
The complexity of populating the Acc_Prof array is O(n^{2}). The array traversal takes O(n). So the total complexity of this algorithm is O(n^{2}).
Now, If we want to find out which jobs were performed to get the maximum profit, we need to traverse the array in reverse order and if the Acc_Prof matches the maxProfit, we will push the name of the job in a stack and subtract Profit of that job from maxProfit. We will do this until our maxProfit > 0 or we reach the beginning point of the Acc_Prof array. The pseudocode will look like:
Procedure FindingPerformedJobs(Job, Acc_Prof, maxProfit):
S = stack()
for i > n down to 0 and maxProfit > 0
if maxProfit is equal to Acc_Prof[i]
S.push(Job[i].name
maxProfit = maxProfit  Job[i].profit
endif
endfor
The complexity of this procedure is: O(n).
One thing to remember, if there are multiple job schedules that can give us maximum profit, we can only find one job schedule via this procedure.