Dynamic Programming (DP) on Arrays Tutorial

We know that Dynamic Programming is a way to reduce the time complexity of a problem using memoization or tabulation of the overlapping states. While applying DP on arrays the array indices act as DP states and transitions occurs between indices.

• : Generally, in Competitive Programming the required solution should perform nearly 10^7 to 10^8 operations, so if the array size N = 1000 to 5000 we can expect to have an N^2 DP solution, and for N = 100 we can have an N^3 DP solution.
• Dynamic Programming can be applied if the problem is to maximize or minimize any value over all possible scenarios.
• : If the question consists of functions f(x) as operations then we can try to think towards Dynamic programming.
• : If you can think of a test case where the Greedy solution fails, then the problem has a high chance of having a DP solution.
• : Determining whether the problem has a recursive solution where an over-lappingsubproblem occurs is a difficult task but if you manage to do that, then DP can be a solution if the constraints allow that.

Before coding the solution, we need to verify whether the time complexity of our solution fits the given constraints or not, we can approximate the time complexity using the below formula:

Time Complexity = Total Number of States * ( 1 + Average number of Transitions per State)

Space Complexity = Average number of States.

The problems that we see on Competitive Programming platforms are mainly made up of two parts:

• Adhoc part
• Standard part

The Adhoc part of the problem is the one that makes each problem unique and also the one that decides the difficulty of the problem. One needs to make several observations to deduce the Adhoc part into the standard solution. Good dp problems will often require you to make adhoc observations to figure out some properties that allow you to apply dp on them.

Let's us see how an Adhoc problem changes a standard DP problem:

: You are given N tiles 1 to N, and initially you are standing at 1. In one move you can make a jump of either 1 or 2 in forward direction. Determine the total number of ways you can reach from tile 1 to tile N.

: N=4

: total 3 ways are possible

: 1->2->3->4

1->2->4

1->3->4

: The Adhoc part of the problem diverts the question towards thinking of moving on tiles from 1 to N in different ways, the programmer will dry run several test cases and make several observations in order to reach to the solution, he might even try to make some mathematical formula to calculate the solution.

: If observed carefully this question is nothing but print the N'th fibonacci number the simplest DP problem that might exist. each tile T can be reached only via previous two tiles that is T is dependent upon T-1 and T-2. Using This we can write the DP states as:

• Number _of_ways[i]= Number _of_ways[i-1] + Number _of_ways[i-2]

• Storing Primality and Smallest Prime factor for First N natural number factors Using Seive
• Storing Factorials and Inverse Factorials for First N natural numbers:

The recurrence relation for the factorial of a number n is given by:

• Factorial [n] = factorial[n-1] * n

The recurrence relation for the inverse factorial of a number n is given by:

• Inverse_fact[n] = (n+1) * (inverse_fact[n+1])

• Prefix Arrays

When tackling DP problems during a contest the most difficult task is to define the states and transitions for the problem. The states and transitions for problems are mostly the variations of existing standard DP problems, let us see how we can try to think towards a particular standard DP problem:

0/1 Knapsack Problem:

• Weight/ Volume/Value Constraints in the problem.
• 0-1 decision problem, i.e. either include or exclude an item to the final result.
• Task is to minimize or maximize the profit.
• O(N^2) solution is permissible

Coin Change:

• Array values can be used as a denomination on a particular sum.
• Minimum array values required to achieve a target.
• Total number of ways required as a result.
• O(N^2) solution is permissible.

Fibonacci:

• For a particular index i, dp[i] depends on some previously calculated indices.
• O(N) solution is required.
• State Transition: F(n)=F(n-1) + F(n-2)

Matrix Chain Multiplication:

• If Each pair of indices {i,j} have to be used as a State then we can try to think towards MCM approach.
• Constraints allow a time complexity of O(N^3)

Longest Increasing Subsequence (LIS):

• dp[i] represents the length of the LIS ending at index i. This idea can be extended to the context where we can calculate the best answer ending at ith index.

Edit Distance:

• In Edit Distance, dp[i][j] represents the minimum number of operations to convert the first i characters of string1 to the first j characters of string2.

Transition:

• If characters s1[i] and s2[j] are the same: dp[i][j] = dp[i-1][j-1].
• Otherwise: dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1.

The transitions in Edit distance are very commonly seen in many questions where a particular dp value depends upon previously calculated adjacent dp cells.

Longest Common Subsequence (LCS):

• In LCS dp[i][j] represents the length of the LCS of the first i characters of string1 and the first j characters of string2.

Transition:

• If characters s1[i] and s2[j] are the same: dp[i][j] = dp[i-1][j-1] + 1.
• Otherwise: dp[i][j] = max(dp[i-1][j], dp[i][j-1]).

Similar to Edit distance, LCS transitions can also be very frequently observed in various problems.