The ultimate guide to dynamic programming interviews. Developed by FAANG engineers, it equips you with strategic DP skills, practice with real-world questions, and patterns for efficient solutions.

Some of the toughest questions in technical interviews require dynamic programming solutions. Dynamic programming (DP) is an advanced optimization technique applied to recursive solutions. However, DP is not a one-size-fits-all technique, and it requires practice to develop the ability to identify the underlying DP patterns. With a strategic approach, coding interview prep for DP problems shouldn’t take more than a few weeks.

This course starts with an introduction to DP and thoroughly discusses five DP patterns. You’ll learn to apply each pattern to several related problems, with a visual representation of the working of the pattern, and learn to appreciate the advantages of DP solutions over naive solutions.

After completing this course, you will have the skills you need to unlock even the most challenging questions, grok the coding interview, and level up your career with confidence.

This course is also available in C++, JavaScript, and Python—with more coming soon!

Grokking Dynamic Programming Interview

# Statement

Suppose you are given a string, `s`. You need to perform the minimum number of cuts to divide the string into substrings, such that all the resulting substrings are palindromes.

Let's say you have the following string:

- "apple"

The resulting substrings after performing the minimum number of cuts are:

- "a"$|$"pp"$|$"l"$|$"e" 

So, a minimum of $3$ cuts are required.

**Constraints:**

- $1 \leq$ `s.length` $\leq 1.5\times10^3$
- `s` consists of only lowercase English characters.

# Examples

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# Try it yourself

Implement your solution in the following coding playground.

import java.util.*;
public class MinCuts {
   public static int minCuts(String arr) {

      // Write your code here
         
      return -1;
   }
}

Help me debug only the function MinCuts.I don't need the corrected code.I don't want explanations or code reviews. Just provide me the exact and to-the-point error that helps me run the code successfully on the test cases provided.

Please help me understand why this code failed for the last failed test case. Please don't give me the corrected code. Just provide me exact and to-the-point advice on how to correct the code so that it runs successfully on the test cases provided. 

Your current task is to review MinCuts function And Answer the following questions: Ensure that the Palindromic Subsequence  pattern used? Does the code follow proper language guidelines. At the end, review the code.

Please provide a brief time complexity analysis of the function MinCuts.While calculating time complexity, include the time complexity of built-in Java functions invoked by the code. The Answer should be to the point and in a few lines.

Please provide a brief Space complexity analysis of the function MinCuts.While calculating space complexity, include the space complexity of built-in Java functions invoked by the code. The Answer should be to the point and in a few lines.

def evaluate_results(self) -> object:
        """
        self.__inputs is the list of inputs that are passed as input to function
        self.__actual_output is a list of outputs that is produced by the learner's code
        return : self.__edu__result object
        """
        # multiple outputs will be added at each index respectively for a single test_case
        expected_outputs = [] 
        # store the 1st expected output that should be displayed to the user 
        expected_outputs.append(self.__inputs[0] + self.__inputs[1])

        if json.dumps(self.__inputs[0] + self.__inputs[1]) == json.dumps(self.__actual_outputs[0]):
            self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
        else:
            self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)

        return self.__edu_result

public class Main{
 public static int minSubArrayLen1(String c) {
    int m = MinCuts.minCuts(c);
    return m;
      
 }
}

testcases:
  - name: "testcase 6"
    inputs:
    - 1 : "radar"
    output: 
    - 1 : 0
  - name: "testcase 7"
    inputs:
    - 1 : "abac"
    output: 
    - 1 : 1
  - name: "testcase 8"
    inputs:
    - 1 : "book"
    output: 
    - 1 : 2
  - name: "testcase 9"
    inputs:
    - 1 : "sleek"
    output: 
    - 1 : 3
  - name: "testcase 10"
    inputs:
    - 1 : "fours"
    output: 
    - 1 : 4
  - name: "testcase 11"
    inputs:
    - 1 : "abcdefghijk"
    output: 
    - 1 : 10
  - name: "testcase 12"
    inputs:
    - 1 : "elwxubtrnarrrjguuqwwoopgwjaaeavczrdubcgfvnxeutcatt"
    output: 
    - 1 : 41
  

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : "radar"
    output: 
    - 1 : 0
  - name: "testcase 2"
    inputs:
    - 1 : "abac"
    output: 
    - 1 : 1
  - name: "testcase 3"
    inputs:
    - 1 : "book"
    output: 
    - 1 : 2
  - name: "testcase 4"
    inputs:
    - 1 : "sleek"
    output: 
    - 1 : 3
  - name: "testcase 5"
    inputs:
    - 1 : "fours"
    output: 
    - 1 : 4
  

def sum(input_1, input_2):
    return input_1+input_2

def validate_inputs(self) -> None:
        """
        self.__inputs is the list of inputs that are passed as input to function
        store the result using the self.store_result(error_msg, status)
        return : void
        """
        # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
        if self.__inputs[0] > 1000:
            self.store_result ("input 1 should be less than 1000", ValidationFunction.FAIL)
        elif self.__inputs[1] > 1000:
            self.store_result("input 2 should be less than 1000", ValidationFunction.FAIL)
        else:
            self.store_result("", ValidationFunction.PASS)


>**Note:** If you clicked the “Submit” button and the code timed out, this means that your solution needs to be optimized in terms of running time.
>
>**Hint:** Use dynamic programming and see the magic.

# Solution

We will first explore the naive recursive solution to this problem and then see how it can be improved using the Palindromic Subsequence dynamic programming pattern.

## Naive approach

A naive approach would be to generate combinations of all possible cuts that can be placed on the string such that all the resulting substrings are palindromes. We will then select the minimum number of cuts.

For example, we have the following string:

**s:** "abac"

To find the minimum number of cuts, we try all possible combinations:

- "a" |"b" |"a" |"c" — $3$ cuts
- "aba" | "c" — $1$ cut

The calculation above shows that we need a recursive solution to make all possible combinations. In other words, we divide the problem into subproblems by placing cuts at every possible position in the string. The position to place a cut is determined using a variable, `k`, that ranges from, `i` to `j - 1`, where `i` and `j` are the start and end indexes of the current substring respectively. We store the minimum number of cuts in a variable, `result`, which is initially set to `j`, i.e., the maximum number of cuts that can be placed on a string to make the resulting substrings (consisting of one character each) palindromes. This is done based on the following rules:

- **Base case:** If we have reached the end of the substring, or the current substring is a palindrome, we return $0$.

- Otherwise, we run a loop from `k = 0` to `k = s.length - 1`. On each iteration, we place a cut after the $k^{th}$ index of the string and divide it into two halves. Since we've placed one cut, the resulting cost of placing more cuts follows the following recurrence relation, where we now re-evaluate these resulting two halves:

   ```
   totalCuts = 1 + minCuts(s, i, k) + minCuts(s, k + 1, j)
   ``` 

   After an iteration for a value of `k` is complete, i.e., we have found the total number of cuts that can be placed for a value of `k`, we take the minimum of the current value of `result` and `totalCuts` and store this in `result`.

- We repeat the above process for all values of `k`. When the loop ends, `result` stores the minimum number of cuts, and we return `result`.

Let’s implement the algorithm as discussed above:



Let's solve the Palindromic Partitioning problem using Dynamic Programming.

No.	s	cuts
1	"sleek"	3
2	"adjacent"	7
3	"radar"	0

Getting Started

0/1 Knapsack

Unbounded Knapsack

Recursive Numbers

Longest Common Substring

Palindromic Subsequence

Conclusion

Palindromic Partitioning

Statement

Examples

Try it yourself