Learn and practice the Dynamic Programming pattern to solve coding interview problems.

Speedrun the Meta Coding Interview: Top Problems in Go

This module covers the most important problem-solving technique in the optimization domain: Dynamic Programming. In computer science, we often come across problems that have a well-defined structure, where the same subproblem is encountered and solved multiple times. Dynamic Programming allows us to compute once and reuse the solutions to recurring subproblems.

Optimization Techniques

# Statement

You are given a string, `s`, and an array of strings, `wordDict`, representing a dictionary. Your task is to add spaces to `s` to break it up into a sequence of valid words from `wordDict`. We are required to return an array of all possible sequences of words (sentences). The order in which the sentences are listed is not significant. 

> **Note:** The same dictionary word may be reused multiple times in the segmentation.


**Constraints:**

- $1 \leq$ `s.length` $\leq 20$

- $1 \leq$ `wordDict.length` $\leq 1000$

- $1 \leq$ `wordDict[i].length` $\leq 10$

- `s` and `wordDict[i]` consist of only lowercase English letters.

- All the strings of `wordDict` are unique.

["pine apple pen apple", "pineapple pen apple", "pine applepen apple"]

["pine apple pen apple", "pine applepen apple"]

What is the output if the following input string and dictionary are provided as input?

s = "pineapplepenapple"

word_dict = ["apple", "pen", "applepen", "pine", "pineapple"]

What is the output if the following input string and dictionary are provided as input?
s = “pineapplepenapple”
word_dict = [“apple”, “pen”, “applepen”, “pine”, “pineapple”]

What is the output if the following input string and dictionary are provided as input?

s = "cattos"

word_dict = ["cats", "cat", "to", "os"]

What is the output if the following input string and dictionary are provided as input?
s = “cattos”
word_dict = [“cats”, “cat”, “to”, “os”]

["super dogs", "super dogs s", "super dog s"]

What is the output if the following input string and dictionary are provided as input?

s = "superdogs"

word_dict = ["super", "dogs", "dog", "s"]

What is the output if the following input string and dictionary are provided as input?
s = “superdogs”
word_dict = [“super”, “dogs”, “dog”, “s”]

Word Break II

Word Break II

# Quick Fix\nYou are Edward, a Senior Software Engineer who specializes in coding interview patterns, data structures and algorithms, and is great at debugging issues in code.\n\n- Edward, please keep in mind that you do not have to print the data of the placeholders. Also, do not mention anything about yourself.\n\nEdward, you are given the following data in the placeholders:\n- Function name: `wordBreak`\n- Candidate's submitted solution code: USER_CODE\n- Error: ERROR_DESCRIPTION\n- Failed test case: FAILED_TEST_CASE\n- Coding language: `Go`\n\n- Edward, your task is to understand the type of error that is occurring in the coding language provided.\n- Once you've understood the error as per the error’s description, and why the test case failed, you will suggest how the candidate can improve their solution code to fix the error.\n- Edward, you will not give the corrected code, just a hint on what the error tells, and how to fix it.\n- Edward, please limit your response strictly to 100 words. No more words than the limit.

# Review Code\nEdward, your current task is to review the `wordBreak` function(s) and the other function(s) invoked in it/them.\n\n-Take your time and review the candidate's solution code.\n- Do not review any extra/helper files or functions not invoked in the solution function `wordBreak`.\n- Do not review the code of data structure objects instantiated like a binary tree, linked list, trie, union find, etc., in the solution functions. Just review the solution/algorithm function of the problem. \n- Also, if the solution code doesn’t use the coding pattern stated above, then please encourage the candidate to reattempt the problem using that pattern corectly.

# Review time complexity\n- Please provide a brief time complexity analysis of the function(s) `wordBreak`. \n- While calculating time complexity, include the time complexity of built-in Go functions invoked by the code. \n- The answer should be to the point in approximately 60 words.

# Review space complexity\n- Please provide a brief space complexity analysis of the function(s) `wordBreak`. \n- While calculating space complexity, include the space complexity of built-in Go functions invoked by the code. \n- Do not include the space used for the input parameters and the space used for just returning an output (not used as a scratch space). \n- The answer should be to the point in approximately 60 words.

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
        """
        def word_break(s, wordDict):
            return helper(s, set(wordDict), {}) #Calling the helper function
            
        def helper(query, dict, result): #Helper Function that breaks down the string into words from subs
            if not query: #If s is empty string
                return []
            
            if query in result:
                return result[query]
            
            res = []
            for word in dict: #Verifying if s can be broken down further
                if not query.startswith(word):
                    continue
                if len(word) == len(query):
                    res.append(word)
                else:
                    resultOfTheRest = helper(query[len(word):], dict, result)
                    for item in resultOfTheRest:
                        item = word + ' ' + item
                        res.append(item)
            result[query] = res
            return res

        expected = word_break(self.__inputs[0], self.__inputs[1])
        actual_output = self.__actual_outputs[0]
        # 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(expected)

        if actual_output != None and sorted(actual_output) == sorted(expected):
            self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
        else:
            self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)

        return self.__edu_result

package main

func wordBreak(s string, wordDict []string) []string {

  	// Replace this placeholder return statement with your code
    return make([]string, 0)
}

word-break-II.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : magiclly
    - 2 : ["ag", "al", "icl", "mag", "magic", "ly", "lly"]
    output: 
    - 1 : ["magic lly", "mag icl ly"]
  - name: "testcase 2"
    inputs:
    - 1 : "raincoats"
    - 2 : ["rain", "oats", "coat", "s", "rains", "oat", "coats", "c"]
    output:
    - 1 : ["rain c oats", "rain c oat s", "rain coats", "rain coat s"]
  - name: "testcase 3"
    inputs:
    - 1 : highway
    - 2 : [ "crash", "cream", "high", "highway", "low", "way"]
    output: 
    - 1 : ["high way", "highway"]
  - name: "testcase 4"
    inputs:
    - 1 : robocat
    - 2 : ["rob", "cat", "robo", "bo", "b"]
    output:
    - 1 : ["robo cat"] 
  - name: "testcase 5"
    inputs:
    - 1 : cocomomo
    - 2 : ["co", "mo", "coco", "momo"]
    output: 
    - 1 : ["co co mo","coco mo"]

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
 """
 inp = self.__inputs[1]

 if not isinstance(inp, list):
 self.store_result ("Input 2: Expected strings array", ValidationFunction.FAIL)

 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 elif len(self.__inputs[0]) < 1 or len(self.__inputs[0]) > 20:
 self.store_result ("The string length should be in range [1, 20]", ValidationFunction.FAIL)
 elif len(self.__inputs[1]) < 1 or len(self.__inputs[1]) > 1000:
 self.store_result("The dictionary length should be in range [1, 1000]", ValidationFunction.FAIL)
 elif any(len(x) < 1 or len(x) > 10 for x in self.__inputs[1]):
 self.store_result("The dictionary words length should be in range [1, 10]", ValidationFunction.FAIL)
 elif not all(ch.islower() for ch in self.__inputs[0]) or not all(all(ch.islower() for ch in word) for word in self.__inputs[1]):
 self.store_result("The dictionary words should be lowercase English characters only [a, z]", ValidationFunction.FAIL)
 elif len(self.__inputs[1]) != len(set(self.__inputs[1])):
 self.store_result("All dictionary words should be unique", ValidationFunction.FAIL)
 else:
 self.store_result("", ValidationFunction.PASS)

Dynamic Programming

Conclusion

Word Break II

Statement

Examples

Understand the problem

Figure it out!

Try it yourself