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
Given a string, `s`, and a dictionary of strings, `wordDict`, check if `s` can be segmented into a space-separated sequence of one or more dictionary words. If yes, return TRUE; else, return FALSE.



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


**Constraints:**

- $1 \leq$ `s.length` $\leq 250$

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

- $1 \leq$ `wordDict[i].length` $\leq 20$

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

- All the strings of `wordDict` are unique.

# Examples

# Understand the problem


Let’s take a moment to make sure you’ve correctly understood the problem. The quiz below helps you check if you’re solving the correct problem:





Can the string "pineapplepenapple" be split into sequences separated by spaces using the following words from the dictionary?

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

Can the string “pineapplepenapple” be split into sequences separated by spaces using the following words from the dictionary?
[“apple”, “pen”, “applepen”, “pine”, “pineapple”]

Can the string "superdogs" be split into sequences separated by spaces using the following words from the dictionary?

["super", "ogs", "dog"]



Can the string “superdogs” be split into sequences separated by spaces using the following words from the dictionary?
[“super”, “ogs”, “dog”]

To split a string into multiple substrings using a dictionary of available words

To merge multiple substrings into a single string using a dictionary of available words

To find the longest word in a string using a dictionary of available words

To find the shortest word in a string using a dictionary of available words

What is the goal of the word break problem?

What is the goal of the word break problem?

# Figure it out!

We have a game for you to play. Rearrange the logical building blocks to develop a clearer understanding of how to solve this problem.





// Importing required functions
import java.util.*;
import java.util.stream.*;

public class NumberFactors{
    // Available numbers are 1, 3 and 4
    public static long countWays (int n) {

        // Replace this placeholder return statement with your code
        return -1;
    }
}

# 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
        """
        # 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(word_break(self.__inputs[0], self.__inputs[1]))

        if json.dumps(word_break(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
# Function to check if a string can be broken down into words from the dictionary
def word_break(s, word_dict):
    n = len(s)
    word_set = set(word_dict)

    dp = [False]*(n+1)

    dp[0] = True

    for i in range(1, n+1):
        for j in range(i):
            if dp[j] and s[j:i] in word_set:
                dp[i] = True
                break
    return dp[n]
    
   

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

word-break.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : magiclly
    - 2 : ["ag", "al", "icl", "mag", "magic", "ly", "lly"]
    output: 
    - 1 : True

  - name: "testcase 2"
    inputs:
    - 1 : raincoats
    - 2 : ["rain", "oats", "coat", "s", "rains", "oat", "coats", "c"]
    output:
    - 1 : True
    
  - name: "testcase 3"
    inputs:
    - 1 : highway
    - 2 : [ "crash", "cream", "high", "low", "away"]
    output: 
    - 1 : False
    
  - name: "testcase 4"
    inputs:
    - 1 : robocat
    - 2 : ["rob", "cat", "obo", "bo", "b"]
    output:
    - 1 : False

  - name: "testcase 5"
    inputs:
    - 1 : cocomomo
    - 2 : ["c", "m", "coc", "o", "momo"]
    output: 
    - 1 : True

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
 def is_valid_1d_array(arr):
 if not isinstance(arr, list):
 return False # Not a valid 1D array if it's not a list
 
 return all(isinstance(elem, str) and not isinstance(elem, bool) for elem in arr)

 if not is_valid_1d_array(self.__inputs[1]):
 self.store_result(("Invalid test case."), ValidationFunction.FAIL)
 
 elif not (1 <= len(self.__inputs[0]) <= 250):
 self.store_result("Length of 's' should be between 1 and 250", ValidationFunction.FAIL)

 elif not (1 <= len(self.__inputs[1]) <= 1000):
 self.store_result("Length of 'word_dict' should be between 1 and 1000", ValidationFunction.FAIL)

 elif any(not (1 <= len(word) <= 20) for word in self.__inputs[1]):
 self.store_result("Length of each word in 'word_dict' should be between 1 and 20", ValidationFunction.FAIL)

 elif not all(c.islower() for c in self.__inputs[0]):
 self.store_result("'s' should consist of only lowercase English letters", ValidationFunction.FAIL)
 
 elif not all(c.islower() and c.isascii() and c.isalpha() for c in self.__inputs[1]):
 self.store_result("'word_dict' should consist of only lowercase English letters", ValidationFunction.FAIL)

 elif len(set(self.__inputs[1])) != len(self.__inputs[1]):
 self.store_result("'word_dict' should consist of unique strings", ValidationFunction.FAIL)

 else:
 self.store_result("", ValidationFunction.PASS)

Dynamic Programming

Conclusion

Word Break

Statement

Examples

Understand the problem

Figure it out!

Try it yourself