Learn and practice the Backtracking and Subsets patterns to solve coding interview problems.

Speedrun the Microsoft Coding Interview: Top Problems in Java

The optimal solution to certain problems requires examining all possibilities. If any possibility is ignored, we are unable to say with mathematical certainty that our solution is optimal. In such situations, we use a technique known as backtracking to exhaustively traverse the entire solution space and collect all the information needed to compute the optimal solution. The Subsets pattern groups together those problems where we need to systematically construct all, or, a mathematically defined subset of all, the possible subsets of a given set. Again, the technique used for the construction of such a set of subsets is, in fact, backtracking.

Go With the Flow

# 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:

["000.000.00", "00.000.000", "000.00.000", "0.0.000.000", "0.00.00.000"]

What is the correct output if the following string is given as input?

IP address = "00000000"

What is the correct output if the following string is given as input?
IP address = “00000000”

["0.0.0.0", "0.0.00", "0.00.0", "00.0.0"]

What is the correct output if the following string is given as input?

IP address = "0000"

What is the correct output if the following string is given as input?
IP address = “0000”

["1.2.1.212", "1.2.12.12", "1.2.121.2", "1.21.2.12, "1.21.21.2, "1.212.1.2", "12.1.2.12", "12.1.21.2", "12.12.1.2", "121.2.1.2"]

What is the correct output if the following string is given as input?

IP address = "121212"

What is the correct output if the following string is given as input?
IP address = “121212”

What is the correct output if the following string is given as input?

IP Address = "25525511335"

What is the correct output if the following string is given as input?
IP Address = “25525511335”

# 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.

> **Note:** As an additional challenge, we have intentionally hidden the solution to this puzzle. 

# Try it yourself

Implement your solution in the following coding playground.

# 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: `restoreIpAddresses`\n- Candidate's submitted solution code: USER_CODE\n- Error: ERROR_DESCRIPTION\n- Failed test case: FAILED_TEST_CASE\n- Coding language: `Java`\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 `restoreIpAddresses` 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 `restoreIpAddresses`.\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) `restoreIpAddresses`. \n- While calculating time complexity, include the time complexity of built-in Java 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) `restoreIpAddresses`. \n- While calculating space complexity, include the space complexity of built-in Java 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 valid(segment):
 segment_length = len(segment) # storing the length of each segment
 if segment_length > 3: # each segment's length should be less than 3
 return False

 # Check if the current segment is valid
 # for either one of following conditions:
 # 1. Check if the current segment is less or equal to 255.
 # 2. Check if the length of the segment is 1. The first character of segment
 # can be `0` only if the length of the segment is 1.
 return int(segment) <= 255 if segment[0] != '0' else len(segment) == 1


 # this function will append the current list of segments to the list of results.
 def update_segment(s, curr_dot, segments, result):
 segment = s[curr_dot + 1:len(s)]

 if valid(segment): # if the segment is acceptable
 segments.append(segment) # add it to the list of segments
 result.append('.'.join(segments))
 segments.pop() # remove the top segment


 def backtrack(s, prev_dot, dots, segments, result):
 # prev_dot : the position of the previously placed dot
 # dots : number of dots to place

 size = len(s)

 # The current dot curr_dot could be placed in
 # a range from prev_dot + 1 to prev_dot + 4.
 # The dot couldn't be placed after the last character in the string.
 for curr_dot in range(prev_dot + 1, min(size - 1, prev_dot + 4)):
 segment = s[prev_dot + 1:curr_dot + 1]
 if valid(segment):
 segments.append(segment)

 # if all 3 dots are placed add the solution to result
 if dots - 1 == 0:
 update_segment(s, curr_dot, segments, result)
 else:
 # continue to place dots
 backtrack(s, curr_dot, dots - 1, segments, result)

 segments.pop() # remove the last placed dot


 def restore_ip_addresses(s):

 # creating empty lists for storing valid IP addresses,
 # and each segment of IP
 result, segments = [], []
 backtrack(s, -1, 3, segments, result)
 return result


 # multiple outputs will be added at each index respectively for a single test_case
 expected = restore_ip_addresses(self.__inputs[0])
 actual = self.__actual_outputs[0]
 # multiple outputs will be added at each index respectively for a single test_case
 expected_outputs = [] 
 
 if actual == None:
 expected_outputs.append(expected)
 self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)
 elif expected == actual:
 expected_outputs.append(expected) 
 self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS) 
 elif actual != expected:
 if sorted(actual) == sorted(expected):
 expected_outputs.append(actual)
 self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
 else: 
 expected_outputs.append(expected)
 self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)
 else: 
 expected_outputs.append(expected)
 self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL) 

 return self.__edu_result

import java.util.*;

public class Main{
 public static List<String> restoreIpAddresses(String s) {

 // Replace this placeholder return statement with your code
 return new ArrayList<String>();
 }
}

restore_ip_addresses (1).yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1: "010010"
    output:
    - 1 : ["0.10.0.10", "0.100.1.0"]
  
  - name: "testcase 2"
    inputs:
    - 1 : "00000000"
    output: 
    - 1 : []
  
  - name: "testcase 3"
    inputs:
    - 1: "255255255255"
    output:
    - 1 : ["255.255.255.255"]
  
  - name: "testcase 4"
    inputs:
    - 1: "201023"
    output:
    - 1: ["2.0.10.23","2.0.102.3","20.1.0.23","20.10.2.3","201.0.2.3"]
  
  - name: "testcase 5"
    inputs:
    - 1: "12121212"
    output:
    - 1: ["1.2.121.212", "1.21.21.212", "1.21.212.12", "1.212.1.212", "1.212.12.12", "1.212.121.2", "12.1.21.212", "12.1.212.12", "12.12.1.212", "12.12.12.12", "12.12.121.2", "12.121.2.12", "12.121.21.2", "121.2.1.212", "121.2.12.12", "121.2.121.2", "121.21.2.12", "121.21.21.2", "121.212.1.2"]

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

def validate_inputs(self) -> None:
 """
 Validate the input string s based on the given constraints.
 Store the result using self.store_result(error_msg, status).
 return: void
 """
 s = self.__inputs[0]

 if not (4 <= len(s) <= 12):
 self.store_result("String length should be in the range 4 <= s.length <= 12", ValidationFunction.FAIL)
 return
 elif not s.isdigit():
 self.store_result("String s should consist of digits only", ValidationFunction.FAIL)
 return
 else:
 self.store_result("", ValidationFunction.PASS)

Try to solve the Restore IP Addresses problem.

Backtracking

Subsets

Conclusion

Restore IP Addresses

Statement

Examples

Understand the problem

Figure it out!

Try it yourself