Learn and practice the Cyclic Sort and Topological Sort patterns to solve coding interview problems.

Educative-99 in Java: Accelerate Your Coding Interview Prep

This module will focus on two advanced sorting algorithms, Cyclic Sort and Topological Sort. These are often used in coding interviews as a way to distinguish between average and outstanding candidates. Cyclic Sort may only be used if the input list meets certain criteria, and in those situations, it’s the most efficient sorting algorithm possible. Topological Sort utilizes graph-based algorithms to deal with partially ordered elements. This module teaches us to recognize whether either of these advanced sorting techniques is applicable, as well as how to use them to solve the given problem.

Advanced Sorting Techniques

# Statement
Given an array, _nums_, containing $n$ distinct numbers in the range $[0, n]$, return the only number in the range that is missing from the array.

**Constraints:**
- $n =$ `nums.length` 
- $1 \leq n \leq 10^3$
- $0 \leq$ `nums[i]` $\leq n$
- There are no duplicates in the array.

# 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:



What would be the output if the following array is given as input?

[0, 2, 4, 6, 7, 8, 3, 9, 5, 10]

What would be the output if the following array is given as input?
[0, 2, 4, 6, 7, 8, 3, 9, 5, 10]

What would be the output if the following array is given as input?

[0, 12, 4, 6, 7, 8, 3, 9, 5, 10, 2, 1]


What would be the output if the following array is given as input?
[0, 12, 4, 6, 7, 8, 3, 9, 5, 10, 2, 1]

What would be the output if the following array is given as input?

[0, 1, 2]

What would be the output if the following array is given as input?
[0, 1, 2]

Missing Number

Missing Number

# 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.


# Try it yourself
Implement your solution in `MissingNumber.java` in the following coding playground. The supporting code template provided in `Traversal.java` is meant to assist in developing your solution to the problem.


import java.util.*;
public class MissingNumber{
   public static int findMissingNumber(int[] arr) {
      
      // Your code will replace this placeholder return statement 
      return 0;
   }
}

import java.util.*;
class Traversal{
 
 static void forwardTraversal(int[] arr){
 for(int i = 0; i < arr.length; i++){
 System.out.print(arr[i]+ "");
 }
 }
 static void backwardTraversal(int[] arr){
 for(int i = arr.length-1; i >= 0; i--){
 System.out.print(arr[i]+ "");
 }
 }
}

Help me debug only the function Missing Number . . I don't need the corrected code.I don't want explanations or code review. 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. Don't give me the corrected code. . If yes, then Just provide me the exact and to-the-point error that helps me run the code successfully on the test cases provided.

Your current task is to review only the findMissingNumber function and the other function(s) invoked in it/them. Do not review extra/helper files or functions not invoked in the solution function findMissingNumber. Do not review the code of data structures 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. Also, Answer the following questions: Is the Cyclic Sort pattern used? Take your time and review the code.

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

CustomDS

import ds_v1.DataStructure.DataStructure;
import java.util.*;
public class CustomDS extends DataStructure {
 int[] values = {};


 public CustomDS() {
 //values = {};
 }
 

 public CustomDS(int[] val){
 this.values = new int [val.length];
 for(int i =0; i < val.length;i++){
 values[i] = val[i];
 }
 
 }

 public String toString() {
 return Arrays.toString(values);
 }

 @Override
 public CustomDS construct(String new_value, Integer type_of_ds) {
 String new_value1 = new_value.substring(1,new_value.length() - 1);
 String[] string = new_value1.split(",");
 int[] arr = new int[string.length];
 
 for (int i = 0; i < string.length; i++) {
 arr[i] = Integer.valueOf(string[i]);
 }
 return new CustomDS(arr);

 }

 @Override
 public String traverse(Object input, Integer type_of_ds) {
 
 return input.toString();
 }

}

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(find_missing_number(self.__inputs[0]))

 if json.dumps(find_missing_number(self.__inputs[0])) == 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
def find_missing_number(nums):
 len_nums = len(nums)
 index = 0

 while index < len_nums:
 value = nums[index]

 if value < len_nums and value != nums[value]:
 nums[index], nums[value] = nums[value], nums[index]

 elif value >= len_nums:
 index+=1

 else:
 index += 1

 for x in range(len_nums):
 if x != nums[x]:
 return x
 return len_nums

public class Main{
 public static int findMissingNumber1(CustomDS c) {
    int m = MissingNumber.findMissingNumber(c.values);
    return m;
      
 }
}

private (1).yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : [0, 1, 2, 4]
    output: 
    - 1 : 3

  - name: "testcase 2"
    inputs:
    - 1: [3, 0, 1, 4]
    output:
    - 1 : 2
        
  - name: "testcase 3"
    inputs:
    - 1: [1, 4, 5, 6, 8, 2, 0, 7]
    output:
    - 1 : 3
        
  - name: "testcase 4"
    inputs:
    - 1: [1, 0, 2, 3, 4, 5, 6, 8, 9, 7, 11]
    output:
    - 1 : 10
        
  - name: "testcase 5"
    inputs:
    - 1: [1]
    output:
    - 1 : 0

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
 nums=self.__inputs[0]
 min_length=0
 max_length=len(nums)
 n=len(nums)
 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, int) and not isinstance(elem, bool) for elem in arr)
 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 if not is_valid_1d_array(nums):
 self.store_result(("Invalid test case."), ValidationFunction.FAIL)
 elif len(nums) < 1 or len(nums) > 1000 :
 self.store_result ("Input length should be between 1 and 1000", ValidationFunction.FAIL)
 elif any(value < 0 or value > n for value in nums):
 self.store_result("All elements in the input set should be between 0 and array length", ValidationFunction.FAIL)
 elif len(nums) != len(set(nums)):
 self.store_result("Input should not have any duplicates", ValidationFunction.FAIL)
 else:
 self.store_result("", ValidationFunction.PASS)

Cyclic Sort

Topological Sort

Conclusion

Missing Number

Statement

Examples

Understand the problem

Figure it out!

Try it yourself