Monday, June 10, 2024

 Problem Statement: Given an integer array arr, in one move you can select a palindromic subsequence arr[i], ..., arr[j] where 0 <= i <= j < arr.length, and remove that subsequence from the given array. Note that after removing a subarray, the elements move to remove the gap between them.

 

Return the minimum number of moves needed to remove all numbers from the array.

 

Solution:

import java.util.*;

class Solution {

    public int minimumMoves(int[] arr) {

        int N = arr.length;

        int max = 1;

        int min = Integer.MAX_VALUE;

        List<Integer> A = new ArrayList<>();

        for (int i = 0; i < arr.length; i++) A.add(arr[i]);

        int count = 0;

        while(A.size() > 0) {

           boolean hasPalindrome = false; 

           List<Integer> elements = new ArrayList<>();

           for (int i = 0; i < (1<<N); i++) { 

               

               List<Integer> combination = new ArrayList<>(); 

                for (int j = 0; j < A.size(); j++) { 

                  if ((i & (1 << j)) > 0) { 

                    combination.add(j); 

                  } 

                } 

                if (isPalindrome(A, combination) && (combination.size() > max) && getCharactersToRemove(A, combination) < min) {

                      hasPalindrome = true;

                      max = combination.size();

                      min = getCharactersToRemove(A, combination);

                      elements = new ArrayList<>(combination);                

                      if (getCharactersToRemove(A, combination) == 0) { break;}

                } else {

                    // System.out.println("A: " + print(A) + " Elements: " + print(elements) + " Combination: " + print(combination) + "isPalindrome=" + String.valueOf(isPalindrome(A, combination)) + " getCharsToRemove=" + getCharactersToRemove(A, combination) + " min = " + min);

                }

           }            

           if (!hasPalindrome) {

               count += 1;

               A.remove(A.size() - 1);

           } else {

               count += getCharactersToRemove(A, elements) + 1;

               A = removeCharacters(A, elements);

               // System.out.println("Removing " + count + " characters at indices:" + print(elements) + " and remaining elements: " + print(A));

               // elements = new ArrayList<>();

               max = 1;

               min = Integer.MAX_VALUE;

           }

        }

        return count;

    }

    public boolean isPalindrome(List<Integer> A, List<Integer> combination) {

        int start = 0;

        int end = combination.size()-1;

        while (start <= end) {

            if (A.get(combination.get(start)) != A.get(combination.get(end))) {

                return false;

            }

            start++;

            end--;

        }

        return true;

    }

    public int getCharactersToRemove(List<Integer> A, List<Integer> combination){

        if (combination.size() < 2) return 0;

        List<Integer> clone = new ArrayList<>(A); 

        return removeCharacters(clone, combination).size();

    }

    public List<Integer> removeCharacters(List<Integer> A, List<Integer> combination) {

     int start = 0;

     int end = combinations.size()-1;

     int last = 0;

     while (start <= end) {

             for (int i = last; i< A.size(); i++) {

                    if (A.get(i) == combination.get(start)) {

                          A.set(I, Integer.MAX_VALUE);

                          last = i+1;

                          start++;

                    }

             }

     }

    List<Integer> result = new ArrayList<>();

    For (int I = 0; I < A.size(); i++) {

         if (A.get(i) != Integer.MAX_VALUE) {

               result.add(A.get(i));

          }

    }

    return result;

    }

    public List<Integer> removeCharacters(List<Integer> A, List<Integer> combination) {

        int start = combination.get(0);

        int end = combination.get(combination.size()-1);

        List<Integer> result = new ArrayList<>();

        if (start > 0){

            result.addAll(A.subList(0, start));

        }

        if (end < A.size() - 1) {

            result.addAll(A.subList(end + 1,A.size()));

        }

        return result;

    }

    public String print(List<Integer> elements){

        StringBuilder sb = new StringBuilder();

        for (int i = 0; i < elements.size(); i++) {

            sb.append(elements.get(i) + " ");

        }

        return sb.toString();

    }

}


Examples:

A = [-1,0,1]           => 3

A = [-1,0,-1]          => 1

A = [-1]                    => 1

A = [-1,0]                => 2

A = [0,0]                 => 1

A = [1,0,1,2,3]     => 3

A = [-2,-1,0,1,0]   => 3


Sunday, June 9, 2024

 Drone Formation Commercial Software:

This is an addendum to the discussion about Drone Formation Commercial Software as described in this document. In this section, we describe the infrastructure for various businesses operating on the software-as-a-service platform described in the document. Each business gets a deployment stamp that comprises of various resources in standard format that can be scaled or customized according to the size and requirements of the business but begin with a standard edition of set of resources. The SaaS offering comes with a high-availability architecture for the businesses by virtue of its performance, scalability and availability. This multi-instance architecture will meet and exceed stringent requirements surrounding data sovereignty, availability and performance. Each stamp comes with redundant components and multiple network paths to avoid single points of failure. The platform itself has a multi-homed network with multiple connections to the internet and comes with pairs of components that maintain continuous replication between replicas. These digital twins can withstand failures and support the combined production loads providing business continuity and disaster recovery. Operations can be transferred from one component of the stamp to another seamlessly. With these assurances, the business operations will never experience a downtime, and transfers will always be successful. A multi-tenant architecture for the platform ensures isolation for each business hosted on the infrastructure. Each business gets its own dashboard to monitor ongoing usage and for troubleshooting. Measurement and billing will make use of tags and labels for the underlying cloud resources. Wherever appropriate, deep monitoring will be involved to provide recommendations above and beyond those available from the public cloud.

Tens of billions of transactions can be expected to come from various drones of different tenants and the databases including the provisioned configuration management database will handle that load. A follow-the-sun model provides continual security, operational monitoring and support of various resources. Traditional data backup and recovery will be supported so that inadvertent deletes can be mitigated with restoration. Full and deferential backups will augment the aging and tiering best practices of the data management. Failovers will not only be planned but also routinely performed or tested. The current passive system is designated as active during the failover and mapping and name resolutions will automatically work out.

Most customers will appreciate the importance of querying abilities provided with this platform from its read-only stack that will serve the position, location and health information of drones in near real-time basis. The querying language will seamlessly integrate with the resource graph queries available on the public cloud for enhanced visibility into decision-making logic and cloud-resource consumption corresponding to a drone or a set of drones. The platform is uniquely positioned to offer both operational as well as management information on the drones for all the businesses.


Saturday, June 8, 2024

  Problem::

Make Array Zero by Subtracting Equal Amounts

You are given a non-negative integer array nums. In one operation, you must:

Choose a positive integer x such that x is less than or equal to the smallest non-zero element in nums.

Subtract x from every positive element in nums.

Return the minimum number of operations to make every element in nums equal to 0.

 

Example 1:

Input: nums = [1,5,0,3,5]

Output: 3

Explanation:

In the first operation, choose x = 1. Now, nums = [0,4,0,2,4].

In the second operation, choose x = 2. Now, nums = [0,2,0,0,2].

In the third operation, choose x = 2. Now, nums = [0,0,0,0,0].

Example 2:

Input: nums = [0]

Output: 0

Explanation: Each element in nums is already 0 so no operations are needed.

 

Constraints:

1 <= nums.length <= 100

0 <= nums[i] <= 100


import java.util.*;

import java.util.stream.*;

class Solution {

    public int minimumOperations(int[] nums) {

        List<Integer> list = Arrays.stream(nums).boxed().collect(Collectors.toList());

        var nonZero = list.stream().filter(x -> x > 0).collect(Collectors.toList());

        int count = 0;

        while(nonZero.size() > 0) {

            var min = nonZero.stream().mapToInt(x -> x).min().getAsInt();

            nonZero = nonZero.stream().map(x -> x - min).filter(x -> x > 0).collect(Collectors.toList());

            count++;

        }

        return count;

    }

}


Input

nums =

[1,5,0,3,5]

Output

3

Expected

3


Input

nums =

[0]

Output

0

Expected

0


Friday, June 7, 2024

 Problem Statement: Given an integer array arr, in one move you can select a palindromic subsequence arr[i], ..., arr[j] where 0 <= i <= j < arr.length, and remove that subsequence from the given array. Note that after removing a subarray, the elements move to remove the gap between them.

 

Return the minimum number of moves needed to remove all numbers from the array.

 

Solution:

import java.util.*;

class Solution {

    public int minimumMoves(int[] arr) {

        int N = arr.length;

        int max = 1;

        int min = Integer.MAX_VALUE;

        List<Integer> A = new ArrayList<>();

        for (int i = 0; i < arr.length; i++) A.add(arr[i]);

        int count = 0;

        while(A.size() > 0) {

           boolean hasPalindrome = false; 

           List<Integer> elements = new ArrayList<>();

           for (int i = 0; i < (1<<N); i++) { 

               

               List<Integer> combination = new ArrayList<>(); 

                for (int j = 0; j < A.size(); j++) { 

                  if ((i & (1 << j)) > 0) { 

                    combination.add(j); 

                  } 

                } 

                if (isPalindrome(A, combination) && (combination.size() > max) && getCharactersToRemove(A, combination) < min) {

                      hasPalindrome = true;

                      max = combination.size();

                      min = getCharactersToRemove(A, combination);

                      elements = new ArrayList<>(combination);                

                      if (getCharactersToRemove(A, combination) == 0) { break;}

                } else {

                    // System.out.println("A: " + print(A) + " Elements: " + print(elements) + " Combination: " + print(combination) + "isPalindrome=" + String.valueOf(isPalindrome(A, combination)) + " getCharsToRemove=" + getCharactersToRemove(A, combination) + " min = " + min);

                }

           }            

           if (!hasPalindrome) {

               count += 1;

               A.remove(A.size() - 1);

           } else {

               count += getCharactersToRemove(A, elements) + 1;

               A = removeCharacters(A, elements);

               // System.out.println("Removing " + count + " characters at indices:" + print(elements) + " and remaining elements: " + print(A));

               // elements = new ArrayList<>();

               max = 1;

               min = Integer.MAX_VALUE;

           }

        }

        return count;

    }

    public boolean isPalindrome(List<Integer> A, List<Integer> combination) {

        int start = 0;

        int end = combination.size()-1;

        while (start <= end) {

            if (A.get(combination.get(start)) != A.get(combination.get(end))) {

                return false;

            }

            start++;

            end--;

        }

        return true;

    }

    public int getCharactersToRemove(List<Integer> A, List<Integer> combination){

        if (combination.size() < 2) return 0;

        List<Integer> clone = new ArrayList<>(A); 

        return removeCharacters(clone, combination).size();

    }

    public List<Integer> removeCharacters(List<Integer> A, List<Integer> combination) {

     int start = 0;

     int end = combinations.size()-1;

     int last = 0;

     while (start <= end) {

             for (int i = last; i< A.size(); i++) {

                    if (A.get(i) == combination.get(start)) {

                          A.set(I, Integer.MAX_VALUE);

                          last = i+1;

                          start++;

                    }

             }

     }

    List<Integer> result = new ArrayList<>();

    For (int I = 0; I < A.size(); i++) {

         if (A.get(i) != Integer.MAX_VALUE) {

               result.add(A.get(i));

          }

    }

    return result;

    }

    public List<Integer> removeCharacters(List<Integer> A, List<Integer> combination) {

        int start = combination.get(0);

        int end = combination.get(combination.size()-1);

        List<Integer> result = new ArrayList<>();

        if (start > 0){

            result.addAll(A.subList(0, start));

        }

        if (end < A.size() - 1) {

            result.addAll(A.subList(end + 1,A.size()));

        }

        return result;

    }

    public String print(List<Integer> elements){

        StringBuilder sb = new StringBuilder();

        for (int i = 0; i < elements.size(); i++) {

            sb.append(elements.get(i) + " ");

        }

        return sb.toString();

    }

}


Examples:

A = [-1,0,1]           => 3

A = [-1,0,-1]          => 1

A = [-1]                    => 1

A = [-1,0]                => 2

A = [0,0]                 => 1

A = [1,0,1,2,3]     => 3

A = [-2,-1,0,1,0]   => 3


Thursday, June 6, 2024

 


Problem::

Make Array Zero by Subtracting Equal Amounts

You are given a non-negative integer array nums. In one operation, you must:

Choose a positive integer x such that x is less than or equal to the smallest non-zero element in nums.

Subtract x from every positive element in nums.

Return the minimum number of operations to make every element in nums equal to 0.

 

Example 1:

Input: nums = [1,5,0,3,5]

Output: 3

Explanation:

In the first operation, choose x = 1. Now, nums = [0,4,0,2,4].

In the second operation, choose x = 2. Now, nums = [0,2,0,0,2].

In the third operation, choose x = 2. Now, nums = [0,0,0,0,0].

Example 2:

Input: nums = [0]

Output: 0

Explanation: Each element in nums is already 0 so no operations are needed.

 

Constraints:

1 <= nums.length <= 100

0 <= nums[i] <= 100


import java.util.*;

import java.util.stream.*;

class Solution {

    public int minimumOperations(int[] nums) {

        List<Integer> list = Arrays.stream(nums).boxed().collect(Collectors.toList());

        var nonZero = list.stream().filter(x -> x > 0).collect(Collectors.toList());

        int count = 0;

        while(nonZero.size() > 0) {

            var min = nonZero.stream().mapToInt(x -> x).min().getAsInt();

            nonZero = nonZero.stream().map(x -> x - min).filter(x -> x > 0).collect(Collectors.toList());

            count++;

        }

        return count;

    }

}


Input

nums =

[1,5,0,3,5]

Output

3

Expected

3


Input

nums =

[0]

Output

0

Expected

0




SQL Schema

 

Table: Books

+----------------+---------+

| Column Name    | Type    |

+----------------+---------+

| book_id        | int     |

| name           | varchar |

| available_from | date    |

+----------------+---------+

book_id is the primary key of this table.

 

Table: Orders

+----------------+---------+

| Column Name    | Type    |

+----------------+---------+

| order_id       | int     |

| book_id        | int     |

| quantity       | int     |

| dispatch_date  | date    |

+----------------+---------+

order_id is the primary key of this table.

book_id is a foreign key to the Books table.

 

Write an SQL query that reports the books that have sold less than 10 copies in the last year, excluding books that have been available for less than one month from today. Assume today is 2019-06-23.

Return the result table in any order.

The query result format is in the following example.

 

Example 1:

Input: 

Books table:

+---------+--------------------+----------------+

| book_id | name               | available_from |

+---------+--------------------+----------------+

| 1       | "Kalila And Demna" | 2010-01-01     |

| 2       | "28 Letters"       | 2012-05-12     |

| 3       | "The Hobbit"       | 2019-06-10     |

| 4       | "13 Reasons Why"   | 2019-06-01     |

| 5       | "The Hunger Games" | 2008-09-21     |

+---------+--------------------+----------------+

Orders table:

+----------+---------+----------+---------------+

| order_id | book_id | quantity | dispatch_date |

+----------+---------+----------+---------------+

| 1        | 1       | 2        | 2018-07-26    |

| 2        | 1       | 1        | 2018-11-05    |

| 3        | 3       | 8        | 2019-06-11    |

| 4        | 4       | 6        | 2019-06-05    |

| 5        | 4       | 5        | 2019-06-20    |

| 6        | 5       | 9        | 2009-02-02    |

| 7        | 5       | 8        | 2010-04-13    |

+----------+---------+----------+---------------+

Output: 

+-----------+--------------------+

| book_id   | name               |

+-----------+--------------------+

| 1         | "Kalila And Demna" |

| 2         | "28 Letters"       |

| 5         | "The Hunger Games" |

+-----------+--------------------+



SELECT DISTINCT b.book_id, b.name

FROM books b 

LEFT JOIN Orders o on b.book_id = o.book_id

GROUP BY b.book_id, b.name, 

DATEDIFF(day, DATEADD(year, -1, '2019-06-23'), o.dispatch_date),  

DATEDIFF(day,  b.available_from, DATEADD(month, -1, '2019-06-23')) 

HAVING SUM(o.quantity) IS NULL OR 

DATEDIFF(day, DATEADD(year, -1, '2019-06-23'), o.dispatch_date) < 0 OR 

(DATEDIFF(day, DATEADD(year, -1, '2019-06-23'), o.dispatch_date) > 0 AND DATEDIFF(day,  b.available_from, DATEADD(month, -1, '2019-06-23')) > 0 AND SUM(o.quantity) < 10);



Case 1

Input

Books =

| book_id | name | available_from |

| ------- | ---------------- | -------------- |

| 1 | Kalila And Demna | 2010-01-01 |

| 2 | 28 Letters | 2012-05-12 |

| 3 | The Hobbit | 2019-06-10 |

| 4 | 13 Reasons Why | 2019-06-01 |

| 5 | The Hunger Games | 2008-09-21 |

Orders =

| order_id | book_id | quantity | dispatch_date |

| -------- | ------- | -------- | ------------- |

| 1 | 1 | 2 | 2018-07-26 |

| 2 | 1 | 1 | 2018-11-05 |

| 3 | 3 | 8 | 2019-06-11 |

| 4 | 4 | 6 | 2019-06-05 |

| 5 | 4 | 5 | 2019-06-20 |

| 6 | 5 | 9 | 2009-02-02 |

| 7 | 5 | 8 | 2010-04-13 |

Output

| book_id | name |

| ------- | ---------------- |

| 2 | 28 Letters |

| 1 | Kalila And Demna |

| 5 | The Hunger Games |

Expected

| book_id | name |

| ------- | ---------------- |

| 1 | Kalila And Demna |

| 2 | 28 Letters |

| 5 | The Hunger Games |




 This is a continuation of previous articles on IaC shortcomings and resolutions. 

As with all automation, it is important to register them in source control so that their maintenance can become easy. It is also important to secure the credentials with which these scripts run. Finally, lockdown of all resources in terms of network access and private planes is just as important as their accessibility for automation.  

Many organizations don’t invest in Azure DevOps  for a variety of reasons, with attachments to on-premises based automation technology or avoiding public cloud automations for misplaced cost concerns. Other reasons can be genuine though. For example, one of the most common tasks is to download and run an executable instead of calling an API. This is convenient for portability and the same executable can be run using various Azure accounts or subscriptions. But if we take an automation account or runbook, then the downloaded executable might not run because the execution policy cannot be changed. The same goes for hybrid worker and the only way to overcome the limitation is to spin up a dedicated compute and modify it  to allow execution. It might be noteworthy to add that compute instance or cluster created on a databricks workspace or Azure machine learning workspace might also not work.  Their pass-through active directory authentication works only for notebooks but not the shell on the compute. This kind of limitation extends to other data oriented automations because they in turn leverage the workspaces and do not allow any activity on the scripts or notebooks that these workspaces cannot run.

Such a limitation, on the other hand, does not hold on the GitOps which has been one of the favorites for code delivery pipelines. In the GitOps, it is not only easy to download the executable from say an object repository such as Artifactory but also easy to pass command line parameters whose values are already known to the pipeline. Organizing the automation in GitOps is also fairly straightforward as pipelines become scoped to the code they push. 

Finally, there is a lot of maintenance work required  with scripts and automations and the use of source control becomes inevitable. Keeping the automations and the code they service as repositories enable them to be shared as appropriate.

These are some of the examples where a cloud native approach might not be straight. When organizations do enable Azure DevOps, they enhance their capabilities which gives them a  more secure, manageable and futuristic platform albeit one that dictates rewriting or rehosting the legacy scripts.


Wednesday, June 5, 2024

 This is  a summary of the book “Grace Under Pressure: Leading through change and crisis” written by John Baldoni and published by Savio Republic in 2023. The author teaches how to maintain composure in stressful situations. The first priority when pressure hits is to make sure our people are okay. Then, make sure we are okay and take a deep breath. We must look as wisely into the future and consider how to prevent today’s stress from compounding.  Any decision must not harm our people. He also stresses the importance of courage, compassion, empathy, hope, resilience and selflessness into our lives.

Grace serves us and our stakeholders. We must practice astute situational assessment and intelligent follow-up.  Leaders with grace under pressure  plan ahead and take care.

Logic helps to find the truth. Our values help to answer the “why” but grace helps to answer “how”. Control what we can. Instead of focusing on “winner takes all” and “winner shares all”. Those who practice this remain resolute.

"Grace under pressure" is a leadership concept that emphasizes maintaining one's coolness in challenging situations. It originated from the words of Ernest Hemingway, who described it as a dependable, composed leader directing a firefight. Grace is a mysterious, spiritual, and magical force that can appear unexpectedly and propel people towards their higher aspirations. High-quality leaders must objectively assess their teams, units, departments, or organizations to plan and implement actions to help their people succeed. They should ask three crucial questions: "What is happening?", "What is not happening?", and "What can I do to influence the outcome?" By designing their investigation to help them and their team take the best steps going forward, leaders can help their teams succeed in challenging situations. However, it is important to remember that sometimes the best group action might be no action at all.

Grace under pressure is a key leadership trait that involves taking care of one's people, taking care of oneself, and planning for the future. These leaders are able to adapt to change and adapt to it, demonstrating integrity, courage, and logical reasoning. They are also humilent, reason-driven, courageous, humorous, and loving.


Grace is essential for leading a meaningful life, as it fuels purpose and becomes the "how" to achieve goals. It enriches connections and creates a sense of community among employees. Leaders who can muster grace under pressure can take their organization in a different direction, fostering connections among their people and building internal cohesion.


Corporate cultures that foster a sense of community can foster connections among their people, helping organizations become a setting for connected communities. Transformations require grace, which involves listening before talking, solving problems, encouraging open communication, instilling hope, banishing fear, and acting forthrightly and with courage.

We must focus on the present, recharge or renew, orient yourself to the future, anchor yourself, and have humility. Leaders must be flexible and adaptable to the unpredictable world they face. We must be careful, deliberate, measured, and reflective in our thoughts and actions, making "mutuality" our watchword. Orienting ourselves to the goal of making things better for those we lead, coaching them to benefit from our values and long-range thinking.

Leaders should also walk behind their people, as stress resilience expert Dr. Sharon Melnick believes leaders don't have time to not do this. Many business-school students are afraid to come across as friendly and civil once they gain authority at work, but a leader with grace knows that they can overcome challenges and become better people. By following these ground rules, leaders can control their actions and achieve success in their organizations.

Grace under pressure leaders maintain resoluteness despite challenges, focusing on dignity and creating a workplace where people feel valued and want to contribute. They communicate effectively, avoid anxiety and fear, remain positive, stress mental health, and engage with team members. To exhibit grace under pressure, leaders should ask three questions: what to do, how to effectively engage with their team, and how their leadership will be portrayed during a crisis.

 

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