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Author
Rikam Palkar
.NET Full Stack, DSA, C#, React, Blazor, WPF, JavaScript
Mumbai (India)
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4.9m
40.4k
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Recognitions & awards
    • Jun 2020
    • Jun 2021
    • Jun 2022
    • Jun 2023
    • Jun 2024
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Awards & Certifications
  • MCA
    VJTI
About

"Microsoft MVP", Five-time "C# Corner MVP", "Verified author" on Medium "Top voice" on LinkedIn, Author behind 'WPF Simplified' and 'Blazor Simplified'.I'm passionate about making the world a better place by writing scalable, efficient code, I'm skilled at Blazor, WPF, C#, and Data Structures & Algorithms.

Activity

DFS and Maths should never be combined!

#Algorithm
- DFS numbers 1 to 9
- Base condition: If the current number exceeds n, stop recursion.
- Else add the current number to the result list.
Multiply the number by 10 to get to the next level (1 - 10, 10 - 19, 100 - 109).
- For the deeper level:
-- If the next level starting number exceeds n, return.
-- Else, recurse for each number in the range [curr number, curr number, + 9].
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Post 12h

Thought of Queue first! Stack approach was easy for edge cases!
#Algorithm
- 26 stacks: one for each letter ('a' to 'z'), to store the indices of those characters.
- Loop input string: Convert the input string into a character array for updates.
- For each character:
-- If it’s a letter, push its index onto the stack corresponding to that letter.
-- If it’s a *, find the smallest letter stack (from 'a' to 'z') that is not empty, pop its top index, and mark that character as '0' - placeholder

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Post 1d

It seems like robots and AI are taking over everything these days!

#Algorithm
- Count the frequency of each character in s.
- Use a stack to hold letters as you process s from left to right.
- Track the smallest character from current index to end of s.
- At each step, move the current letter to the stack and update frequency.
- Pop from the stack to the result whenever the top of the stack is ≤ the smallest left character.

Code: https://lnkd.in/ebraFu_M

#DSA

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Post 2d
useRef: The React Hook You’re Sleeping On

useRef: The React Hook You’re Sleeping On

2d 360 0 Article

DSU Was Right There, But Not in My Head

Disjoint Set Algorithm!


#Algorithm

- Convert each char to num and form DSU of both strings

- When Union, always choose the smaller character as the group’s parent to ensure the smallest lexicographical representative.

- Return parent of each node


#Code: https://github.com/RikamPalkar/DSA-Simplified/blob/main/LeetCode/1061. Lexicographically Smallest Equivalent String.cs


#DSA #Programming #Leetcode


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Post 3d
Understanding useState in React

Understanding useState in React

4d 501 0 Article

Who Hid the Candies, deep down the graph?


#Algorithm

- Process boxes in the order you discover them, so BFS hence queue

- If a box is unopened and can be opened (status == 1), open it, collect candies, and mark it as opened.

- Add all the contained boxes and newly unlocked keys to your queue, update their status = 1.

- Repeat until there are no more boxes in the queue.
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Post 5d

After a Bunch of Trial and Error!

#Leetcode 135


#Algorithm:

1 - left-to-right pass:

If the next child’s rating is higher, they get 1 more candy than the previous.

2 - Do a right-to-left pass:

If the previous child’s rating is higher, ensure they have at least 1 more candy than the next child.

3. Sum up the candies.


#Code: https://github.com/RikamPalkar/DSA-Simplified/blob/main/LeetCode/135. Candy.cs


#DSA #Programming

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Post 6d
React Children! Yeah, That’s What They Called

React Children! Yeah, That’s What They Called

1w 509 2 Article

As a kid, Snakes and Ladders was fun, didn’t realize I’d be coding the shortest path for it one day!

#Algorithm:
1. BFS hence Queue, add 1 to begin with.
2. For each move, simulate all 6-sided dice rolls (loop from 1 => 6).
3. Use Helper method to map square numbers to (row, col) on the board.
4. If there’s a snake or ladder (board[row][col] != -1), jump to that destination square.
5. Track visited squares and increment steps at each BFS level.
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Post 1w
Interfaces vs Types in TypeScript

Interfaces vs Types in TypeScript

1w 473 0 Article

Math > Loop

Time, Space: O(1)

#Algorithm

1. Add all numbers from 1 to n

- Use formula: totalSum = n × (n + 1) / 2


2. Add all numbers from 1 to n that are divisible by m

- Count how many: count = n / m

- Then: divisibleSum = m × count × (count + 1) / 2


3. Return nonDivisibleSum - divisibleSum


#Code: https://github.com/RikamPalkar/DSA-Simplified/blob/main/LeetCode/2894. Divisible and Non-divisible Sums Difference.cs


#Programming #DSA

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Post 1w
How to Pass Functions & Params in React + TypeScript

How to Pass Functions & Params in React + TypeScript

1w 448 0 Article
One Cool Trick in React: Memo

One Cool Trick in React: Memo

1w 414 0 Article

Nope, not the frequency array!

#Algorithm
- map to count how many times each word appears.
- Loop through each word:
If its reverse (e.g., "ab" → "ba") is already in the map : res +4
and decrease the count of the reverse word.
- Else, store this word in the dictionary or increase its count.
- Check the dictionary for any word like "cc", "dd" (same letter twice):
If there's at least one such word left, we can use one of them in the middle of the palindrome (adds 2 characters).
- Return the

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Post 2w
Let's Start with React Components

Let's Start with React Components

2w 300 0 Article
Create React TS Apps with CRA & Vite

Create React TS Apps with CRA & Vite

2w 396 0 Article

In-place & in pain.


#Algorithm

1. Check 0’s in first row and column

2. If value == 0, flag first row/column

3. Loop and zero marked rows/columns

4. Zero first row/column if flagged


#Code: https://github.com/RikamPalkar/DSA-Simplified/blob/main/LeetCode/73. Set Matrix Zeroes.cs


#DSA #Programming

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Post 2w
Sir, This Is a TypeScript Function

Sir, This Is a TypeScript Function

2w 286 0 Article

Brute force TLE > optimized with diff array!


#Code: https://github.com/RikamPalkar/DSA-Simplified/blob/main/LeetCode/3355. Zero Array Transformation I.cs


#Algorithm

Create a difference array of len + 1.

For each query [l, r]:

- diff[l] += 1 > we start decrementing from index l

- diff[r + 1] -= 1 > we stop decrementing after index r

Build the prefix sum of diff array, //how many times each index was decremented.

For each index i:

If nums[i] - prefixSumAt[i] > 0, means nums[i] w

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Post 2w
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