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Day 3 ✅ #LeetCodeChallenge
Solved LeetCode 3075 – Maximum Happiness Sum
• Used a min-heap of size k to keep only the top k happiness values
• Removed smaller values to maintain best candidates
• Ignored negative contributions
TC= O(n log k)
#DSA #Heap #LeetCodeDaily
Day 109 of my #LeetCodeChallenge 🔥
Today’s problem:
961. N-Repeated Element in Size 2N Array
Sounds like a counting problem at first 🤔
But brute force isn’t the point here.
You’re given a 2n sized array where one element appears n times.
The real trick 👇
👉 That element is everywhere.
Key observation:
• The repeated element can’t stay far apart
• In any small window, it must appear at least twice
• So no hash map needed — just compare nearby elements
How it works:
• Traverse the array left to right
• Compare nums[i] with i+1, i+2, i+3
• The first match you find is the answer
That’s it.
Simple logic > extra space.
Complexity:
O(n) time
O(1) space
Takeaway:
Constraints often hide the real solution.
Observation can beat brute force — and even beat hash maps.
Full write-up + JavaScript solution 👇
https://t.co/iCtYPT1WXZ
#100DaysOfCode #LeetCode #Day109 #Arrays #Observation #JavaScript #CodingJourney
![nitinahirwal_in's tweet photo. Day 109 of my #LeetCodeChallenge 🔥
Today’s problem:
961. N-Repeated Element in Size 2N Array
Sounds like a counting problem at first 🤔
But brute force isn’t the point here.
You’re given a 2n sized array where one element appears n times.
The real trick 👇
👉 That element is everywhere.
Key observation:
• The repeated element can’t stay far apart
• In any small window, it must appear at least twice
• So no hash map needed — just compare nearby elements
How it works:
• Traverse the array left to right
• Compare nums[i] with i+1, i+2, i+3
• The first match you find is the answer
That’s it.
Simple logic > extra space.
Complexity:
O(n) time
O(1) space
Takeaway:
Constraints often hide the real solution.
Observation can beat brute force — and even beat hash maps.
Full write-up + JavaScript solution 👇
https://t.co/iCtYPT1WXZ
#100DaysOfCode #LeetCode #Day109 #Arrays #Observation #JavaScript #CodingJourney](https://pbs.twimg.com/media/G9q6UhdbcAAqklg.jpg)
Day 88 of my #LeetCodeChallenge 🔥
Today’s problem:
3433. Count Mentions Per User
This one surprised me. It looks simple at first… until you realize how much the order of events changes everything 😅
The setup feels like building a mini chat app:
• Users can go offline for 60 time units
• Messages can tag ALL users
• Or only online users (HERE)
• Or specific users (idX)
And the kicker:
If OFFLINE and MESSAGE share the same timestamp, the user goes offline before the message is processed.
One small rule that completely shifts the results.
My approach was to simulate the timeline:
Sort events by timestamp
Always handle OFFLINE before MESSAGE
Track when each user becomes online again
Parse every message token and update mention counts
Once that part fell into place, the problem became surprisingly clean.
Complexity:
O(n * users) time
O(users) space
Takeaway:
Even simple simulations depend heavily on processing order. One tiny rule can flip the entire logic — which is why I enjoyed this one more than expected.
Full write-up + JavaScript solution:
https://t.co/9Rw3ZX3Rbn
#100DaysOfCode #LeetCode #Day88 #JavaScript #Simulation #CodingJourney 🚀
#100DaysofCode -> #80Days of DSA
Today I solved one question at #leetcodechallenge.
1611. Minimum One Bit Operations to Make Integers Zero.
Day 51 of my #LeetCodeChallenge 🚀
Today’s problem:
3321. Find X-Sum of All K-Long Subarrays II 🧮
🧩 Approach
• Sliding window of size k
• Frequency map to track counts
• Two heaps: best (top X), rest (others)
• Lazy updates + rebalancing to maintain top-X by (freq ↓, value ↓)
⚡ Complexity
O(n log D) time | O(D) space (D = distinct in window)
✨ Takeaways
Priority queues + lazy deletion = efficient dynamic top-K over a moving window. Scales cleanly to n ≤ 1e5 💡
👉 Full write-up + JS solution:
https://t.co/zlGRm67t7h
#100DaysOfCode #LeetCode #JavaScript #SlidingWindow #Heaps #CodingChallenge
Day 50 of my #LeetCodeChallenge 🚀
Today’s problem:
3318. Find X-Sum of All K-Long Subarrays I 🧮
🧩 Approach
- Use a sliding window of size k
- Maintain frequency counts of elements (since nums[i] ∈ [1..50])
- For each window:
- Sort by frequency ↓ then value ↓
- Take top x elements → compute sum = count * value
- Slide the window efficiently by updating counts
⚡ Complexity
O(n × 50 log 50) ≈ O(n) time | O(1) space
✨ Takeaways
A clean mix of frequency counting + sliding window — simple yet powerful for constrained inputs 💡
Perfect warm-up for mastering window-based problems ✅
Day 50 done ✅ streak alive and strong 🔥
👉 Full write-up + JS solution:
https://t.co/B7nL4KPLl2
#100DaysOfCode #LeetCode #JavaScript #SlidingWindow #HashMap #CodingChallenge
![nitinahirwal_in's tweet photo. Day 50 of my #LeetCodeChallenge 🚀
Today’s problem:
3318. Find X-Sum of All K-Long Subarrays I 🧮
🧩 Approach
- Use a sliding window of size k
- Maintain frequency counts of elements (since nums[i] ∈ [1..50])
- For each window:
- Sort by frequency ↓ then value ↓
- Take top x elements → compute sum = count * value
- Slide the window efficiently by updating counts
⚡ Complexity
O(n × 50 log 50) ≈ O(n) time | O(1) space
✨ Takeaways
A clean mix of frequency counting + sliding window — simple yet powerful for constrained inputs 💡
Perfect warm-up for mastering window-based problems ✅
Day 50 done ✅ streak alive and strong 🔥
👉 Full write-up + JS solution:
https://t.co/B7nL4KPLl2
#100DaysOfCode #LeetCode #JavaScript #SlidingWindow #HashMap #CodingChallenge](https://pbs.twimg.com/media/G46qmcXXcAAHGr3.jpg)
Day 48 of my #LeetCodeChallenge 🚀
Today’s problem:
2257. Count Unguarded Cells in the Grid 🔗
🧩 Approach
- Represent the grid with states (0: empty, 1: guard, 2: wall, 3: guarded)
- For each guard, sweep in 4 directions (⬆️⬇️⬅️➡️)
- Stop when hitting a wall or another guard
- Count remaining unguarded cells (value = 0)
⚡ Complexity
O(m × n) time | O(m × n) space
✨ Takeaways
Simulating guard visibility is trickier than it looks 👀
A clean grid-based approach keeps it simple & robust 💡
Day 48 done ✅ streak continues 🔥
👉 Full write-up + JS solution:
https://t.co/MS7nXFNIIk
#100DaysOfCode #LeetCode #JavaScript #Grid #Simulation #CodingChallenge
Day 47 of my #LeetCodeChallenge
Today’s problem:
3217. Delete Nodes From Linked List Present in Array 🔗
🧩 Approach
- Convert nums into a Set for O(1) lookups
- Use a dummy node to simplify deletions
- Traverse & skip nodes whose values exist in the set
⚡ Complexity
O(n + m) time | O(m) space
✨ Takeaways
Efficient linked list cleanup using hashing 💡
Dummy node pattern = cleaner, safer deletions 🧠
📊 My submission:
⏱ Runtime: 78 ms (beats 50.63%)
💾 Memory: 100.10 MB (beats 46.84%)
Day 47 done ✅ streak continues 🚀
👉 Full write-up + JS solution:
https://t.co/9xfm04hVlI
#100DaysOfCode #LeetCode #JavaScript #LinkedList #Hashing #CodingChallenge
Day 46 of my #LeetCodeChallenge
Today’s problem:
3289. Get Sneaky Numbers 🕵️♂️
🧩 Approach
- Use a seen array to track visited numbers
- If a number appears again → push to result
- Stop when both duplicates are found
⚡ Complexity
O(n) time | O(n) space
✨ Takeaways
Simple hashing trick for detecting duplicates ⚙️
Clean, fast, and satisfying 💡
📊 My submission:
⏱ Runtime: 2 ms (beats 56.25%)
💾 Memory: 57.82 MB (beats 22.08%)
Day 46 done ✅ streak continues 🚀
👉 Full write-up + JS solution:
https://t.co/PhvtJjyvCk
#100DaysOfCode #LeetCode #JavaScript #DSA #CodingChallenge
Day 91 of 100 Days of Leetcode Challenge - Problem 3354: Make Array Elements Equal to Zero
#100DaysOfLeetcode #LeetcodeChallenge #ProblemSolving #CodingLife #TechCareer #AlgorithmDesign #LeetcodeSolutions #CodingCommunity
Day 25/100 of #100DaysOfCode – Java DSA
Solved “Maximum Number of Distinct Elements After Operations” 🧩 🔹 Runtime: 19ms | 🔹 Memory: 58MB #JavaCoding #LeetCodeChallenge #CodeEveryday #zigbee
🚀 Day 29 of my #LeetCodeChallenge
Ever wondered how to detect two adjacent strictly increasing subarrays of length k? 🤔
That’s today’s problem:
3349. Adjacent Increasing Subarrays Detection I 🔎
🧩 Approach
- Precompute inc[i] = length of increasing run ending at i.
- A subarray [a..a+k-1] is valid if inc[a+k-1] ≥ k.
- Just check two consecutive windows [a..a+k-1] & [a+k..a+2k-1].
⚡ Complexity
O(n) time, O(n) space.
✨ Takeaways
Precomputing run lengths is super handy for sequence checks — avoids re-checking every subarray 👌
Day 29 done ✅ streak continues 🚀
👉 Full write-up + JS solution:
https://t.co/65ADq6FhTz
![nitinahirwal_in's tweet photo. 🚀 Day 29 of my #LeetCodeChallenge
Ever wondered how to detect two adjacent strictly increasing subarrays of length k? 🤔
That’s today’s problem:
3349. Adjacent Increasing Subarrays Detection I 🔎
🧩 Approach
- Precompute inc[i] = length of increasing run ending at i.
- A subarray [a..a+k-1] is valid if inc[a+k-1] ≥ k.
- Just check two consecutive windows [a..a+k-1] & [a+k..a+2k-1].
⚡ Complexity
O(n) time, O(n) space.
✨ Takeaways
Precomputing run lengths is super handy for sequence checks — avoids re-checking every subarray 👌
Day 29 done ✅ streak continues 🚀
👉 Full write-up + JS solution:
https://t.co/65ADq6FhTz](https://pbs.twimg.com/media/G3OP_E2WcAACsbd.jpg)
🚀 Day 27 of my #LeetCodeChallenge
It’s Sunday, so I picked a hard one to challenge myself 🧠🔥
Problem: 3539. Find Sum of Array Product of Magical Sequences ✨
The twist? We form sequences of length m, but only keep them if
Σ 2^seq[i] has exactly k set bits in binary.
🧩 Approach
- Think in terms of counts: how many times each index is chosen.
- Use combinatorics (nCr) to distribute slots.
- Simulate binary addition (carry + set bits).
- DP state = (remaining slots, carry, ones so far).
- At the end → accept if ones + popcount(carry) == k.
⚡ Complexity
Time → O(n * m³ * k) (works since m ≤ 30, n ≤ 50, k ≤ 30)
Space → O(m² * k) with rolling DP.
✨ Takeaways
This one felt like blending binomial coefficients + DP over binary carry.
Pretty elegant once you spot the reduction 👌
Day 27 done, hard Sunday cracked 🚀
👉 Full write-up + JS solution:
https://t.co/IXzUxF4x7d
![nitinahirwal_in's tweet photo. 🚀 Day 27 of my #LeetCodeChallenge
It’s Sunday, so I picked a hard one to challenge myself 🧠🔥
Problem: 3539. Find Sum of Array Product of Magical Sequences ✨
The twist? We form sequences of length m, but only keep them if
Σ 2^seq[i] has exactly k set bits in binary.
🧩 Approach
- Think in terms of counts: how many times each index is chosen.
- Use combinatorics (nCr) to distribute slots.
- Simulate binary addition (carry + set bits).
- DP state = (remaining slots, carry, ones so far).
- At the end → accept if ones + popcount(carry) == k.
⚡ Complexity
Time → O(n * m³ * k) (works since m ≤ 30, n ≤ 50, k ≤ 30)
Space → O(m² * k) with rolling DP.
✨ Takeaways
This one felt like blending binomial coefficients + DP over binary carry.
Pretty elegant once you spot the reduction 👌
Day 27 done, hard Sunday cracked 🚀
👉 Full write-up + JS solution:
https://t.co/IXzUxF4x7d](https://pbs.twimg.com/media/G3D8Cn1W4AA9D5j.jpg)
🚀 Day 24 of my #LeetCodeChallenge
Today’s problem: Find the Minimum Amount of Time to Brew Potions (3494) ⚗️🧙♂️
The task: multiple wizards brew potions in order under a no-wait rule. Each potion passes through all wizards sequentially.
🧩 Approach
- Compute prefix sums of wizard skills
- Brewing time of last potion = total skill × last mana
- For each potion pair, calculate the safe gap using:
gap(x,y) = max( x*A[i] − y*A[i−1] )
- Total = last potion time + all gaps
⚡ Complexity
Time → O(n·m)
Space → O(n)
✨ Takeaways
Think of it like a conveyor belt. Prefix sums + gap formula ensure potions don’t collide.
Day 24 done, brewing efficiently 🚀
👉 Full write-up + JS solution:
https://t.co/3soswqrvRu
![nitinahirwal_in's tweet photo. 🚀 Day 24 of my #LeetCodeChallenge
Today’s problem: Find the Minimum Amount of Time to Brew Potions (3494) ⚗️🧙♂️
The task: multiple wizards brew potions in order under a no-wait rule. Each potion passes through all wizards sequentially.
🧩 Approach
- Compute prefix sums of wizard skills
- Brewing time of last potion = total skill × last mana
- For each potion pair, calculate the safe gap using:
gap(x,y) = max( x*A[i] − y*A[i−1] )
- Total = last potion time + all gaps
⚡ Complexity
Time → O(n·m)
Space → O(n)
✨ Takeaways
Think of it like a conveyor belt. Prefix sums + gap formula ensure potions don’t collide.
Day 24 done, brewing efficiently 🚀
👉 Full write-up + JS solution:
https://t.co/3soswqrvRu](https://pbs.twimg.com/media/G20xlIEWsAA9tGg.jpg)
1401 days of daily coding on @LeetCode — consistency is key! 🚀
Proud of this streak and excited to keep leveling up every day.
#CodingJourney #LeetCodeChallenge #100DaysOfCode #DevCommunity #CodeEveryday
🚀 Day 12 of my #LeetCodeChallenge
Today’s problem: Largest Triangle Area 🔺
Given points on a 2D plane, find the max possible triangle area.
🧩 Approach
Use Shoelace Formula to compute area
Try all triplets of points (O(n³), totally fine for n ≤ 50)
Track the maximum area
⚡ Complexity
Time → O(n³) (manageable with ≤ 50 points)
Space → O(1)
✨ Takeaways
Sometimes brute force + the right math trick is all you need.
Shoelace formula makes geometry problems super simple 🚀
👉 Full detailed write-up + solution:
https://t.co/xeoIXe4pGq
Hit Day 40 on my #100DaysOfLeetCode! 🎉 Today's problem: "Find Closest Person" (3516). Loved this one for its simplicity. It's all about distance ∣z−x∣ vs ∣z−y∣. O(1)
#Coding #Algorithms #DSA #SoftwareDevelopment #LeetCodeChallenge
🌟 500 Days of Code — Badge Unlocked!
🙌 Discipline > Motivation
📈 Every submission is a step forward, no matter how small
#LeetCode #LeetCodeDaily #LeetCodeChallenge #CodeGrind #LeetCodeProblems #LeetCodePractice #LeetCodeCommunity #CodeEveryday #100DaysOfCode #DSA
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