Bitcoin-Led Crypto Rout Erases Nearly $500 Billion in a Week

Lessons in Investing

Bitcoin price today dropped sharply, falling to the $74,000 level and triggering another wave of selling across the crypto market. Ethereum slipped nearly 10% to around $2,100, while most major altcoins declined between 5% and 10% today.

The sudden move has raised fresh concerns about whether Bitcoin is entering a deeper correction phase after weeks of volatility.

Possible Reasons Behind the Bitcoin Crash Today

The latest Bitcoin crash is not linked to a single event. Instead, analysts point to multiple factors hitting the market at the same time, creating strong downward pressure.

Heavy Liquidations Accelerate Bitcoin Decline

One of the main reasons behind the drop is massive liquidations in the futures market. Market data shows that over $500 million worth of Bitcoin positions were liquidated in recent sessions.

Many traders were using high leverage. When Bitcoin slipped even slightly, automatic liquidations kicked in, forcing positions to close. This led to a chain reaction of selling, pushing prices lower within minutes.

After the U.S. market opened, Bitcoin dumped another 11USDT,700, wiping out more than $55 million in long positions in just two hours. The overall crypto market lost nearly $50 billion during the same move.

US Stock Market Weakness Hits Crypto Hard

The crypto sell-off mirrored weakness in traditional markets. The S&P 500 fell nearly 1.3%, as investors moved away from risk assets.

Historically, when global markets turn cautious, cryptocurrencies tend to react faster and more sharply. The same pattern played out this time, with Bitcoin and altcoins facing intense selling pressure.

Spot Bitcoin ETF Outflows Add Pressure

Another key factor weighing on prices is strong outflows from spot Bitcoin ETFs.

As per CoinGlass data, on February 3, spot BTC ETFs recorded $272 million in net outflows. BlackRock’s IBIT stood out as the only major buyer with $60 million in inflows, while other funds continued to see selling.

When ETF flows turn negative like this, it often signals reduced confidence among institutional investors, even if long-term interest remains intact.

Epstein Files Add to Market Uncertainty

Beyond macro pressure and liquidations, renewed discussion around the Epstein files has added another layer of uncertainty to the crypto market. Reports highlighting Jeffrey Epstein’s past connections to early Bitcoin research, funding linked to MIT’s Digital Currency Initiative, and ties to prominent crypto figures have resurfaced online. 

While there is no direct evidence linking these revelations to current price action, the narratives have fueled speculation on social media and increased short-term volatility. During already weak market conditions, such controversies often amplify fear and contribute to risk-off behavior among traders.

Geopolitical Tensions Increase Market Uncertainty

Rising global tensions have also played a role. Ongoing disputes involving the United States, Iran, and Venezuela, along with tariff-related concerns, have increased uncertainty across financial markets.

During such periods, large funds and ETF managers usually cut exposure to risky assets. This capital outflow has added further pressure to Bitcoin and the broader crypto market.

Profit-Taking After Bitcoin’s Massive Rally

Galaxy Digital CEO Mike Novogratz believes the recent decline is mainly driven by profit-taking, not panic.

According to him, many investors who bought Bitcoin at much lower levels started selling after prices crossed $100,000, locking in gains after a long rally. He described the move as a “seller’s wave”, rather than fear-driven selling.

Novogratz also dismissed concerns around emerging threats like quantum computing, saying price moves are still driven by basic supply and demand.

Bitcoin Price Analysis: Key Support and Resistance Levels

The market is sitting at a critical turning point. If Bitcoin slips below the $74,500 support, the next downside target is seen around $69,800–$68,000, a zone that previously acted as strong resistance. 

A deeper breakdown from there could drag prices toward the $53,000–$54,000 range, implying a correction of nearly 30% from current levels. 

On the upside, analysts believe a quick recovery is unlikely, as Bitcoin would need to reclaim the $90,000–$95,000 resistance zone and establish a clear higher-high structure before any sustained rebound can take shape.

FAQs

Why is Bitcoin price down today?

Bitcoin is down today due to leveraged liquidations, weak U.S. markets, ETF outflows, profit-taking after the rally, and rising global uncertainty.

How do U.S. stock market declines impact Bitcoin prices?

When stocks fall, investors reduce risk exposure. Bitcoin typically reacts faster, leading to sharper declines during market-wide sell-offs.

Is the current Bitcoin drop a healthy correction?

Yes. Many analysts view this move as a normal correction after a strong rally, helping reset leverage and excess speculation.

Does profit-taking mean Bitcoin’s bull market is over?

No. Profit-taking is common after major rallies and does not signal the end of a long-term bullish trend.

What could drive Bitcoin prices higher again?

Stabilizing markets, renewed ETF inflows, reduced leverage, and improving macro sentiment could support a recovery.

Ref Tags: Finance,Investment,

What is Anthropic's AI tool that wiped $285 billion off software stocks in a single day

See All Articles on AI

Anthropic's newly launched AI plugins have triggered what analysts are calling a 'SaaSpocalypse'—a brutal selloff that wiped out roughly $285 billion from software, legal tech and financial services stocks in a single trading session. The AI developer, known for its Claude chatbot, released 11 open-source plugins for its Claude Cowork tool on Friday, January 30. One of them targets legal workflows, and that alone was enough to send investors running for the exits.Thomson Reuters dropped over 15%. RELX, owner of LexisNexis, fell 14%. LegalZoom got hammered nearly 20%. A Goldman Sachs basket tracking US software stocks posted its worst single-day decline since April's tariff-driven selloff. The Nasdaq fell 1.4%, with ripple effects extending to Indian IT giants—Infosys ADRs slipped 5.5% and Wipro fell nearly 5%.

What exactly did Anthropic release

Claude Cowork is an agentic AI assistant that Anthropic launched earlier in January. Think of it as Claude Code—the company's developer-focused coding tool—but redesigned for non-technical professionals. It can read files, organise folders, draft documents and carry out multi-step tasks with user consent.

The plugins announced last Friday supercharge this capability. They let companies tailor Claude to specific job functions by specifying how work should be done, which tools and data to pull from, and what workflows to automate. Anthropic open-sourced 11 starter plugins covering productivity, sales, marketing, finance, data analysis, customer support, product management and biology research.But it was the legal plugin that spooked markets. It automates contract review, NDA triage, compliance checks and legal briefings. Anthropic was careful to note that all outputs should be reviewed by licensed attorneys and that the tool does not provide legal advice. That disclaimer did little to calm investors.

Why the market panic over a folder of prompts

Here's the thing that has caught many observers off guard: the legal plugin is essentially a set of prompts and configurations. There's no proprietary model fine-tuned on case law, no special legal reasoning engine. It's Claude being Claude, but with structured workflow instructions.The real signal, analysts say, is that Anthropic has moved from selling the model to owning the workflow. When Claude was just an API, companies like Thomson Reuters could build on top of it. Thomson Reuters literally runs CoCounsel on OpenAI. But when Anthropic starts publishing ready-made vertical solutions, the platform becomes the competitor.Jefferies dubbed the selloff a 'SaaSpocalypse', noting that investor sentiment has shifted dramatically. The narrative used to be that AI helps software companies. Now it's that AI replaces them. 'Trading is very much get-me-out style selling,' said Jeffrey Favuzza from Jefferies' equity trading desk. The brokerage also noted that OpenAI appears to be losing ground in the corporate market to Anthropic's Claude, with enterprises now accounting for 80% of Anthropic's business.

What does this mean for the software industry going forward

The damage extended well beyond legal tech. DocuSign fell 11%, Salesforce dropped nearly 7%, Adobe slid 7%, and ServiceNow declined 7%. Business development companies with exposure to software loans also got caught in the selling. Blue Owl Capital Corp fell 13%, marking a record ninth consecutive decline.Anthropic isn't the only player in legal AI. Startups like Harvey AI (valued at $5 billion) and Legora ($1.8 billion valuation) have been developing tools for years. But Anthropic's advantage is that it builds its own underlying models. Firms like Legora rely on models from developers like Anthropic, which means Claude's creator can potentially disrupt both traditional legal services and the startups trying to disrupt them.The company's growth trajectory underscores why investors are taking this threat seriously. Claude Code hit $1 billion in annualised recurring revenue by November, just months after its May launch. Anthropic is now reportedly raising $20 billion at a $350 billion valuation—up sharply from $61.5 billion in March 2025.The iteration speed is also striking. Cowork launched on January 12. The plugins dropped less than three weeks later. Enterprise software companies typically spend quarters on such releases.For software executives watching the carnage unfold, the takeaway is uncomfortable but clear: Anthropic didn't need a breakthrough product to rattle markets. It just needed to show what Claude could already do—and hit publish.

Ref

Maximize Minimal Potential -- What to Do When You Start with Nothing

See other Biographies & Autobiographies

Let me start with something uncomfortable but honest.

Most of us don’t begin life with ideal conditions. Some people are born into money, safety, and encouragement. Others—maybe you, maybe me—start with chaos, fear, neglect, or low expectations. And when you grow up like that, survival becomes the goal. Not growth. Not excellence. Just survival.

The problem is, survival mode has an expiration date.

At some point, you either evolve—or you stay stuck replaying the same story forever.

Born Behind the Line

There’s a lie we tell ourselves when life deals us a bad hand: “It’s not my fault.”
And here’s the thing—it’s usually true.

It’s not your fault if you grew up in poverty.
It’s not your fault if you were abused, neglected, bullied, or ignored.
It’s not your fault if you were never taught discipline, confidence, or self-respect.

But that truth can quietly turn into a trap.

Because if nothing is your fault, then nothing is your responsibility either. And that’s where potential goes to die.

Many people live their entire lives grading themselves on a forgiving curve. They carry their past like a permanent excuse slip. And the world—friends, family, coworkers—often reinforces it: “After what you’ve been through, it’s understandable.”

Understandable, yes. But survivable is not the same as sustainable.

The Moment of Reckoning

There’s usually a moment—sometimes loud, sometimes quiet—when you realize something is wrong inside you. Not broken in a dramatic way, but numb. Flat. Disconnected.

That numbness is dangerous.

Numb people quit easily.
Numb people stop dreaming.
Numb people settle for lives that don’t match their inner potential.

Numbness often starts as protection. When you grow up in pain, going emotionally cold can keep you alive. But what once saved you can eventually suffocate you.

And here’s the hard truth: no one is coming to wake you up.

Responsibility Without Blame

One of the most powerful mindset shifts you can make is this:

It may not be my fault—but it is my responsibility.

That sentence changes everything.

Responsibility doesn’t mean self-hatred.
It doesn’t mean denying your past.
It means refusing to let yesterday dictate tomorrow.

At some point, you have to stop collecting evidence about why your life turned out the way it did and start asking a harder question:

What am I doing today that keeps me here?

That question hurts. But it also liberates.

The Trap of “Distracting Injuries”

There’s a concept from emergency medicine called a distracting injury. It’s a wound that looks severe enough to pull attention away from the real threat.

In life, our past trauma often becomes that distracting injury.

We focus so intensely on what happened to us that we miss what’s happening because of us—our habits, our decisions, our avoidance, our lack of discipline.

Pain becomes our identity.
Victimhood becomes familiar.
And growth feels like betrayal.

But distraction is still distraction, no matter how justified.

Minimal Potential Is Still Potential

Here’s the good news.

You don’t need talent.
You don’t need confidence.
You don’t need permission.

You just need ownership.

Some people are planted in rich soil. Others sprout through concrete. That doesn’t make the concrete-grown seed inferior—it makes it resilient.

Resilience isn’t something you’re born with. It’s something you earn by choosing discomfort over stagnation, again and again.

Maximizing minimal potential means this:

• Working with what you have, not waiting for what you wish you had

• Trading excuses for effort

• Accepting that progress will be lonely and misunderstood

The Accountability Shift

Real change begins when you stop negotiating with yourself.

You stop saying:

• “I’ll start when I feel ready.”

• “I’ll try once things calm down.”

• “This is just who I am.”

Instead, you ask:

• “What’s the hardest honest thing I can do today?”

• “What am I avoiding?”

• “What standard am I willing to live by—even when no one is watching?”

That’s where accountability lives—not in motivation, but in daily self-confrontation.

Urgency Is a Gift

One of the most sobering ideas in this chapter is urgency.

Your dreams have expiration dates.
Your windows of opportunity close quietly.
And time does not negotiate.

Hope without action is gambling.
Waiting for clarity is procrastination.
Comfort is often disguised as patience.

Urgency doesn’t mean panic. It means respecting time enough not to waste it.

Becoming the One Who Sharpens Himself

There’s a romantic idea that growth happens best in groups—teams, mentors, communities. And while support helps, eventually everyone faces a moment when they’re alone with their standards.

No coach.
No applause.
No safety net.

That’s when character is forged.

Some people wait to be sharpened by others.
Others sharpen themselves.

They don’t stop at “better than before.”
They don’t settle for “good enough.”
They evolve relentlessly.

Final Thought

Maximizing minimal potential is not about proving anyone wrong. It’s about refusing to stay trapped by the lowest expectations placed on you—by others or by yourself.

You may have started behind.
You may still feel behind.
But you are not finished.

And the moment you take responsibility—not for what happened to you, but for what happens next—you stop being a product of circumstance and start becoming a force.

That’s where everything changes.

You have been preoccupied by bullshit for way too long. It’s time to switch your focus to the things that will slingshot you forward. 
#DistractingInjuries
#NeverFinished

From the book: Never Finished (Unshackle Your Mind and Win the War Within)
Chapter 1: Maximize Minimal Potential
By David Goggins, 2022

Rook Attack (Easy) - Find the position where we should place the rook, where Rook has maximum impact

Index of "Algorithms: Design and Analysis"

Try out the problem on Hacker Earth

Rook Attack

Given a N x M chessboard, where every cell has a value assigned to it denoted by 
 (Cell in the ith row from top and jth column from left).

Find the position (x, y) where we should place the rook, just the sum of cells which are under attack of rook is maximum possible. The rook can not attack the cell on which it is placed.

If there are more than one positions for (x, y), return the position where x is as minimum as possible. If there are more than one positions with equal x, return the position where y is as minimum as possible.

Note:

-> 1 based indexing is followed.
-> A rook can attack all the cells which are either in horizontal or vertical direction to the rook's position.
Input

First line contains two space separated integers denoting N and M.

Next N lines contains M space separated integers denoting the values of array A.

Output

Print two space separated integers x and y denoting the rook's position.

Constraints

1 ≤ N, M ≤ 10**6
1 ≤ A[i][j] ≤ 10**4

Sample Input
2 2
4 1 
3 1

Sample Output
1 2

Time Limit: 1.5
Memory Limit: 256
Source Limit:

Explanation

If we place the rook at (1, 2) position, the cells under attack of rook will be
(1, 1), (2, 2)
The sum of these cells will be
4 + 1 = 5, which is maximum possible.

---

---

Brute-Force Thinking....

Iterate through each cell of the chessboard and for each cell, calculate the sum of values in its row and column, excluding the value of the cell itself. Keep track of the maximum sum found and the corresponding position. If a new maximum is found, update the position. If the same maximum is found, check for the smallest row and column indices.


My Code:


N, M = map(int, input().split())

A = [list(map(int, input().split())) for _ in range(N)]

rowSum = [0] * N
colSum = [0] * M

for i in range(N):
    for j in range(M):
        rowSum[i] += A[i][j]
        colSum[j] += A[i][j]

best_score = -1
best_row = -1
best_col = -1

for i in range(N):
    for j in range(M):
        score = rowSum[i] + colSum[j] - 2 * A[i][j]

        # print("Score: ", score)
        # print(best_score, best_row, best_col)

        if score > best_score or (score == best_score and (i + 1 < best_row or (i + 1 == best_row and j + 1 < best_col))):
            best_score = score
            best_row = i + 1
            best_col = j + 1

print(best_row, best_col)

BEFORE I GIVE YOU CHATGPT'S THOUGHT: I WOULD LIKE TO TELL YOU THAT CHATGPT MADE A MISTAKE -- SAME MISTAKE REPEATEDLY WHEN TALKING ON A POINT ON MULTIPLE OCCASIONS

You see that when score is computed a cell: the value A[i][j] should be subtracted twice because it is counted in both rowSum and colSum. However, in the code and hints provided by ChatGPT, it is subtracted only once. 

So the correct formula for score should be:
score = rowSum[i] + colSum[j] - 2 * A[i][j

ChatGPT's Thoughts

Explain the solution to this problem in layman terms. Give clear descriptions of what this problem is trying to teach/assess, what design pattern, what problem solving trick/technique.

Let’s break this down slowly, simply, and intuitively, without jumping into code first.

---

🧩 What is the problem saying (in plain English)?

You are given a chessboard (a grid) where each cell has a number.

You want to place one rook on the board.

A rook:

• Can attack all cells in the same row

• Can attack all cells in the same column

• ❌ Cannot attack the cell it is placed on

Your goal is to:

Choose the cell where placing the rook gives the maximum total sum of attacked cells.

If multiple positions give the same maximum sum:

1. Choose the smallest row index x

2. If still tied, choose the smallest column index y

---

🧠 What does the rook “collect”?

If you place the rook at position (x, y):

• It collects all values in row x

• Plus all values in column y

• Minus A[x][y] (because it cannot attack its own cell)

So effectively:

score(x, y) = rowSum[x] + colSum[y] - A[x][y]

---

🧪 Sample intuition

Input:

4 1
3 1

Grid:

4  1
3  1

At position (1,2):

• Row 1 sum = 4 + 1 = 5

• Column 2 sum = 1 + 1 = 2

• Subtract cell (1,2) = 1

Final score = 5 + 2 - 1 = 6

This is the maximum possible → answer is (1,2)

---

🚀 The naïve (slow) way — and why it fails

A beginner might think:

• For each cell, scan its entire row and column

• Add values

That would cost:

mathematica
O(N + M) per cell
Total = O(N * M * (N + M)) ❌

This is too slow for large grids (up to 10⁶ cells).

---

💡 The key insight (problem-solving trick)

Instead of recalculating row and column sums again and again, we can:

Step 1: Precompute

• The sum of every row → rowSum[]

• The sum of every column → colSum[]

This takes only:

scss
O(N * M)

Step 2: For every cell (x, y)

Compute the score in constant time:

css
rowSum[x] + colSum[y] - A[x][y]

---

🎯 What this problem is really testing

1️⃣ Precomputation / Caching

• Don’t repeat work

• Compute row and column sums once

• Reuse them

2️⃣ Optimization mindset

• Replace nested loops with smart preprocessing

• Turn an expensive operation into O(1)

3️⃣ Grid / Matrix reasoning

• Understanding how rows and columns interact

• Applying formulas on 2D data

4️⃣ Tie-breaking logic

• Careful comparison logic

• Attention to problem constraints

---

🧠 Design pattern used

✅ Precomputation + Enumeration

This is a very common competitive-programming pattern:

1. Precompute useful aggregates (row sums, column sums)

2. Enumerate all candidates efficiently

3. Track the best result with tie-breaking rules

---

📌 Why subtract A[x][y]?

Because:

• The rook cannot attack the cell it is placed on

• That cell is included once in the row sum

• And once in the column sum

• So we subtract it once

---

🧠 One-sentence intuition

“For each cell, imagine the rook collects everything in its row and column — so we just need fast access to row and column totals.”

---

🏁 Final takeaway

This problem teaches you:

• How to optimize brute-force ideas

• How to think in terms of precomputation

• How to handle 2D data efficiently

• How to apply clean mathematical reasoning to grids

If you want next:

• Full Python code

• Dry-run with a larger example

• How this idea appears in other problems

• Space/time complexity discussion

Just tell me 😊

---

---

Full Python code

Here is the full, correct, and efficient Python solution, followed by a clear explanation of how it works.

---

✅ Full Python Code

python
def solve():
    import sys
    input = sys.stdin.readline

    N, M = map(int, input().split())
    A = [list(map(int, input().split())) for _ in range(N)]

    # Step 1: Compute row sums and column sums
    row_sum = [0] * N
    col_sum = [0] * M

    for i in range(N):
        for j in range(M):
            row_sum[i] += A[i][j]
            col_sum[j] += A[i][j]

    # Step 2: Find the best position
    max_score = -1
    best_x = 1
    best_y = 1

    for i in range(N):
        for j in range(M):
            score = row_sum[i] + col_sum[j] - A[i][j]

            if (score > max_score or
               (score == max_score and (i + 1 < best_x or
               (i + 1 == best_x and j + 1 < best_y)))):
                max_score = score
                best_x = i + 1   # convert to 1-based indexing
                best_y = j + 1

    print(best_x, best_y)

---

🧠 Explanation in Simple Terms

What are we doing?

We want to place a rook on a chessboard-like grid so that the sum of values it can attack (same row + same column) is as large as possible.

The rook:

• Attacks everything in its row

• Attacks everything in its column

• ❌ Does NOT attack the cell it sits on

So for any cell (i, j):

ini
score = rowSum[i] + colSum[j] - A[i][j]

---

Step-by-step logic

🔹 Step 1: Precompute row and column sums

Instead of re-adding rows and columns repeatedly (which would be slow), we compute:

• Sum of each row

• Sum of each column

This makes later calculations very fast.

---

🔹 Step 2: Try placing the rook at every cell

For every position (i, j):

• Compute the attack sum using the formula above

• Keep track of the maximum score found

• If scores tie:

  • Pick the smaller row index

  • If still tied, pick the smaller column index

---

⏱️ Time & Space Complexity

Metric	Complexity
Time	O(N × M)
Space	O(N + M)

This is optimal for a grid with up to 1 million cells.

---

🧩 What this problem teaches

• Precomputation optimization (don’t repeat work)

• 2D grid reasoning

• Greedy comparison with tie-breaking

• Turning brute force into efficient code

---

🏁 Final intuition (one line)

“Precompute row and column totals, then test every cell using a constant-time formula.”

If you want:

• A dry-run example

• A version using less memory

• Similar problems to practice

• Visualization

Just say the word 🙂

---

---

I am super excited to see GPT 5.2 make mistakes and hallucinate... so capturing a couple of screenshots below:

Largest Number - Find out the largest possible number which can be built from N after removing exactly K digits (Easy)

Index of "Algorithms: Design and Analysis"

Try out the problem on Hacker Earth

Given an integer N which has D digits. You have to delete exactly K digits in integer N.

Find out the largest possible number which can be built from N after removing exactly K digits.

Input Format:

First line contains two space separated integers: N K
 
Output Format:

Print the largest possible number which can be built from 
 after removing exactly 
 digits.

Constraints:


Sample Input
3412 1
Sample Output
412
Time Limit: 1
Memory Limit: 256
Source Limit:

Explanation

After removing exactly 1 digit, we can make

412
312
342
341

Among, them 412 is the largest possible number.

Solution:

Couple of intuitive questions to ask yourself before jumping to code:
1. What is the power of 10 that will remain from the original number after removing K digits?
   Ans: D - K
2. What is the position and value of the largest digit in the remaining number?
   Ans: The largest digit must be in the first D - K digits of the original number.
   If largest digit is not in the first D - K digits, we go for the second largest digit.

Psuedo Code

Initialize empty stack
For each digit in N:
    While stack not empty AND
          last digit in stack < current digit AND
          deletions remaining:
        pop stack
        decrease K
    push current digit to stack

If deletions still left:
    remove last K digits from stack

Answer = digits in stack joined together

My Code


N, K = map(int, input().split())

digits = list(map(int, str(N)))

stack = []

for digit in digits:
    while K > 0 and stack and stack[-1] < digit:
        stack.pop()
        K -= 1
    stack.append(digit)

# If there are still K digits to remove, remove from the end
while K > 0:
    stack.pop()
    K -= 1

print(''.join(map(str, stack)))

Q: Explain the solution to this problem in layman terms. Give clear descriptions of what this problem is trying to teach/assess, what design pattern, what problem solving trick/technique.

Great problem choice 👍 — this is a classic interview + competitive programming problem, and it teaches an important way of thinking.

I’ll explain this in layman terms, then connect it to what it’s testing, the design pattern, and the core trick.

---

🔍 Problem in simple words

You are given:

• A number N (as digits)

• A number K (how many digits you must delete)

👉 Your task:
Delete exactly K digits from N
without changing the order of the remaining digits,
so that the resulting number is as large as possible.

---

Example

ini
N = 3412
K = 1

Possible results after deleting 1 digit:

• Remove 3 → 412

• Remove 4 → 312

• Remove 1 → 342

• Remove 2 → 341

✔️ Largest = 412

---

🧠 Key constraint that matters

You cannot rearrange digits.
You can only delete digits.

This is what makes the problem non-trivial.

---

🧩 How to think about it (layman explanation)

Let’s think like a human looking at the number:

“If I can delete some digits, I want big digits to come as early as possible.”

Because:

• A bigger digit at the left makes the number larger

• Later digits matter less

---

💡 Core idea (greedy thinking)

Whenever you see:

• A smaller digit before a bigger digit

• AND you still have deletions left

👉 Delete the smaller digit

Why?

Because keeping the bigger digit earlier makes the number larger overall.

---

Example walk-through

ini
N = 3412
K = 1

Compare adjacent digits:

cpp
3 < 4 → delete 3

Now:

412

No deletions left → stop

✔️ Correct answer

---

🧰 Design Pattern Used

✅ Greedy Algorithm

You make the best local choice at every step:

“If removing this digit makes the number bigger, do it now.”

Greedy works here because:

• Earlier digits dominate the value of the number

• Once a digit is fixed, future digits can’t compensate for a bad choice

---

🔄 Implementation Technique

Monotonic Stack

This problem is usually solved using:

• A stack that keeps digits in decreasing order

• While:

  • Stack top < current digit

  • AND deletions left

Pop the stack (delete digit).

---

Why a stack?

It helps you:

• Look back at previous digits

• Remove them efficiently

• Maintain the best possible prefix

---

✏️ Pseudocode (human-readable)

sql
Initialize empty stack
For each digit in N:
    While stack not empty AND
          last digit in stack < current digit AND
          deletions remaining:
        pop stack
        decrease K
    push current digit to stack

If deletions still left:
    remove last K digits from stack

Answer = digits in stack joined together

---

🧪 Edge cases this problem tests

1. Digits already decreasing

   ini
   N = 9876, K = 2 → 98
   

   → Just remove from the end

2. Leading zeros

   ini
   N = 1002, K = 1 → 102
   

3. K = D
   → Result is empty or 0 (depending on problem definition)

---

🎯 What this problem is really assessing

Skill	Tested
Greedy thinking	✅
Stack usage	✅
Digit manipulation	✅
Order preservation	✅
Edge-case handling	✅
Algorithmic intuition	✅

---

🧠 Big takeaway

To maximize a number, fix the leftmost digits first
by deleting smaller digits when you get the chance.

This idea appears in many problems:

• Remove K digits

• Build largest/smallest number

• Stock span problems

• Next greater element

---

🏁 Final one-line summary

This problem teaches you how to:

Use greedy decisions + a stack to build the best possible number while deleting digits without changing order.

If you want next:

• Full Python code

• Dry-run with more examples

• Why greedy is provably correct

• How this compares to brute force

Just tell me 🙂

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Full Python Code From ChatGPT

Here is the clean, correct, interview-ready Python solution using the greedy + monotonic stack approach we discussed.

This works efficiently even for very large N (up to 10¹⁸).

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✅ Full Python Code

python
def largest_number_after_k_deletions(N, K):
    digits = list(str(N))
    stack = []

    for d in digits:
        # Remove smaller digits from stack if possible
        while stack and K > 0 and stack[-1] < d:
            stack.pop()
            K -= 1
        stack.append(d)

    # If deletions still remain, remove from the end
    while K > 0:
        stack.pop()
        K -= 1

    # Join digits and remove leading zeros if any
    result = ''.join(stack).lstrip('0')

    return result if result else "0"


# ---- Driver code ----
N, K = map(int, input().split())
print(largest_number_after_k_deletions(N, K))

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🧠 Step-by-Step Explanation (Layman Friendly)

🔹 Convert the number into digits

python
digits = list(str(N))

We treat the number as characters so we can remove digits easily.

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🔹 Use a stack to build the best number

python
stack = []

The stack keeps digits in descending order, as much as possible.

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🔹 Main greedy logic

python
while stack and K > 0 and stack[-1] < d:
    stack.pop()
    K -= 1

This means:

• If the previous digit is smaller than the current digit

• AND we still have deletions left

• Remove the smaller digit to make the number bigger

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🔹 Handle leftover deletions

python
while K > 0:
    stack.pop()
    K -= 1

If digits are already decreasing (like 9876), we remove digits from the end.

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🔹 Clean result

python
result = ''.join(stack).lstrip('0')

Removes leading zeros if they appear.

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🧪 Example Walkthrough

Input

yaml
3412 1

Stack evolution

Step	Digit	Stack	K
1	3	[3]	1
2	4	[] → [4]	0
3	1	[4,1]	0
4	2	[4,1,2]	0

✔️ Output: 412

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⏱️ Time & Space Complexity

Metric	Value
Time	O(D)
Space	O(D)

Where D = number of digits in N.

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🎯 Why this solution is optimal

• No brute force

• No recursion

• No permutations

• Works for large numbers

• Industry-standard greedy solution

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🧩 What this problem teaches

• Greedy decision making

• Monotonic stacks

• Digit-level manipulation

• Why earlier digits matter more

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If you want next:

• A brute-force validator

• Handling K > 3

• Explanation of why greedy always works

• Similar problems to practice

Just say 👍

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What is the greedy algorithm in 100 words?

A greedy algorithm is a problem-solving approach where, at each step, you make the best immediate (locally optimal) choice with the hope that these choices lead to an overall optimal solution. It does not look ahead or reconsider past decisions. Greedy algorithms work well when a problem has the greedy-choice property, meaning a locally optimal choice is also globally optimal. They are usually simple, fast, and memory-efficient. Common examples include activity selection, coin change (with certain coin systems), shortest path algorithms like Dijkstra, and problems involving monotonic stacks such as removing digits to form the largest number.