Ugly Number II (Smooth Numbers)

🎯

Ugly Number II (Smooth Numbers)

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Imagine you want to generate numbers whose only prime factors are 2, 3, or 5 - these 'smooth' numbers appear in computer graphics and signal processing.

💡 This problem asks for the nth number whose prime factors are limited to 2, 3, and 5. Beginners often struggle because the naive approach is slow and the optimal solution requires merging multiple sequences efficiently.

📋

Problem Statement

Given an integer n, return the nth ugly number. Ugly numbers are positive numbers whose prime factors only include 2, 3, and 5. The sequence starts with 1 as the first ugly number.

1 ≤ n ≤ 10^5

💡

Example

Input"n = 10"

Output12

The first 10 ugly numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12. The 10th is 12.

• n = 1 → output 1 (smallest ugly number)
• n = 2 → output 2 (first prime factor multiple)
• n = 100 → output 1536 (larger input to test efficiency)
• n = 0 → invalid input, expect handling or defined behavior

⚠️

Common Mistakes

Not handling duplicates when multiples overlap

Output contains repeated ugly numbers, incorrect nth number

✅ Increment all pointers whose multiples equal the chosen minimum

Starting count from 0 instead of 1 for ugly numbers

Off-by-one errors in output

✅ Initialize first ugly number as 1 and count accordingly

Using brute force for large n without optimization

Time limit exceeded or very slow runtime

✅ Use pointer or heap approach for efficiency

Not checking for invalid inputs like n=0

Undefined behavior or runtime errors

✅ Add input validation or clarify constraints

Incorrectly implementing min-heap or forgetting to use a set

Duplicates in heap cause incorrect results or infinite loops

✅ Use a set to track seen numbers and prevent duplicates

🧠

Brute Force (Check Each Number Sequentially)

💡 This approach exists to build intuition by directly checking each number for the ugly property, helping understand the problem's definition before optimizing.

Intuition

Start from 1 and check every number incrementally to see if it only has prime factors 2, 3, and 5. Count how many such numbers appear until reaching the nth.

Algorithm

1. Initialize count and current number to 1.
2. While count is less than n, increment current number.
3. Check if current number is ugly by dividing out factors 2, 3, and 5 repeatedly.
4. If after division the number becomes 1, increment count.
5. Return the current number when count reaches n.

💡 The main difficulty is understanding how to check if a number is ugly by factor division and realizing the inefficiency of checking every number.

</>

Code

def is_ugly(num):
    if num <= 0:
        return False
    for p in [2, 3, 5]:
        while num % p == 0:
            num //= p
    return num == 1

def nthUglyNumber(n):
    count = 1  # 1 is the first ugly number
    current = 1
    while count < n:
        current += 1
        if is_ugly(current):
            count += 1
    return current

# Driver code
if __name__ == '__main__':
    print(nthUglyNumber(10))  # Output: 12

Line Notes

def is_ugly(num):Defines helper to check if a number is ugly

if num <= 0:Negative or zero can't be ugly numbers

while num % p == 0:Divide out all factors of p to test primality

return num == 1If fully divided by 2,3,5, leftover must be 1

count = 1 # 1 is the first ugly numberStart counting from 1 as the first ugly number

while count < n:Loop until we find nth ugly number

if is_ugly(current):Check if current number qualifies

return currentReturn the nth ugly number found

public class Solution {
    public static boolean isUgly(int num) {
        if (num <= 0) return false;
        int[] primes = {2, 3, 5};
        for (int p : primes) {
            while (num % p == 0) {
                num /= p;
            }
        }
        return num == 1;
    }

    public static int nthUglyNumber(int n) {
        int count = 1; // 1 is the first ugly number
        int current = 1;
        while (count < n) {
            current++;
            if (isUgly(current)) {
                count++;
            }
        }
        return current;
    }

    public static void main(String[] args) {
        System.out.println(nthUglyNumber(10)); // Output: 12
    }
}

Line Notes

public static boolean isUgly(int num) {Helper method to check ugly number

if (num <= 0) return false;Exclude non-positive numbers

while (num % p == 0) {Remove all prime factors p

return num == 1;Check if fully factored by 2,3,5

int count = 1; // 1 is the first ugly numberInitialize count with first ugly number

while (count < n) {Loop until nth ugly number found

if (isUgly(current)) {Check if current number is ugly

return current;Return the nth ugly number

#include <iostream>
using namespace std;

bool isUgly(int num) {
    if (num <= 0) return false;
    int primes[3] = {2, 3, 5};
    for (int p : primes) {
        while (num % p == 0) {
            num /= p;
        }
    }
    return num == 1;
}

int nthUglyNumber(int n) {
    int count = 1; // 1 is the first ugly number
    int current = 1;
    while (count < n) {
        current++;
        if (isUgly(current)) {
            count++;
        }
    }
    return current;
}

int main() {
    cout << nthUglyNumber(10) << endl; // Output: 12
    return 0;
}

Line Notes

bool isUgly(int num) {Helper function to check ugly number

if (num <= 0) return false;Exclude zero or negative numbers

while (num % p == 0) {Divide out prime factors 2,3,5

return num == 1;Confirm number fully factored by 2,3,5

int count = 1; // 1 is the first ugly numberStart counting from 1

while (count < n) {Loop until nth ugly number found

if (isUgly(current)) {Check if current number is ugly

return current;Return nth ugly number

function isUgly(num) {
    if (num <= 0) return false;
    const primes = [2, 3, 5];
    for (let p of primes) {
        while (num % p === 0) {
            num /= p;
        }
    }
    return num === 1;
}

function nthUglyNumber(n) {
    let count = 1; // 1 is the first ugly number
    let current = 1;
    while (count < n) {
        current++;
        if (isUgly(current)) {
            count++;
        }
    }
    return current;
}

// Test
console.log(nthUglyNumber(10)); // Output: 12

Line Notes

function isUgly(num) {Helper to check if number is ugly

if (num <= 0) return false;Exclude invalid numbers

while (num % p === 0) {Remove all factors of p

return num === 1;Confirm fully factored by 2,3,5

let count = 1; // 1 is the first ugly numberInitialize count with 1

while (count < n) {Loop until nth ugly number found

if (isUgly(current)) {Check if current number is ugly

return current;Return nth ugly number

⏱

Complexity

TimeO(n * log n) approximately

SpaceO(1)

For each number, we divide by 2,3,5 repeatedly which takes O(log n) in worst case. We may check many numbers until nth ugly number is found, so total is roughly O(n log n).

💡 For n=20, this means checking many numbers and dividing factors repeatedly, which is slow and inefficient.

Interview Verdict: TLE / Use only to introduce

This approach is too slow for large n but helps understand the problem and why optimization is needed.

🧠

Better Approach (Merge Multiples Using Three Pointers)

💡 This approach improves efficiency by generating ugly numbers in order using pointers, avoiding checking every number.

Intuition

Maintain three pointers for multiples of 2, 3, and 5. At each step, pick the smallest next multiple to generate the next ugly number, ensuring sorted order.

Algorithm

1. Initialize an array ugly with first element 1.
2. Set three pointers i2, i3, i5 to 0, representing indices for multiples of 2, 3, and 5.
3. For each position from 1 to n-1, compute next multiples: next2 = ugly[i2]*2, next3 = ugly[i3]*3, next5 = ugly[i5]*5.
4. Choose the minimum among next2, next3, next5 as the next ugly number.
5. Increment the pointer(s) whose multiple was chosen to avoid duplicates.
6. Return ugly[n-1] after filling the array.

💡 The challenge is to keep track of multiples and pointers correctly to generate numbers in ascending order without missing or duplicating.

</>

Code

def nthUglyNumber(n):
    ugly = [0] * n
    ugly[0] = 1
    i2 = i3 = i5 = 0
    for i in range(1, n):
        next2 = ugly[i2] * 2
        next3 = ugly[i3] * 3
        next5 = ugly[i5] * 5
        ugly[i] = min(next2, next3, next5)
        if ugly[i] == next2:
            i2 += 1
        if ugly[i] == next3:
            i3 += 1
        if ugly[i] == next5:
            i5 += 1
    return ugly[-1]

# Driver code
if __name__ == '__main__':
    print(nthUglyNumber(10))  # Output: 12

Line Notes

ugly = [0] * nPreallocate array to store ugly numbers

ugly[0] = 1First ugly number is 1 by definition

i2 = i3 = i5 = 0Initialize pointers for multiples of 2,3,5

next2 = ugly[i2] * 2Next candidate multiple of 2

ugly[i] = min(next2, next3, next5)Choose smallest next ugly number

if ugly[i] == next2:Increment pointer to avoid duplicates

return ugly[-1]Return nth ugly number

public class Solution {
    public static int nthUglyNumber(int n) {
        int[] ugly = new int[n];
        ugly[0] = 1;
        int i2 = 0, i3 = 0, i5 = 0;
        for (int i = 1; i < n; i++) {
            int next2 = ugly[i2] * 2;
            int next3 = ugly[i3] * 3;
            int next5 = ugly[i5] * 5;
            int nextUgly = Math.min(next2, Math.min(next3, next5));
            ugly[i] = nextUgly;
            if (nextUgly == next2) i2++;
            if (nextUgly == next3) i3++;
            if (nextUgly == next5) i5++;
        }
        return ugly[n - 1];
    }

    public static void main(String[] args) {
        System.out.println(nthUglyNumber(10)); // Output: 12
    }
}

Line Notes

int[] ugly = new int[n];Array to store ugly numbers

ugly[0] = 1;Initialize first ugly number

int i2 = 0, i3 = 0, i5 = 0;Pointers for multiples

int next2 = ugly[i2] * 2;Next multiple of 2 candidate

int nextUgly = Math.min(next2, Math.min(next3, next5));Choose smallest next ugly number

if (nextUgly == next2) i2++;Increment pointer to avoid duplicates

return ugly[n - 1];Return nth ugly number

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int nthUglyNumber(int n) {
    vector<int> ugly(n);
    ugly[0] = 1;
    int i2 = 0, i3 = 0, i5 = 0;
    for (int i = 1; i < n; i++) {
        int next2 = ugly[i2] * 2;
        int next3 = ugly[i3] * 3;
        int next5 = ugly[i5] * 5;
        int nextUgly = min(next2, min(next3, next5));
        ugly[i] = nextUgly;
        if (nextUgly == next2) i2++;
        if (nextUgly == next3) i3++;
        if (nextUgly == next5) i5++;
    }
    return ugly[n - 1];
}

int main() {
    cout << nthUglyNumber(10) << endl; // Output: 12
    return 0;
}

Line Notes

vector<int> ugly(n);Vector to store ugly numbers

ugly[0] = 1;Initialize first ugly number

int i2 = 0, i3 = 0, i5 = 0;Pointers for multiples

int next2 = ugly[i2] * 2;Next candidate multiple of 2

int nextUgly = min(next2, min(next3, next5));Choose smallest next ugly number

if (nextUgly == next2) i2++;Increment pointer to avoid duplicates

return ugly[n - 1];Return nth ugly number

function nthUglyNumber(n) {
    const ugly = new Array(n);
    ugly[0] = 1;
    let i2 = 0, i3 = 0, i5 = 0;
    for (let i = 1; i < n; i++) {
        const next2 = ugly[i2] * 2;
        const next3 = ugly[i3] * 3;
        const next5 = ugly[i5] * 5;
        const nextUgly = Math.min(next2, next3, next5);
        ugly[i] = nextUgly;
        if (nextUgly === next2) i2++;
        if (nextUgly === next3) i3++;
        if (nextUgly === next5) i5++;
    }
    return ugly[n - 1];
}

// Test
console.log(nthUglyNumber(10)); // Output: 12

Line Notes

const ugly = new Array(n);Array to hold ugly numbers

ugly[0] = 1;Initialize first ugly number

let i2 = 0, i3 = 0, i5 = 0;Pointers for multiples

const next2 = ugly[i2] * 2;Next multiple candidate of 2

const nextUgly = Math.min(next2, next3, next5);Choose smallest next ugly number

if (nextUgly === next2) i2++;Increment pointer to avoid duplicates

return ugly[n - 1];Return nth ugly number

⏱

Complexity

TimeO(n)

SpaceO(n)

We generate each ugly number exactly once, and each step involves constant time operations, so total time is O(n). Space is O(n) for the array.

💡 For n=100000, this means 100000 operations, which is efficient and feasible.

Interview Verdict: Accepted / Efficient for large n

This approach is optimal for this problem and commonly expected in interviews.

🧠

Optimal Approach (Using Min-Heap / Priority Queue)

💡 This approach uses a min-heap to always pick the next smallest ugly number, ensuring no duplicates and efficient generation.

Intuition

Start with 1 in a min-heap. Pop the smallest number, then push its multiples by 2, 3, and 5 if not already seen. Repeat until nth number is popped.

Algorithm

1. Initialize a min-heap with 1 and a set to track seen numbers.
2. Pop the smallest number from the heap as the current ugly number.
3. Push its multiples by 2, 3, and 5 into the heap if not already in the set.
4. Repeat until the nth ugly number is popped.
5. Return the last popped number.

💡 The key is to avoid duplicates using a set and efficiently get the next smallest number using a heap.

</>

Code

import heapq

def nthUglyNumber(n):
    heap = [1]
    seen = {1}
    ugly = 1
    for _ in range(n):
        ugly = heapq.heappop(heap)
        for factor in [2, 3, 5]:
            new_ugly = ugly * factor
            if new_ugly not in seen:
                seen.add(new_ugly)
                heapq.heappush(heap, new_ugly)
    return ugly

# Driver code
if __name__ == '__main__':
    print(nthUglyNumber(10))  # Output: 12

Line Notes

heap = [1]Initialize min-heap with first ugly number

seen = {1}Set to avoid duplicates

ugly = heapq.heappop(heap)Pop smallest ugly number

for factor in [2, 3, 5]:Generate next candidates

if new_ugly not in seen:Avoid duplicates

heapq.heappush(heap, new_ugly)Add new candidate to heap

return uglyReturn nth ugly number after n pops

import java.util.*;

public class Solution {
    public static int nthUglyNumber(int n) {
        PriorityQueue<Long> heap = new PriorityQueue<>();
        Set<Long> seen = new HashSet<>();
        heap.offer(1L);
        seen.add(1L);
        long ugly = 1;
        for (int i = 0; i < n; i++) {
            ugly = heap.poll();
            for (int factor : new int[]{2, 3, 5}) {
                long newUgly = ugly * factor;
                if (!seen.contains(newUgly)) {
                    seen.add(newUgly);
                    heap.offer(newUgly);
                }
            }
        }
        return (int) ugly;
    }

    public static void main(String[] args) {
        System.out.println(nthUglyNumber(10)); // Output: 12
    }
}

Line Notes

PriorityQueue<Long> heap = new PriorityQueue<>();Min-heap to get smallest number

Set<Long> seen = new HashSet<>();Set to avoid duplicates

heap.offer(1L);Start with 1

ugly = heap.poll();Pop smallest ugly number

for (int factor : new int[]{2, 3, 5}) {Generate next candidates

if (!seen.contains(newUgly)) {Avoid duplicates

heap.offer(newUgly);Add new candidate to heap

return (int) ugly;Return nth ugly number

#include <iostream>
#include <queue>
#include <unordered_set>
using namespace std;

int nthUglyNumber(int n) {
    priority_queue<long, vector<long>, greater<long>> heap;
    unordered_set<long> seen;
    heap.push(1);
    seen.insert(1);
    long ugly = 1;
    for (int i = 0; i < n; i++) {
        ugly = heap.top();
        heap.pop();
        for (int factor : {2, 3, 5}) {
            long newUgly = ugly * factor;
            if (seen.find(newUgly) == seen.end()) {
                seen.insert(newUgly);
                heap.push(newUgly);
            }
        }
    }
    return (int)ugly;
}

int main() {
    cout << nthUglyNumber(10) << endl; // Output: 12
    return 0;
}

Line Notes

priority_queue<long, vector<long>, greater<long>> heap;Min-heap for smallest number

unordered_set<long> seen;Set to avoid duplicates

heap.push(1);Initialize heap with 1

ugly = heap.top();Get smallest ugly number

for (int factor : {2, 3, 5}) {Generate next candidates

if (seen.find(newUgly) == seen.end()) {Check duplicates

heap.push(newUgly);Add new candidate

return (int)ugly;Return nth ugly number

class MinHeap {
    constructor() {
        this.heap = [];
    }
    swap(i, j) {
        [this.heap[i], this.heap[j]] = [this.heap[j], this.heap[i]];
    }
    push(val) {
        this.heap.push(val);
        this.bubbleUp();
    }
    bubbleUp() {
        let idx = this.heap.length - 1;
        while (idx > 0) {
            let parent = Math.floor((idx - 1) / 2);
            if (this.heap[parent] <= this.heap[idx]) break;
            this.swap(parent, idx);
            idx = parent;
        }
    }
    pop() {
        if (this.heap.length === 1) return this.heap.pop();
        const top = this.heap[0];
        this.heap[0] = this.heap.pop();
        this.bubbleDown();
        return top;
    }
    bubbleDown() {
        let idx = 0;
        const length = this.heap.length;
        while (true) {
            let left = 2 * idx + 1;
            let right = 2 * idx + 2;
            let smallest = idx;
            if (left < length && this.heap[left] < this.heap[smallest]) smallest = left;
            if (right < length && this.heap[right] < this.heap[smallest]) smallest = right;
            if (smallest === idx) break;
            this.swap(idx, smallest);
            idx = smallest;
        }
    }
}

function nthUglyNumber(n) {
    const heap = new MinHeap();
    const seen = new Set();
    heap.push(1);
    seen.add(1);
    let ugly = 1;
    for (let i = 0; i < n; i++) {
        ugly = heap.pop();
        for (const factor of [2, 3, 5]) {
            const newUgly = ugly * factor;
            if (!seen.has(newUgly)) {
                seen.add(newUgly);
                heap.push(newUgly);
            }
        }
    }
    return ugly;
}

// Test
console.log(nthUglyNumber(10)); // Output: 12

Line Notes

class MinHeap {Custom min-heap implementation

heap.push(1);Initialize heap with 1

const seen = new Set();Set to avoid duplicates

ugly = heap.pop();Pop smallest ugly number

for (const factor of [2, 3, 5]) {Generate next candidates

if (!seen.has(newUgly)) {Avoid duplicates

heap.push(newUgly);Add new candidate to heap

return ugly;Return nth ugly number

⏱

Complexity

TimeO(n log n)

SpaceO(n)

Each ugly number is pushed and popped once from the heap, each heap operation is O(log n), so total time is O(n log n). Space is O(n) for heap and set.

💡 For n=100000, this is slower than the pointer approach but still efficient enough.

Interview Verdict: Accepted / Alternative approach

This approach is acceptable but less optimal than the pointer method; useful to demonstrate heap knowledge.

📊

All Approaches - One-Glance Tradeoffs

💡 The pointer approach is the best choice for interviews due to its optimal time and space. Brute force is only for explanation, and heap is a good alternative to show data structure knowledge.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n log n) approx.	O(1)	No	N/A	Mention only - never code
2. Pointer Merge	O(n)	O(n)	No	N/A	Code this approach
3. Min-Heap	O(n log n)	O(n)	No	N/A	Optional alternative

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice coding all approaches, and prepare to explain tradeoffs clearly in interviews.

How to Present

Step 1: Clarify problem constraints and definition of ugly numbers.Step 2: Present brute force approach to show understanding of problem.Step 3: Explain inefficiency and introduce pointer-based merging approach.Step 4: Code the pointer approach carefully, explaining pointers and duplicates.Step 5: Optionally mention heap-based approach if time permits.Step 6: Test with sample and edge cases.

Time Allocation

Clarify: 2min → Approach: 5min → Code: 10min → Test: 3min. Total ~20min

What the Interviewer Tests

Ability to understand prime factorization, optimize naive solutions, use multiple pointers or heaps, and write clean, bug-free code.

Common Follow-ups

• What if primes are different? → Use generalized heap or pointer approach with given primes.
• Can you optimize space? → Use pointer approach with O(n) space; further optimization is limited.

💡 These follow-ups test your ability to generalize and optimize beyond the base problem.

🔍

Pattern Recognition

When to Use

1) Asked for nth number with restricted prime factors, 2) sequence generated by multiples of fixed primes, 3) need to merge sorted sequences, 4) avoid duplicates in merged sequence.

Signature Phrases

'prime factors only include 2, 3, and 5''return the nth ugly number'

NOT This Pattern When

Problems asking for general factorization or arbitrary prime factorization without fixed primes

Similar Problems

Super Ugly Number - generalizes to arbitrary prime setsNth Ugly Number - simpler version with fewer constraintsMerge k Sorted Lists - similar merging technique

Practice

1. What is the time complexity of the optimized Sieve of Eratosthenes algorithm that counts primes less than n by marking multiples of odd primes only, and why?

medium

A. O(n * sqrt(n)) because each number is checked against primes up to sqrt(n)

B. O(n) because each number is visited only once during marking

C. O(n log n) because each prime's multiples are marked up to n

D. O(n log log n) because the sieve marks multiples efficiently up to sqrt(n)

Solution

1. Step 1: Identify outer loop complexity

   The outer loop runs over odd numbers up to sqrt(n), roughly O(sqrt(n)/2).

2. Step 2: Analyze inner marking loop

   Each composite number is marked only once by its smallest prime factor, leading to total marking cost O(n log log n).

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Classic sieve time complexity is O(n log log n) [OK]

Hint: Sieve complexity is famously O(n log log n) [OK]

Common Mistakes:

• Confusing with brute force O(n sqrt(n))
• Assuming linear O(n) due to single visits
• Overestimating to O(n log n)

2. Examine the following buggy Python code intended to compute the LCM safely without overflow:

def gcd_iterative(a, b):
    while b != 0:
        a, b = b, a % b
    return a

def lcm_safe(a, b):
    gcd = gcd_iterative(a, b)
    return a * b // gcd

if __name__ == '__main__':
    a, b = 1000000, 1000000
    print("LCM =", lcm_safe(a, b))

What is the subtle bug in this code?

medium

A. Line 7: Multiplying a and b before dividing by gcd can cause integer overflow.

B. Line 3: The Euclidean algorithm lacks a base case for recursion.

C. Line 9: The input values are not validated for being positive integers.

D. Line 2: The while loop condition should be 'while a != 0' instead of 'while b != 0'.

Solution

1. Step 1: Analyze lcm_safe function

   The function computes LCM as a * b // gcd. Multiplying a and b first can overflow for large inputs before division.

2. Step 2: Identify the bug

   To avoid overflow, division by gcd should happen before multiplication: (a // gcd) * b.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Multiplying large numbers before division risks overflow [OK]

Hint: Divide before multiply to avoid overflow in LCM calculation [OK]

Common Mistakes:

• Ignoring overflow risk in LCM calculation
• Misunderstanding Euclidean algorithm base case

3. Consider this buggy recursive reverse integer code snippet:

def reverse(x: int) -> int:
    INT_MAX = 2**31 - 1
    INT_MIN = -2**31
    sign = -1 if x < 0 else 1
    x = abs(x)

    def helper(num, rev):
        if num == 0:
            return rev
        pop = num % 10
        # Bug: Missing overflow check here
        rev = rev * 10 + pop
        return helper(num // 10, rev)

    rev = helper(x, 0)
    return sign * rev if rev != 0 else 0

What is the root cause of incorrect behavior in this code?

medium

A. Line 'rev = rev * 10 + pop' lacks overflow check before multiplication

B. Line with 'sign = -1 if x < 0 else 1' does not handle zero correctly

C. Line 'pop = num % 10' incorrectly extracts digits for negative numbers

D. Line 'return sign * rev if rev != 0 else 0' returns zero incorrectly for valid reversed zero

Solution

1. Step 1: Identify missing overflow check

   The code multiplies rev by 10 and adds pop without checking if rev will overflow 32-bit integer limits.

2. Step 2: Consequence of missing check

   Without overflow check before multiplication, the reversed integer can exceed INT_MAX causing incorrect results or runtime errors.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Overflow check must precede rev update to avoid errors [OK]

Hint: Overflow check must happen before rev update [OK]

Common Mistakes:

• Ignoring overflow before multiplying rev by 10
• Assuming sign handling fixes overflow
• Misunderstanding digit extraction for negatives

4. Suppose you want to count primes less than n but now the input n can be up to 10^12, too large for a standard sieve. Which approach is best suited to handle this scenario efficiently?

hard

A. Use a segmented sieve that processes ranges of size sqrt(n) to reduce memory usage.

B. Use the standard sieve of Eratosthenes with a larger memory allocation.

C. Use brute force primality checks for each number up to n.

D. Use a dynamic programming approach to store prime counts up to n.

Solution

1. Step 1: Understand memory constraints

   Standard sieve requires O(n) space, which is infeasible for n=10^12.

2. Step 2: Identify suitable algorithm

   Segmented sieve divides the range into manageable blocks, using O(sqrt(n)) memory and still marking primes efficiently.

3. Step 3: Compare alternatives

   Brute force is too slow, DP is not applicable, and standard sieve is too large.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Segmented sieve is standard for large n prime counting [OK]

Hint: Segmented sieve handles large n with limited memory [OK]

Common Mistakes:

• Trying standard sieve for huge n
• Using brute force for large ranges
• Misapplying DP to prime counting

5. Suppose you want to compute the LCM of multiple numbers (more than two). Which of the following approaches correctly extends the iterative Euclidean algorithm and avoids overflow?

hard

A. Multiply all numbers together first, then divide by the GCD of all numbers at the end.

B. Compute LCM pairwise from left to right using lcm_safe(a, b) repeatedly on the list.

C. Use brute force to find the smallest number divisible by all inputs by checking multiples of the max number.

D. Compute GCD of all numbers first, then multiply all numbers and divide by that GCD.

Solution

1. Step 1: Understand LCM of multiple numbers

   LCM(a,b,c) = LCM(LCM(a,b), c). Computing pairwise LCMs iteratively is standard and safe.

2. Step 2: Evaluate options

   Multiplying all numbers first risks overflow. Brute force is inefficient. Computing GCD of all and dividing product is incorrect because GCD of all does not directly give LCM.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Pairwise LCM with overflow-safe method is correct and efficient [OK]

Hint: Compute LCM pairwise to avoid overflow and maintain correctness [OK]

Common Mistakes:

• Multiplying all numbers first causing overflow
• Misusing GCD of all numbers to compute LCM