Ugly Number II (Merge 3 Sequences)

🎯

Ugly Number II (Merge 3 Sequences)

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Imagine you want to generate a sequence of numbers whose only prime factors are 2, 3, or 5, like a special recipe that only allows these ingredients. Finding the nth such number efficiently is the challenge.

💡 This problem is about generating a sequence defined by multiplicative factors. Beginners often struggle because the naive approach tries to check every number, which is inefficient. Understanding how to merge multiple sequences or use heaps to generate the sequence efficiently is key.

📋

Problem Statement

An ugly number is a positive integer whose prime factors only include 2, 3, and 5. Given an integer n, return the nth ugly number. Input: Integer n Output: Integer representing the nth ugly number

1 ≤ n ≤ 10^5The answer is guaranteed to fit in a 32-bit signed integer

💡

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 multiple of 2)
• n = 15 → output 24 (checks correctness beyond first 10)
• n = 100000 → output fits in 32-bit integer (tests efficiency and overflow handling)

⚠️

Common Mistakes

Not handling duplicates when multiple pointers produce the same number

Sequence contains duplicates, causing incorrect nth number

✅ Advance all pointers whose multiples equal the chosen minimum

Using brute force for large n without optimization

Code times out or runs too slowly

✅ Switch to three-pointer or heap approach

Forgetting to initialize dp[0] = 1

Sequence starts incorrectly, leading to wrong output

✅ Explicitly set dp[0] = 1 before loops

Not using a set to avoid duplicates in heap approach

Heap grows unnecessarily large, causing performance issues

✅ Use a set to track seen numbers before pushing into heap

🧠

Brute Force (Check Every Number)

💡 This approach exists to build intuition by directly checking each number for the ugly property. It is simple but inefficient, highlighting why optimization is necessary.

Intuition

Start from 1 and check every number incrementally to see if it only has prime factors 2, 3, and 5. Count how many ugly numbers have been found until reaching n.

Algorithm

1. Initialize count = 1 (since 1 is ugly) and current number = 1
2. While count < n, increment current number by 1
3. Check if current number is ugly by dividing out factors 2, 3, and 5 repeatedly
4. If after division the number is 1, increment count
5. Return the current number when count == n

💡 The main difficulty is understanding how to check if a number is ugly by factor division and why this approach is slow for large n.

</>

Code

def is_ugly(num):
    for p in [2, 3, 5]:
        while num > 0 and 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))  # Expected output: 12

Line Notes

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

while num > 0 and num % p == 0:Repeatedly divides out prime factor p to test if only these factors remain; order ensures num > 0 before modulo

return num == 1If after division the number is 1, it is ugly

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

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

if is_ugly(current):Check if current number is ugly before incrementing count

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

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

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

Line Notes

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

while (num > 0 && num % p == 0) {Divide out prime factors repeatedly; order ensures num > 0 before modulo

return num == 1;If reduced to 1, number is ugly

int count = 1; // 1 is uglyInitialize count with 1 since 1 is ugly

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

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

#include <iostream>
using namespace std;

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

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

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

Line Notes

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

while (num > 0 && num % p == 0) {Divide out prime factors 2,3,5; order ensures num > 0 before modulo

return num == 1;If reduced to 1, number is ugly

int count = 1; // 1 is uglyStart counting from 1

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

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

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

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

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

Line Notes

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

while (num > 0 && num % p === 0) {Divide out prime factors repeatedly; order ensures num > 0 before modulo

return num === 1;If reduced to 1, number is ugly

let count = 1; // 1 is uglyInitialize count with 1

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

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

⏱

Complexity

TimeO(n * log n) approximately, since each number requires factorization which can be up to O(log n) in worst case

SpaceO(1) extra space

For each number, we divide out factors 2, 3, and 5 repeatedly which takes O(log n) in worst case (number of divisions proportional to log of the number). We do this until we find n ugly numbers, so roughly O(n log n).

💡 For n=20, this means checking about 20 numbers, each with multiple divisions, which is manageable but grows quickly for large n.

Interview Verdict: TLE for large n

This approach is too slow for large inputs, showing the need for more efficient methods.

🧠

Three Pointer Merge (Dynamic Programming Style)

💡 This approach introduces the idea of merging three sorted sequences generated by multiplying previous ugly numbers by 2, 3, and 5. It is more efficient and a common interview solution.

Intuition

Maintain three pointers for multiples of 2, 3, and 5. At each step, pick the smallest next candidate ugly number from these sequences and advance pointers accordingly.

Algorithm

1. Initialize an array dp with dp[0] = 1 as the first ugly number
2. Initialize three pointers i2, i3, i5 to 0
3. For each position from 1 to n-1, compute next multiples: dp[i2]*2, dp[i3]*3, dp[i5]*5
4. Choose the minimum of these as dp[i]
5. Increment the pointer(s) whose multiple matched the minimum
6. Return dp[n-1]

💡 The challenge is to correctly update pointers to avoid duplicates and maintain sorted order.

</>

Code

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

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

Line Notes

dp = [0] * nCreate array to store ugly numbers

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

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

next2, next3, next5 = dp[i2]*2, dp[i3]*3, dp[i5]*5Calculate next candidates from each pointer

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

if dp[i] == next2: i2 += 1Advance pointer if its candidate was chosen

if dp[i] == next3: i3 += 1Advance pointer if its candidate was chosen

if dp[i] == next5: i5 += 1Advance pointer if its candidate was chosen

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

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

Line Notes

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

dp[0] = 1;Initialize first ugly number

int i2 = 0, i3 = 0, i5 = 0;Pointers for multiples of 2, 3, 5

int next2 = dp[i2] * 2;Next candidate from 2's sequence

dp[i] = Math.min(next2, Math.min(next3, next5));Pick smallest candidate

if (dp[i] == next2) i2++;Advance pointer if chosen

if (dp[i] == next3) i3++;Advance pointer if chosen

if (dp[i] == next5) i5++;Advance pointer if chosen

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

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

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

Line Notes

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

dp[0] = 1;Initialize first ugly number

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

int next2 = dp[i2] * 2;Next candidate from 2's sequence

dp[i] = min({next2, next3, next5});Pick smallest candidate

if (dp[i] == next2) i2++;Advance pointer if chosen

if (dp[i] == next3) i3++;Advance pointer if chosen

if (dp[i] == next5) i5++;Advance pointer if chosen

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

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

Line Notes

const dp = new Array(n).fill(0);Array to store ugly numbers

dp[0] = 1;Initialize first ugly number

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

const next2 = dp[i2] * 2;Next candidate from 2's sequence

dp[i] = Math.min(next2, next3, next5);Pick smallest candidate

if (dp[i] === next2) i2++;Advance pointer if chosen

if (dp[i] === next3) i3++;Advance pointer if chosen

if (dp[i] === next5) i5++;Advance pointer if chosen

⏱

Complexity

TimeO(n)

SpaceO(n)

We generate each ugly number once, and each step involves constant time operations with three pointers.

💡 For n=100000, this means 100000 iterations, which is efficient and feasible in typical interview time constraints.

Interview Verdict: Accepted

This approach is efficient and commonly accepted in interviews for this problem.

🧠

Min-Heap Based Merge

💡 This approach uses a min-heap to always pick the next smallest ugly number candidate from the sequences generated by multiplying by 2, 3, and 5. It is intuitive and generalizes well to k-way merges.

Intuition

Start with 1 in a min-heap. At each step, pop the smallest number, add its multiples by 2, 3, and 5 back into the heap if not seen before, until the nth ugly number is popped.

Algorithm

1. Initialize a min-heap and a set to track seen numbers
2. Push 1 into the heap and mark it seen
3. Repeat n times: pop the smallest number from the heap
4. For each factor in [2,3,5], multiply popped number and if not seen, push into heap and mark seen
5. Return the last popped number

💡 The main challenge is managing duplicates and ensuring efficient insertion and extraction from the heap.

</>

Code

import heapq

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

# Driver code
if __name__ == '__main__':
    print(nthUglyNumber(10))  # Expected 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 candidate

for f in [2, 3, 5]:Generate next candidates by multiplying factors

if nxt not in seen:Avoid duplicates before pushing

heapq.heappush(heap, nxt)Push new candidate into heap

import java.util.*;

public class UglyNumberHeap {
    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;
        int[] factors = {2, 3, 5};
        for (int i = 0; i < n; i++) {
            ugly = heap.poll();
            for (int f : factors) {
                long nxt = ugly * f;
                if (!seen.contains(nxt)) {
                    seen.add(nxt);
                    heap.offer(nxt);
                }
            }
        }
        return (int) ugly;
    }

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

Line Notes

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

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

heap.offer(1L);Start with 1

ugly = heap.poll();Pop smallest number

for (int f : factors) {Generate next candidates

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

heap.offer(nxt);Push new candidate

#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;
    int factors[3] = {2, 3, 5};
    for (int i = 0; i < n; i++) {
        ugly = heap.top();
        heap.pop();
        for (int f : factors) {
            long nxt = ugly * f;
            if (seen.find(nxt) == seen.end()) {
                seen.insert(nxt);
                heap.push(nxt);
            }
        }
    }
    return (int)ugly;
}

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

Line Notes

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

unordered_set<long> seen;Track duplicates

heap.push(1);Start with 1

ugly = heap.top();Pop smallest number

for (int f : factors) {Generate next candidates

if (seen.find(nxt) == seen.end()) {Avoid duplicates

heap.push(nxt);Push new candidate

class MinHeap {
    constructor() {
        this.heap = [];
    }
    push(val) {
        this.heap.push(val);
        this.bubbleUp();
    }
    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;
    }
    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.heap[parent], this.heap[idx]] = [this.heap[idx], this.heap[parent]];
            idx = parent;
        }
    }
    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.heap[smallest], this.heap[idx]] = [this.heap[idx], this.heap[smallest]];
            idx = smallest;
        }
    }
}

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

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

Line Notes

class MinHeap {Custom min-heap implementation

heap.push(1);Initialize heap with 1

seen.add(1);Track duplicates

ugly = heap.pop();Pop smallest number

for (const f of factors) {Generate next candidates

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

heap.push(nxt);Push new candidate

⏱

Complexity

TimeO(n log n)

SpaceO(n)

Each of the n iterations involves heap operations which take O(log n). The set ensures no duplicates are pushed.

💡 For n=100000, this approach is slower than the three-pointer method but still acceptable in many cases.

Interview Verdict: Accepted

This approach is accepted and demonstrates knowledge of heaps and duplicate management.

📊

All Approaches - One-Glance Tradeoffs

💡 In interviews, the three-pointer merge approach is the best balance of clarity and efficiency and should be coded unless asked otherwise.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n log n)	O(1)	No	N/A	Mention only - never code due to inefficiency
2. Three Pointer Merge	O(n)	O(n)	No	Yes	Code this approach for best performance and clarity
3. Min-Heap Based Merge	O(n log n)	O(n)	No	Yes	Mention or code if asked about general k-way merges

💼

Interview Strategy

💡 Use this guide to understand the problem deeply before interviews. Start by clarifying the problem, then explain the brute force approach to show understanding. Progress to the efficient three-pointer method, and optionally mention the heap approach to demonstrate breadth.

How to Present

Step 1: Clarify the definition of ugly numbers and input/output formatStep 2: Describe the brute force approach and its inefficiencyStep 3: Introduce the three-pointer merge method as an efficient solutionStep 4: Optionally discuss the heap-based approach for generalizationStep 5: Code the three-pointer method with clear pointer updatesStep 6: Test with sample inputs and edge cases

Time Allocation

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

What the Interviewer Tests

The interviewer tests your ability to recognize the sequence generation pattern, optimize naive solutions, and implement pointer or heap-based merges correctly without duplicates.

Common Follow-ups

• What if the prime factors change? → Use heap or generalized pointers
• How to handle very large n? → Use 64-bit integers or optimized data structures

💡 These follow-ups test your ability to generalize the solution and handle edge cases like overflow or different factor sets.

🔍

Pattern Recognition

When to Use

1) Problem asks for nth number with prime factors from a fixed set; 2) Sequence can be generated by merging multiple sorted sequences; 3) Need to avoid duplicates; 4) Efficient generation required for large n.

Signature Phrases

'prime factors only 2, 3, and 5''find the nth ugly number''merge sequences generated by multiples'

NOT This Pattern When

Problems that ask for counting numbers with factors but do not require sequence generation or merging.

Similar Problems

Super Ugly Number - generalizes to arbitrary prime factorsMerge k Sorted Lists - similar k-way merge techniqueNth Magical Number - similar factor-based sequence generation

Practice

1. You are given two sorted arrays and need to find the k pairs with the smallest sums formed by taking one element from each array. Which approach guarantees an optimal solution with efficient time complexity?

easy

A. Use dynamic programming to build a table of sums and select the smallest pairs from it.

B. Use a brute force approach by enumerating all pairs, then sorting them to find the smallest sums.

C. Use a min-heap to lazily generate pairs starting from the smallest possible sums, expanding only as needed.

D. Use a greedy approach by always pairing the smallest elements from each array without backtracking.

Solution

1. Step 1: Understand problem constraints

   The problem requires finding the k smallest pairs by sum from two sorted arrays efficiently.

2. Step 2: Evaluate approaches

   Brute force enumerates all pairs, which is inefficient. DP is not suitable since pairs are not independent subproblems. Greedy fails because pairing smallest elements once misses other smaller sums. The min-heap approach lazily expands pairs starting from the smallest sums, ensuring optimality and efficiency.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Min-heap approach matches the known optimal solution [OK]

Hint: Min-heap lazily expands pairs for optimal efficiency [OK]

Common Mistakes:

• Assuming greedy pairing always works
• Thinking DP applies directly
• Believing brute force is efficient enough

2. Given the following code snippet, what is the final output when called with nums1 = [1,7,11], nums2 = [2,4,6], and k = 3?

def countPairs(nums1, nums2, mid):
    count = 0
    j = len(nums2) - 1
    for i in range(len(nums1)):
        while j >= 0 and nums1[i] + nums2[j] > mid:
            j -= 1
        count += (j + 1)
    return count

def kthSmallestSum(nums1, nums2, k):
    low = nums1[0] + nums2[0]
    high = nums1[-1] + nums2[-1]
    while low < high:
        mid = (low + high) // 2
        if countPairs(nums1, nums2, mid) < k:
            low = mid + 1
        else:
            high = mid
    return low

print(kthSmallestSum(nums1, nums2, k))

easy

A. 6

B. 7

C. 8

D. 9

Solution

1. Step 1: Trace binary search bounds

   low=3 (1+2), high=17 (11+6). Mid values checked: 10, 6, 4, 3, 7.

2. Step 2: Count pairs ≤ mid

   For mid=7, countPairs returns 3 pairs ≤7, which matches k=3, so high=7.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Output matches expected k-th smallest sum [OK]

Hint: Binary search narrows sum range to find k-th smallest sum [OK]

Common Mistakes:

• Off-by-one in binary search bounds
• Miscounting pairs in countPairs

3. Consider the following Python code implementing the optimal approach for the smallest range covering k lists. Given the input nums = [[4,10,15,24,26],[0,9,12,20],[5,18,22,30]], what is the final returned range?

easy

A. [20, 24]

B. [5, 20]

C. [0, 24]

D. [4, 20]

Solution

1. Step 1: Initialize heap and current_max

   Heap contains (4,0,0), (0,1,0), (5,2,0); current_max = 5.

2. Step 2: Iterate popping min and pushing next elements

   Pop (0,1,0), push (9,1,1), current_max=9; range = 9-0=9. Pop (4,0,0), push (10,0,1), current_max=10; range=10-4=6. Pop (5,2,0), push (18,2,1), current_max=18; range=18-5=13. Pop (9,1,1), push (12,1,2), current_max=18; range=18-9=9. Pop (10,0,1), push (15,0,2), current_max=18; range=18-10=8. Pop (12,1,2), push (20,1,3), current_max=20; range=20-12=8. Pop (15,0,2), push (24,0,3), current_max=24; range=24-15=9. Pop (18,2,1), push (22,2,2), current_max=24; range=24-18=6. Pop (20,1,3), no next element, break.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Final minimal range is [4,20] covering all lists [OK]

Hint: Track min and max updates carefully [OK]

Common Mistakes:

• Forgetting to update current_max after pushing new element
• Stopping too early before all lists exhausted
• Miscomputing range after each pop

4. The following code attempts to find the smallest range covering k sorted lists using a min-heap. Identify the line containing the subtle bug that can cause incorrect results or runtime errors.

medium

A. Line popping the smallest element from the heap

B. Line checking if ele_idx + 1 == len(nums[list_idx]) to break the loop

C. Line updating current_max after pushing next_val into the heap

D. Line initializing min_range to float('inf')

Solution

1. Step 1: Identify the role of current_max

   current_max tracks the maximum element in the current window; it must be updated when pushing a new element.

2. Step 2: Locate missing update

   The code comment indicates current_max is not updated after pushing next_val, causing incorrect range calculation.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Without updating current_max, range may be underestimated or incorrect [OK]

Hint: Always update current_max after pushing new heap element [OK]

Common Mistakes:

• Forgetting to update current_max after pushing new element
• Breaking loop too early or too late
• Misusing heap pop or initialization

5. Suppose the problem changes so that elements from the arrays can be reused multiple times to form sums (i.e., pairs can include the same element multiple times). Which modification to the binary search approach is necessary to correctly find the k-th smallest sum?

hard

A. No modification needed; reuse does not affect counting pairs.

B. Use a min-heap approach with a visited set to handle duplicates and reuse correctly.

C. Modify countPairs to count pairs with replacement by resetting pointer j for each i.

D. Use binary search but countPairs must be replaced with a formula counting sums with repetition.

Solution

1. Step 1: Understand reuse effect on counting

   With reuse allowed, pairs are not limited to unique indices; counting must consider all j for each i.

2. Step 2: Modify countPairs

   Resetting j for each i allows counting all pairs with sums ≤ mid including repeated elements.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Resetting j ensures correct counting with reuse [OK]

Hint: Reusing elements breaks monotonic pointer j; must reset per i [OK]

Common Mistakes:

• Assuming original countPairs works with reuse
• Using heap without visited set causes duplicates