Reorganize String (No Two Adjacent Same)

🎯

Reorganize String (No Two Adjacent Same)

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Imagine you are organizing a playlist where no two songs by the same artist should play consecutively to keep the audience engaged.

💡 This problem is about rearranging characters so that no two identical ones are adjacent. Beginners often struggle because they try naive rearrangements without considering frequency or the need to separate the most frequent characters carefully.

📋

Problem Statement

Given a string s, rearrange the characters of s so that any two adjacent characters are not the same. If such an arrangement is not possible, return an empty string. Otherwise, return any valid rearrangement.

1 ≤ s.length ≤ 10^5s consists of lowercase English letters only

💡

Example

Input"aab"

Outputaba

One possible rearrangement is 'aba' where no two adjacent characters are the same.

Input"aaab"

Output

No valid rearrangement exists because 'a' occurs too frequently.

• Single character string → always valid, output same character
• All characters are the same → no valid rearrangement, output empty string
• String with two characters both same → no valid rearrangement, output empty string
• String with two characters different → valid rearrangement is original or swapped

⚠️

Common Mistakes

Not checking if the most frequent character count exceeds (n+1)/2

Returns invalid rearrangement or crashes due to impossible placement

✅ Add a feasibility check before attempting rearrangement

Placing characters without alternating indices leading to adjacent duplicates

Output string has adjacent identical characters, failing problem requirements

✅ Place characters first at even indices, then odd indices to separate duplicates

In heap approach, forgetting to hold back the previously used character

Same character may be placed consecutively, violating constraints

✅ Store previous character and frequency, push back into heap only after placing a different character

Using brute force or generating all permutations for large inputs

Time limit exceeded or program crashes due to combinatorial explosion

✅ Use greedy or heap-based approaches for efficiency

Not handling edge cases like single character or all identical characters

Incorrect output or runtime errors

✅ Add explicit checks and test edge cases thoroughly

🧠

Brute Force (Backtracking / Permutation Generation)

💡 This approach exists to understand the problem deeply by trying all possible rearrangements. It is impractical for large inputs but helps grasp the constraints and why smarter methods are needed.

Intuition

Try every possible permutation of the string and check if any permutation has no two adjacent identical characters.

Algorithm

1. Generate all permutations of the input string.
2. For each permutation, check if any two adjacent characters are the same.
3. If a valid permutation is found, return it immediately.
4. If no valid permutation exists, return an empty string.

💡 The main challenge is generating permutations efficiently and checking adjacency, which is straightforward but computationally expensive.

</>

Code

from itertools import permutations

def reorganizeString(s: str) -> str:
    for perm in permutations(s):
        if all(perm[i] != perm[i+1] for i in range(len(perm)-1)):
            return ''.join(perm)
    return ""

# Driver code
if __name__ == "__main__":
    test_cases = ["aab", "aaab", "abc", "a"]
    for s in test_cases:
        print(f"Input: {s} -> Output: {reorganizeString(s)}")

Line Notes

from itertools import permutationsImport permutations to generate all possible orderings of characters

for perm in permutations(s):Try every possible arrangement of characters

if all(perm[i] != perm[i+1] for i in range(len(perm)-1))Check if no two adjacent characters are the same

return ''.join(perm)Return the first valid permutation found

return ""If no valid permutation exists, return empty string

import java.util.*;

public class ReorganizeString {
    public static String reorganizeString(String s) {
        List<String> permutations = new ArrayList<>();
        permute(s.toCharArray(), 0, permutations);
        for (String perm : permutations) {
            if (isValid(perm)) return perm;
        }
        return "";
    }

    private static void permute(char[] arr, int l, List<String> res) {
        if (l == arr.length) {
            res.add(new String(arr));
            return;
        }
        for (int i = l; i < arr.length; i++) {
            swap(arr, l, i);
            permute(arr, l + 1, res);
            swap(arr, l, i);
        }
    }

    private static boolean isValid(String s) {
        for (int i = 0; i < s.length() - 1; i++) {
            if (s.charAt(i) == s.charAt(i + 1)) return false;
        }
        return true;
    }

    private static void swap(char[] arr, int i, int j) {
        char temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }

    public static void main(String[] args) {
        String[] tests = {"aab", "aaab", "abc", "a"};
        for (String s : tests) {
            System.out.println("Input: " + s + " -> Output: " + reorganizeString(s));
        }
    }
}

Line Notes

permute(char[] arr, int l, List<String> res)Generate all permutations by swapping characters recursively

if (l == arr.length)Base case: one permutation generated, add to results

swap(arr, l, i)Backtrack by swapping back to restore original array

permute(arr, l + 1, res)Recursively generate permutations for next position

isValid(String s)Check if no two adjacent characters are the same

return ""Return empty string if no valid permutation found

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

bool isValid(const string& s) {
    for (int i = 0; i < (int)s.size() - 1; i++) {
        if (s[i] == s[i + 1]) return false;
    }
    return true;
}

void permute(string& s, int l, vector<string>& res) {
    if (l == (int)s.size()) {
        res.push_back(s);
        return;
    }
    for (int i = l; i < (int)s.size(); i++) {
        swap(s[l], s[i]);
        permute(s, l + 1, res);
        swap(s[l], s[i]);
    }
}

string reorganizeString(string s) {
    vector<string> permutations;
    permute(s, 0, permutations);
    for (auto& perm : permutations) {
        if (isValid(perm)) return perm;
    }
    return "";
}

int main() {
    vector<string> tests = {"aab", "aaab", "abc", "a"};
    for (auto& s : tests) {
        cout << "Input: " << s << " -> Output: " << reorganizeString(s) << endl;
    }
    return 0;
}

Line Notes

void permute(string& s, int l, vector<string>& res)Recursively generate all permutations by swapping characters

if (l == (int)s.size())Base case: add current permutation to results

swap(s[l], s[i])Backtrack by swapping back to restore original string

permute(s, l + 1, res)Recursively generate permutations for next position

isValid(const string& s)Check if no two adjacent characters are the same

return ""Return empty string if no valid permutation found

function reorganizeString(s) {
    const results = [];
    const arr = s.split('');

    function permute(l) {
        if (l === arr.length) {
            results.push(arr.join(''));
            return;
        }
        for (let i = l; i < arr.length; i++) {
            [arr[l], arr[i]] = [arr[i], arr[l]];
            permute(l + 1);
            [arr[l], arr[i]] = [arr[i], arr[l]];
        }
    }

    function isValid(str) {
        for (let i = 0; i < str.length - 1; i++) {
            if (str[i] === str[i + 1]) return false;
        }
        return true;
    }

    permute(0);
    for (const perm of results) {
        if (isValid(perm)) return perm;
    }
    return "";
}

// Test cases
const tests = ["aab", "aaab", "abc", "a"];
tests.forEach(s => console.log(`Input: ${s} -> Output: ${reorganizeString(s)}`));

Line Notes

function permute(l)Generate all permutations recursively by swapping characters

if (l === arr.length)Base case: add current permutation to results

[arr[l], arr[i]] = [arr[i], arr[l]]Backtrack by swapping back to restore original array

permute(l + 1)Recursively generate permutations for next position

function isValid(str)Check if no two adjacent characters are the same

return ""Return empty string if no valid permutation found

⏱

Complexity

TimeO(n! * n)

SpaceO(n! * n)

Generating all permutations is O(n!) and checking adjacency is O(n) per permutation, making this approach infeasible for large n.

💡 For n=10, this means over 3 million permutations, each checked for adjacency, which is too slow for interviews.

Interview Verdict: TLE

This approach is too slow for large inputs but is useful to understand the problem and motivate better solutions.

🧠

Greedy Approach Using Sorting and Two Pointers

💡 This approach improves efficiency by sorting characters by frequency and placing the most frequent characters first, then filling gaps with others to avoid adjacency.

Intuition

Sort characters by frequency descending, place the most frequent characters at even indices first, then fill remaining characters in the gaps to avoid adjacency.

Algorithm

1. Count frequency of each character.
2. Sort characters by descending frequency.
3. Place the most frequent characters at even indices (0, 2, 4, ...).
4. Place remaining characters at odd indices (1, 3, 5, ...).
5. If any character frequency is more than (n+1)/2, return empty string.

💡 The key insight is that the most frequent character cannot occupy more than half the string rounded up, or adjacency is unavoidable.

</>

Code

def reorganizeString(s: str) -> str:
    from collections import Counter
    n = len(s)
    freq = Counter(s)
    max_freq = max(freq.values())
    if max_freq > (n + 1) // 2:
        return ""
    sorted_chars = sorted(freq.items(), key=lambda x: -x[1])
    res = [None] * n
    idx = 0
    for ch, count in sorted_chars:
        for _ in range(count):
            res[idx] = ch
            idx += 2
            if idx >= n:
                idx = 1
    return "".join(res)

# Driver code
if __name__ == "__main__":
    tests = ["aab", "aaab", "abc", "a"]
    for s in tests:
        print(f"Input: {s} -> Output: {reorganizeString(s)}")

Line Notes

from collections import CounterUse Counter to efficiently count character frequencies

max_freq = max(freq.values())Find the highest frequency to check feasibility

if max_freq > (n + 1) // 2If any character is too frequent, no valid rearrangement exists

sorted_chars = sorted(freq.items(), key=lambda x: -x[1])Sort characters by descending frequency

res = [None] * nPrepare result array to place characters

idx = 0Start placing characters at even indices

for ch, count in sorted_charsPlace each character count times

if idx >= n: idx = 1Switch to odd indices after filling even positions

import java.util.*;

public class ReorganizeString {
    public static String reorganizeString(String s) {
        int n = s.length();
        int[] freq = new int[26];
        for (char c : s.toCharArray()) freq[c - 'a']++;
        int maxFreq = 0;
        for (int f : freq) maxFreq = Math.max(maxFreq, f);
        if (maxFreq > (n + 1) / 2) return "";

        Character[] chars = new Character[26];
        for (int i = 0; i < 26; i++) chars[i] = (char) (i + 'a');
        Arrays.sort(chars, (a, b) -> freq[b - 'a'] - freq[a - 'a']);

        char[] res = new char[n];
        int idx = 0;
        for (char ch : chars) {
            int count = freq[ch - 'a'];
            for (int i = 0; i < count; i++) {
                if (idx >= n) idx = 1;
                res[idx] = ch;
                idx += 2;
            }
        }
        return new String(res);
    }

    public static void main(String[] args) {
        String[] tests = {"aab", "aaab", "abc", "a"};
        for (String s : tests) {
            System.out.println("Input: " + s + " -> Output: " + reorganizeString(s));
        }
    }
}

Line Notes

int[] freq = new int[26]Frequency array for lowercase letters

if (maxFreq > (n + 1) / 2) return ""Check if rearrangement is possible

Arrays.sort(chars, (a, b) -> freq[b - 'a'] - freq[a - 'a'])Sort characters by descending frequency

char[] res = new char[n]Result array to place characters

int idx = 0Start placing characters at even indices

for (char ch : chars)Iterate over characters sorted by frequency

if (idx >= n) idx = 1Switch from even to odd indices when even positions filled

res[idx] = chPlace character at current index

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

string reorganizeString(string s) {
    int n = s.size();
    vector<int> freq(26, 0);
    for (char c : s) freq[c - 'a']++;
    int maxFreq = *max_element(freq.begin(), freq.end());
    if (maxFreq > (n + 1) / 2) return "";

    vector<char> chars(26);
    for (int i = 0; i < 26; i++) chars[i] = 'a' + i;
    sort(chars.begin(), chars.end(), [&](char a, char b) {
        return freq[a - 'a'] > freq[b - 'a'];
    });

    string res(n, ' ');
    int idx = 0;
    for (char ch : chars) {
        int count = freq[ch - 'a'];
        for (int i = 0; i < count; i++) {
            if (idx >= n) idx = 1;
            res[idx] = ch;
            idx += 2;
        }
    }
    return res;
}

int main() {
    vector<string> tests = {"aab", "aaab", "abc", "a"};
    for (auto& s : tests) {
        cout << "Input: " << s << " -> Output: " << reorganizeString(s) << endl;
    }
    return 0;
}

Line Notes

vector<int> freq(26, 0)Frequency vector for all lowercase letters

if (maxFreq > (n + 1) / 2) return ""Check feasibility of rearrangement

sort(chars.begin(), chars.end(), ... )Sort characters by descending frequency

string res(n, ' ')Initialize result string with spaces

int idx = 0Start placing characters at even indices

if (idx >= n) idx = 1Switch from even to odd indices after filling even positions

res[idx] = chPlace character at current index

function reorganizeString(s) {
    const n = s.length;
    const freq = new Array(26).fill(0);
    for (const ch of s) freq[ch.charCodeAt(0) - 97]++;
    const maxFreq = Math.max(...freq);
    if (maxFreq > Math.floor((n + 1) / 2)) return "";

    const chars = Array.from({ length: 26 }, (_, i) => String.fromCharCode(i + 97));
    chars.sort((a, b) => freq[b.charCodeAt(0) - 97] - freq[a.charCodeAt(0) - 97]);

    const res = new Array(n);
    let idx = 0;
    for (const ch of chars) {
        let count = freq[ch.charCodeAt(0) - 97];
        for (let i = 0; i < count; i++) {
            if (idx >= n) idx = 1;
            res[idx] = ch;
            idx += 2;
        }
    }
    return res.join('');
}

// Test cases
const tests = ["aab", "aaab", "abc", "a"];
tests.forEach(s => console.log(`Input: ${s} -> Output: ${reorganizeString(s)}`));

Line Notes

const freq = new Array(26).fill(0)Frequency array for lowercase letters

if (maxFreq > Math.floor((n + 1) / 2)) return ""Check if rearrangement is possible

chars.sort((a, b) => freq[b.charCodeAt(0) - 97] - freq[a.charCodeAt(0) - 97])Sort characters by descending frequency

let idx = 0Start placing characters at even indices

if (idx >= n) idx = 1Switch from even to odd indices after filling even positions

res[idx] = chPlace character at current index

⏱

Complexity

TimeO(n log n)

SpaceO(n)

Counting frequencies is O(n), sorting 26 letters is O(1) practically, and placing characters is O(n). Overall efficient for large inputs.

💡 For n=10^5, this approach runs quickly because sorting is on fixed alphabet size and placement is linear.

Interview Verdict: Accepted

This approach is efficient and accepted in interviews; it balances simplicity and performance well.

🧠

Greedy Approach Using Max Heap (Priority Queue)

💡 This approach uses a max heap to always pick the character with the highest remaining frequency, ensuring no two adjacent characters are the same by temporarily holding the last used character.

Intuition

Use a max heap to pick the most frequent character available, append it to the result, then hold it aside until the next character is placed to avoid adjacency.

Algorithm

1. Count frequency of each character.
2. Build a max heap of (frequency, character).
3. Pop the top character and append it to the result.
4. Hold the previously used character aside if it still has remaining frequency.
5. Repeat until heap is empty.
6. If the result length equals input length, return result; else return empty string.

💡 The key is to never reuse the same character immediately by holding it until another character is placed.

</>

Code

import heapq
from collections import Counter

def reorganizeString(s: str) -> str:
    freq = Counter(s)
    max_heap = [(-count, ch) for ch, count in freq.items()]
    heapq.heapify(max_heap)
    prev_count, prev_char = 0, ''
    result = []

    while max_heap:
        count, ch = heapq.heappop(max_heap)
        result.append(ch)
        if prev_count < 0:
            heapq.heappush(max_heap, (prev_count, prev_char))
        prev_count, prev_char = count + 1, ch

    res_str = ''.join(result)
    if len(res_str) != len(s):
        return ""
    return res_str

# Driver code
if __name__ == "__main__":
    tests = ["aab", "aaab", "abc", "a"]
    for s in tests:
        print(f"Input: {s} -> Output: {reorganizeString(s)}")

Line Notes

freq = Counter(s)Count frequency of each character

max_heap = [(-count, ch) for ch, count in freq.items()]Create max heap by negating counts

heapq.heapify(max_heap)Transform list into a heap structure

prev_count, prev_char = 0, ''Track previously used character and its remaining count

while max_heap:Process characters until none left

count, ch = heapq.heappop(max_heap)Pop character with highest frequency

result.append(ch)Append chosen character to result

if prev_count < 0: heapq.heappush(max_heap, (prev_count, prev_char))Push back previous character if still available

prev_count, prev_char = count + 1, chUpdate previous character with decremented count

res_str = ''.join(result)Convert list of characters to string

if len(res_str) != len(s)Check if rearrangement is complete and valid

import java.util.*;

public class ReorganizeString {
    public static String reorganizeString(String s) {
        int[] freq = new int[26];
        for (char c : s.toCharArray()) freq[c - 'a']++;

        PriorityQueue<int[]> maxHeap = new PriorityQueue<>((a, b) -> b[0] - a[0]);
        for (int i = 0; i < 26; i++) {
            if (freq[i] > 0) maxHeap.offer(new int[]{freq[i], i});
        }

        StringBuilder result = new StringBuilder();
        int[] prev = {0, -1}; // frequency, char index

        while (!maxHeap.isEmpty()) {
            int[] curr = maxHeap.poll();
            result.append((char) (curr[1] + 'a'));
            curr[0]--;
            if (prev[0] > 0) maxHeap.offer(prev);
            prev = curr;
        }

        if (result.length() != s.length()) return "";
        return result.toString();
    }

    public static void main(String[] args) {
        String[] tests = {"aab", "aaab", "abc", "a"};
        for (String s : tests) {
            System.out.println("Input: " + s + " -> Output: " + reorganizeString(s));
        }
    }
}

Line Notes

int[] freq = new int[26]Frequency array for characters

PriorityQueue<int[]> maxHeap = new PriorityQueue<>((a, b) -> b[0] - a[0])Max heap based on frequency

if (freq[i] > 0) maxHeap.offer(new int[]{freq[i], i})Add characters with positive frequency

StringBuilder result = new StringBuilder()Build result string efficiently

int[] prev = {0, -1}Track previously used character and remaining frequency

while (!maxHeap.isEmpty())Process characters until heap is empty

int[] curr = maxHeap.poll()Pop character with highest frequency

result.append((char) (curr[1] + 'a'))Append character to result

curr[0]--Decrement frequency after use

if (prev[0] > 0) maxHeap.offer(prev)Push back previous character if still available

prev = currUpdate previous character to current

if (result.length() != s.length()) return ""Check if rearrangement is valid

#include <iostream>
#include <string>
#include <queue>
#include <vector>

using namespace std;

string reorganizeString(string s) {
    vector<int> freq(26, 0);
    for (char c : s) freq[c - 'a']++;

    priority_queue<pair<int, char>> maxHeap;
    for (int i = 0; i < 26; i++) {
        if (freq[i] > 0) maxHeap.push({freq[i], (char)(i + 'a')});
    }

    string result;
    pair<int, char> prev = {0, '#'};

    while (!maxHeap.empty()) {
        auto curr = maxHeap.top(); maxHeap.pop();
        result += curr.second;
        curr.first--;
        if (prev.first > 0) maxHeap.push(prev);
        prev = curr;
    }

    if ((int)result.size() != (int)s.size()) return "";
    return result;
}

int main() {
    vector<string> tests = {"aab", "aaab", "abc", "a"};
    for (auto& s : tests) {
        cout << "Input: " << s << " -> Output: " << reorganizeString(s) << endl;
    }
    return 0;
}

Line Notes

vector<int> freq(26, 0)Frequency vector for characters

priority_queue<pair<int, char>> maxHeapMax heap storing frequency and character

if (freq[i] > 0) maxHeap.push(...)Add characters with positive frequency

string resultBuild result string

pair<int, char> prev = {0, '#'}Track previously used character and remaining frequency

while (!maxHeap.empty())Process characters until heap is empty

auto curr = maxHeap.top(); maxHeap.pop()Pop character with highest frequency

result += curr.secondAppend character to result

curr.first--Decrement frequency after use

if (prev.first > 0) maxHeap.push(prev)Push back previous character if still available

prev = currUpdate previous character to current

if ((int)result.size() != (int)s.size()) return ""Check if rearrangement is valid

function reorganizeString(s) {
    const freq = new Array(26).fill(0);
    for (const ch of s) freq[ch.charCodeAt(0) - 97]++;

    const maxHeap = [];
    for (let i = 0; i < 26; i++) {
        if (freq[i] > 0) maxHeap.push([freq[i], String.fromCharCode(i + 97)]);
    }
    maxHeap.sort((a, b) => b[0] - a[0]);

    const result = [];
    let prev = [0, ''];

    while (maxHeap.length > 0) {
        const [count, ch] = maxHeap.shift();
        result.push(ch);
        if (prev[0] > 0) {
            maxHeap.push(prev);
            maxHeap.sort((a, b) => b[0] - a[0]);
        }
        prev = [count - 1, ch];
    }

    const resStr = result.join('');
    return resStr.length === s.length ? resStr : "";
}

// Test cases
const tests = ["aab", "aaab", "abc", "a"];
tests.forEach(s => console.log(`Input: ${s} -> Output: ${reorganizeString(s)}`));

Line Notes

const freq = new Array(26).fill(0)Frequency array for characters

for (const ch of s) freq[ch.charCodeAt(0) - 97]++Count frequency of each character

const maxHeap = []Initialize array to simulate max heap

if (freq[i] > 0) maxHeap.push([freq[i], String.fromCharCode(i + 97)])Add characters with positive frequency

maxHeap.sort((a, b) => b[0] - a[0])Sort heap to simulate max heap behavior

const result = []Build result array

let prev = [0, '']Track previously used character and remaining frequency

while (maxHeap.length > 0)Process characters until heap is empty

const [count, ch] = maxHeap.shift()Pop character with highest frequency

result.push(ch)Append character to result

if (prev[0] > 0)Push back previously used character if still available

prev = [count - 1, ch]Update previous character with decremented count

const resStr = result.join('')Convert result array to string

return resStr.length === s.length ? resStr : ""Return result if valid, else empty string

⏱

Complexity

TimeO(n log k)

SpaceO(k)

Where k is the number of unique characters (up to 26). Each character is pushed and popped from the heap multiple times, each heap operation is O(log k).

💡 For English lowercase letters, k=26, so heap operations are very fast, making this approach efficient for large inputs.

Interview Verdict: Accepted

This is the optimal greedy approach using a heap, commonly expected in interviews for this problem.

📊

All Approaches - One-Glance Tradeoffs

💡 In 95% of interviews, code the heap-based greedy approach for optimal balance of clarity and efficiency.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n! * n)	O(n! * n)	Yes (deep recursion)	Yes	Mention only - never code
2. Greedy Sorting + Two Pointers	O(n log n)	O(n)	No	Yes	Good alternative if heap not allowed
3. Greedy Max Heap	O(n log k)	O(k)	No	Yes	Optimal approach to implement

💼

Interview Strategy

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

How to Present

Step 1: Clarify the problem and constraints with the interviewer.Step 2: Explain the brute force approach to show understanding.Step 3: Discuss the greedy approach using sorting and two pointers.Step 4: Present the optimal heap-based greedy approach.Step 5: Code the heap-based approach and test with examples.

Time Allocation

Clarify: 2min → Approach discussion: 5min → Coding: 10min → Testing & optimization: 3min. Total ~20min

What the Interviewer Tests

Interviewer tests your ability to identify greedy patterns, handle edge cases, and implement efficient data structures like heaps.

Common Follow-ups

• What if the input contains uppercase letters or other characters? → Extend frequency array or map accordingly.
• Can you generalize this to avoid k adjacent characters? → Use a similar heap approach with k-1 holding buffer.

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

🔍

Pattern Recognition

When to Use

1) Need to rearrange characters to avoid adjacency; 2) Frequency counts matter; 3) Greedy selection of most frequent elements; 4) Use of heap or sorting to manage frequencies.

Signature Phrases

no two adjacent characters are the samerearrange stringfrequency of characters

NOT This Pattern When

Problems that require exact order preservation or dynamic programming for subsequence counts are different.

Similar Problems

Task Scheduler - scheduling tasks with cooldowns to avoid adjacencyRearrange String k Distance Apart - generalizes adjacency constraint to distance k

Practice

1. You are given an array where each element represents the maximum jump length from that position. You need to determine if you can reach the last index starting from the first index. Which algorithmic approach guarantees an optimal solution with linear time complexity?

easy

A. Dynamic Programming with bottom-up tabulation to check reachability for each index

B. Breadth-first search using a queue to explore reachable indices level by level

C. Depth-first search exploring all possible jump paths recursively without memoization

D. Greedy algorithm tracking the maximum reachable index while iterating through the array

Solution

1. Step 1: Understand problem constraints

   The problem requires checking if the last index is reachable from the first index using jumps defined by array values.

2. Step 2: Identify optimal approach

   Greedy approach efficiently tracks the furthest reachable index in one pass, guaranteeing O(n) time complexity, unlike exhaustive search or DP which are slower.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Greedy approach is linear and optimal for this problem [OK]

Hint: Greedy tracks max reachable index in one pass [OK]

Common Mistakes:

• Confusing DP with greedy, thinking recursion is needed

2. Consider the following BFS-based jump function. What is the returned value when calling jump([2,1,1,1,4])?

easy

A. 3

B. 2

C. 4

D. 1

Solution

1. Step 1: Trace first BFS level

   Start at index 0 with jump length 2, enqueue indices 1 and 2, jumps = 1.

2. Step 2: Trace second BFS level

   From indices 1 and 2, enqueue reachable indices: from 1 (jump 1) -> 2 (visited), from 2 (jump 1) -> 3, jumps = 2.

3. Step 3: Trace third BFS level

   From index 3 (jump 1), enqueue index 4 (last index), jumps = 3, return 3.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Minimum jumps to reach end is 3 [OK]

Hint: Count BFS levels until last index reached [OK]

Common Mistakes:

• Off-by-one error in counting jumps
• Returning jumps too early
• Miscounting reachable indices

3. The following code attempts to solve the Jump Game problem. Identify the line that contains a subtle bug that causes incorrect results on some inputs.

def canJump(nums):
    maxReach = 0
    for i, jump in enumerate(nums):
        # Bug: missing check if current index is beyond maxReach
        maxReach = max(maxReach, i + jump)
        if maxReach >= len(nums) - 1:
            return True
    return False

medium

A. Line 2: Initialization of maxReach

B. Line 3: for loop header enumerating nums

C. Line 4: Missing check if i > maxReach before updating maxReach

D. Line 6: Checking if maxReach reaches or exceeds last index

Solution

1. Step 1: Understand the missing condition

   The code does not check if the current index i is beyond maxReach, which means it may continue even when stuck.

2. Step 2: Identify the bug line

   Line 4 updates maxReach without verifying if i is reachable, causing false positives on inputs with unreachable indices.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Adding "if i > maxReach: return False" fixes the bug [OK]

Hint: Check if current index is reachable before updating maxReach [OK]

Common Mistakes:

• Forgetting to check i > maxReach
• Assuming maxReach update alone suffices

4. What is the worst-case time complexity of the BFS-based solution for Jump Game II on an input array of length n?

medium

A. O(n^2)

B. O(n)

C. O(n log n)

D. O(n^3)

Solution

1. Step 1: Identify outer and inner loops

   The BFS processes each index once, but for each index, it may enqueue up to nums[pos] next positions, which can be up to n.

2. Step 2: Analyze nested iteration

   In worst case (e.g., nums[i] = n for all i), inner loop runs O(n) times for each of O(n) indices -> O(n^2) total.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Nested loops over n indices and jumps -> O(n^2) [OK]

Hint: Nested loops over n indices and jumps -> O(n^2) [OK]

Common Mistakes:

• Assuming BFS is always O(n)
• Confusing with greedy linear time
• Ignoring inner loop over jump range

5. Suppose the problem is modified so that digits can be removed multiple times (i.e., digits can be reused after removal) to form the smallest number after k removals. Which of the following algorithmic changes correctly adapts the solution?

hard

A. Use dynamic programming to consider all sequences with repeated digits allowed

B. Sort digits and pick the smallest k digits to remove, ignoring order

C. Use the same greedy stack approach but allow re-inserting popped digits later

D. Use a priority queue to always remove the largest digit available at each step

Solution

1. Step 1: Understand reuse of digits breaks greedy assumptions

   Allowing reuse means the problem is no longer about removing fixed digits but about sequences with repetition, invalidating the greedy stack approach.

2. Step 2: Why dynamic programming is needed

   DP can explore all subsequences with repeated digits and track minimal results, handling reuse correctly.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Greedy fails with reuse; DP handles repeated digit choices [OK]

Hint: Reuse breaks greedy; DP needed for repeated choices [OK]

Common Mistakes:

• Trying to reuse greedy stack with re-insertion
• Sorting digits ignores order and reuse constraints
• Using priority queue removes largest digit but ignores reuse