#358HardPremium on LC~50 min

Rearrange String k Distance Apart

Time O(n) · Space O(c) · Official statement on LeetCode

Solutions

// Time:  O(n)
// Space: O(c)

class Solution {
public:
    string rearrangeString(string s, int k) {
        if (!k) {
            return s;
        }
        unordered_map<char, int> cnts;
        for (const auto& c : s) {
            ++cnts[c];
        }
        const int bucket_cnt = max_element(cbegin(cnts), cend(cnts), [](const auto& a, const auto& b) {
            return a.second < b.second;
        })->second;
        if (!((bucket_cnt - 1) * k + count_if(cbegin(cnts), cend(cnts), [&](const auto& x) { return x.second == bucket_cnt; }) <= size(s))) {
            return "";
        }
        vector<char> partial_sorted_cnts;
        for (const auto& [c, v] : cnts) {
            if (v == bucket_cnt) {
                partial_sorted_cnts.emplace_back(c);
            }
        }
        for (const auto& [c, v] : cnts) {
            if (v != bucket_cnt) {
                partial_sorted_cnts.emplace_back(c);
            }
        }
        string result(size(s), 0);
        int i = (size(s) - 1) % k;
        for (const auto& c : partial_sorted_cnts) {
            for (int _ = 0; _ < cnts[c]; ++_) {
                result[i] = c;
                i += k;
                if (i >= size(result)) {
                    i = (i - 1) % k;
                }
            }
        }
        return result;
    }
};

// Time:  O(n)
// Space: O(c)
// reference: https://codeforces.com/blog/entry/110184 1774B - Coloring
class Solution2 {
public:
    string rearrangeString(string s, int k) {
        if (!k) {
            return s;
        }
        unordered_map<char, int> cnts;
        for (const auto& c : s) {
            ++cnts[c];
        }
        const int bucket_cnt = (size(s) + k - 1) / k;
        const int mx = max_element(cbegin(cnts), cend(cnts), [](const auto& a, const auto& b) {
            return a.second < b.second;
        })->second;
        if (!(mx <= bucket_cnt && count_if(cbegin(cnts), cend(cnts), [&](const auto& x) { return x.second == bucket_cnt; }) <= (size(s) - 1) % k + 1)) {
            return "";
        }
        vector<char> partial_sorted_cnts;
        for (const auto& [c, v] : cnts) {
            if (v == bucket_cnt) {
                partial_sorted_cnts.emplace_back(c);
            }
        }
        for (const auto& [c, v] : cnts) {
            if (v <= bucket_cnt - 2) {
                partial_sorted_cnts.emplace_back(c);
            }
        }
        for (const auto& [c, v] : cnts) {
            if (v == bucket_cnt - 1) {
                partial_sorted_cnts.emplace_back(c);
            }
        }
        string result(size(s), 0);
        int i = 0;
        for (const auto& c : partial_sorted_cnts) {
            for (int _ = 0; _ < cnts[c]; ++_) {
                result[i] = c;
                i += k;
                if (i >= size(result)) {
                    i = i % k + 1;
                }
            }
        }
        return result;
    }
};

// Time:  O(n)
// Space: O(n)
class Solution3 {
public:
    string rearrangeString(string s, int k) {
        unordered_map<char, int> cnts;
        for (const auto& c : s) {
            ++cnts[c];
        }
        const int bucket_cnt = max_element(cbegin(cnts), cend(cnts), [](const auto& a, const auto& b) {
            return a.second < b.second;
        })->second;
        vector<char> partial_sorted_cnts;
        for (const auto& [c, v] : cnts) {
            if (v == bucket_cnt) {
                partial_sorted_cnts.emplace_back(c);
            }
        }
        for (const auto& [c, v] : cnts) {
            if (v == bucket_cnt - 1) {
                partial_sorted_cnts.emplace_back(c);
            }
        }
        for (const auto& [c, v] : cnts) {
            if (v <= bucket_cnt - 2) {
                partial_sorted_cnts.emplace_back(c);
            }
        }
        vector<string> buckets(bucket_cnt);
        int i = 0;
        for (const auto& c : partial_sorted_cnts) {
            for (int _ = 0; _ < cnts[c]; ++_) {
                buckets[i].push_back(c);
                i = (i + 1) % max(cnts[c], bucket_cnt - 1);
            }
        }
        string result;
        for (int i = 0; i < size(buckets) - 1; ++i) {
            if (size(buckets[i]) < k) {
                return "";
            } else {
                result += buckets[i];
            }
        }
        result += buckets[bucket_cnt - 1];
        return result;
    }
};

// Time:  O(nlogc), c is the count of unique characters.
// Space: O(c)
class Solution4 {
public:
    string rearrangeString(string s, int k) {
        if (k == 0) {
            return s;
        }

        unordered_map<char, int> cnts;
        for (const auto& c : s) {
            ++cnts[c];
        }

        priority_queue<pair<int, char>> heap;
        for (const auto& kvp : cnts) {
            heap.emplace(kvp.second, kvp.first);
        }

        string result;
        while (!heap.empty()) {
            vector<pair<int, char>> used_cnt_chars;
            int cnt = min(k, static_cast<int>(s.length() - result.length()));
            for (int i = 0; i < cnt; ++i) {
                if (heap.empty()) {
                    return "";
                }
                auto cnt_char = heap.top();
                heap.pop();
                result.push_back(cnt_char.second);
                if (--cnt_char.first > 0) {
                    used_cnt_chars.emplace_back(move(cnt_char));
                }
            }
            for (auto& cnt_char: used_cnt_chars) {
                heap.emplace(move(cnt_char));
            }
        }
        return result;
    }
};

Beginner Explanation

What is Rearrange String k Distance Apart?

Rearrange String k Distance Apart (LeetCode #358) is a Hard problem that primarily trains binary heap.

How to think about it

Restate the goal in your own words before coding.
Work a tiny example by hand so the invariant becomes obvious.
Identify the pattern — this problem aligns with greedy and heap.
Only then translate the idea into code.

Why this problem matters

Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: Greedy, Heap.

AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.

Interview Walkthrough

Interview approach for Rearrange String k Distance Apart

Opening (30–60 seconds)

Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
State a brute force so the interviewer knows you can solve it naively.
Propose the optimal direction tied to greedy and heap.

Core solution narrative

Define the state you track (pointers, DP cell, set membership, stack top, etc.).
Explain the transition when you process the next element.
Call out time (O(n)) and space (O(c)) before coding.
Code cleanly; narrate variable names.

What interviewers listen for

Correctness on edge cases
Complexity honesty
Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)

Follow-up questions they may ask

Can you solve it with less memory?
What if the input stream is infinite / doesn't fit in RAM?
How would tests look for adversarial inputs?

Optimized Approach

Optimized solution notes

The reference solutions on AlgoForge target O(n) time and O(c) space.

Pattern focus: greedy and heap

Use the pattern as a checklist:

greedy — confirm the invariant holds after each step
heap — confirm the invariant holds after each step

Multiple methods appear in the source solutions — compare them and explain when each is preferable.

Implementation tips

Prefer readable names over micro-optimizations in interviews.
Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
After AC-level logic, re-scan for off-by-one and null checks.

Complexity Analysis

Complexity

Measure	Bound
Time	O(n)
Space	O(c)

How to justify this in an interview

Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
Space: include hash maps, recursion stack, and output allocation when the problem asks for it.

If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.

Common Mistakes

Common mistakes on Rearrange String k Distance Apart

Skipping edge cases — empty collections, single-element inputs, max constraints.
Wrong invariant for greedy and heap — updating state too early or too late.
Mutating input unexpectedly when the problem forbids it.
Off-by-one in windows, ranges, or binary search bounds.
Ignoring overflow / precision for integer arithmetic problems.
Overengineering — jumping to an advanced structure when a simpler approach works.

Alternative Approaches

Alternatives

The source file includes more than one method. Compare:

Primary optimized path — best complexity for typical interviews.
Secondary approach — often brute force, sorting-based, or space-optimized variant.

Practice articulating when you would pick each (constraints, readability, follow-ups).

Edge Cases

Edge cases checklist

Minimum input size
Maximum input size / time limits
Duplicates and already-sorted input
Negative numbers / zeros (if applicable)
Disconnected structures (graphs/trees)
Single path vs branching recursion depth

Pattern Recognition

Spotting this pattern

Signal phrases that point to greedy and heap:

Sorted input or ability to sort without changing the answer class
Need for contiguous subarray / substring → consider sliding window
Need for O(1) membership → hash set/map
Optimal substructure + overlapping subproblems → DP
Connectivity / components → graph DFS/BFS or Union-Find

Primary topics: binary heap.

Follow-up Interview Questions

Follow-ups

How does the solution change if the input is a stream?
Can you solve it in-place?
What if duplicates must be handled differently?
How would you parallelize the approach?
Design tests that would break a buggy implementation.

Practice Recommendations

What to practice next

Re-solve Rearrange String k Distance Apart in a second language (cpp, python).
Drill 3–5 more problems tagged binary heap.
Teach the solution out loud in under 5 minutes.
Add this problem to your revision calendar in 3 days and 14 days.

Visualization

Conceptual diagram for Rearrange String k Distance Apart: show input structure (binary heap), highlight the moving parts of the greedy and heap approach, and annotate each step with the maintained invariant and complexity.

Study checklist

Read the official problem statement on LeetCode
Solve on paper / whiteboard first
Implement the greedy and heap approach
Verify edge cases from the checklist
State time and space complexity aloud
Compare with the AlgoForge reference solution
Schedule a revision session

Revision notes

Rearrange String k Distance Apart (#358) — Hard. Pattern: greedy and heap. Complexity: O(n) time / O(c) space. Re-derive the invariant before coding.

FAQs

What is the time complexity of Rearrange String k Distance Apart?+

The reference solutions aim for O(n) time and O(c) space. Always re-derive complexity from the code you write in the interview.

What pattern does Rearrange String k Distance Apart use?+

It primarily maps to greedy and heap, within the broader topic of binary heap.

Is Rearrange String k Distance Apart good for interviews?+

Yes — as a Hard problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.

Where can I read the official statement?+

Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/rearrange-string-k-distance-apart/

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