#3264Easy~20 min

Final Array State After K Multiplication Operations I

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

cpppython

Solutions

// Time:  O(n + (n + logr) + nlog(logr) + nlogn) = O(nlogn), assumed log(x) takes O(1) time
// Space: O(n)

// sort, two pointers, sliding window, fast exponentiation
class Solution {
public:
    vector<int> getFinalState(vector<int>& nums, int k, int multiplier) {
        static const double EPS = 1e-15;
        const auto& count = [](const auto& x, int target) {
            return static_cast<int>(target - x + EPS);
        };

        if (multiplier == 1) {
            return nums;
        }
        using P = pair<double, int>;
        vector<P> vals;
        for (int i = 0; i < size(nums); ++i) {
            vals.emplace_back(log(nums[i]) / log(multiplier), i);
        }
        sort(begin(vals), end(vals));
        int right = 1;
        for (int left = 0; right <= static_cast<int>(vals.back().first) + 1; ++right) {
            for (; left < size(vals) && count(vals[left].first, right) >= 1; ++left);
            if (k - left < 0) {
                --right;
                break;
            }
            k -= left;
        }
        if (right == static_cast<int>(vals.back().first) + 2) {
            --right;
        }
        for (int idx = 0; idx < size(vals); ++idx) {
            const auto& [x, i] = vals[idx];
            const int c = count(x, right);
            if (c <= 0) {
                break;
            }
            nums[i] *= pow(multiplier, c);
        }
        vals.clear();
        for (int i = 0; i < size(nums); ++i) {
            vals.emplace_back(nums[i], i);
        }
        sort(begin(vals), end(vals));
        const int q =  k / size(nums), r = k % size(nums);
        const int m = pow(multiplier, q);
        vector<int> result(size(nums));
        for (int idx = 0; idx < size(vals); ++idx) {
            const auto& [x, i] = vals[idx];
            result[i] = x * m * (idx < r ? multiplier : 1);
        }
        return result;
    }
};

// Time:  O(n + min(n, k) * log(logr) + nlog(logr) + nlogn) = O(nlogr), assumed log(x) takes O(1) time
// Space: O(n)
// binary search, sort, fast exponentiation
class Solution2 {
public:
    vector<int> getFinalState(vector<int>& nums, int k, int multiplier) {
        static const double EPS = 1e-15;
        const auto& binary_search_right = [](auto left, auto right, const auto& check) {
            while (left <= right) {
                const auto mid = left + (right - left) / 2;
                if (!check(mid)) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
            return right;
        };

        const auto& count = [](const auto& x, int target) {
            return static_cast<int>(target - x + EPS);
        };

        if (multiplier == 1) {
            return nums;
        }
        using P = pair<double, int>;
        vector<P> vals;
        for (int i = 0; i < size(nums); ++i) {
            vals.emplace_back(log(nums[i]) / log(multiplier), i);
        }
        sort(begin(vals), end(vals));
        const auto& check = [&](const auto& target) {
            int result = 0;
            for (const auto& [x, i] : vals) {
                const int c = count(x, target);
                if (c <= 0) {
                    break;
                }
                result += c;
            }
            return result <= k;
        };

        const int target = binary_search_right(1, static_cast<int>(vals.back().first) + 1, check);
        for (int idx = 0; idx < size(vals); ++idx) {
            const auto& [x, i] = vals[idx];
            const int c = count(x, target);
            if (c <= 0) {
                break;
            }
            k -= c;
            nums[i] *= pow(multiplier, c);
        }
        vals.clear();
        for (int i = 0; i < size(nums); ++i) {
            vals.emplace_back(nums[i], i);
        }
        sort(begin(vals), end(vals));
        const int q =  k / size(nums), r = k % size(nums);
        const int m = pow(multiplier, q);
        vector<int> result(size(nums));
        for (int idx = 0; idx < size(vals); ++idx) {
            const auto& [x, i] = vals[idx];
            result[i] = x * m * (idx < r ? multiplier : 1);
        }
        return result;
    }
};

// Time:  O(min(nlogr, k) * logn + nlogn) = O(nlogn * logr)
// Space: O(n)
// heap, sort, fast exponentiation
class Solution3 {
public:
    vector<int> getFinalState(vector<int>& nums, int k, int multiplier) {
        if (multiplier == 1) {
            return nums;
        }
        using P = pair<int, int>;
        vector<P> vals;
        for (int i = 0; i < size(nums); ++i) {
            vals.emplace_back(nums[i], i);
        }
        priority_queue<P, vector<P>, greater<P>> min_heap(cbegin(vals), cend(vals));
        const int mx = ranges::max(nums);
        for (; k; --k) {
            const auto [x, i] = min_heap.top(); min_heap.pop();
            if (x >= mx) {
                break;
            }
            nums[i] *= multiplier;
            min_heap.emplace(nums[i], i);
        }
        vals.clear();
        for (int i = 0; i < size(nums); ++i) {
            vals.emplace_back(nums[i], i);
        }
        sort(begin(vals), end(vals));
        const int q =  k / size(nums), r = k % size(nums);
        const int m = pow(multiplier, q);
        vector<int> result(size(nums));
        for (int idx = 0; idx < size(vals); ++idx) {
            const auto& [x, i] = vals[idx];
            result[i] = x * m * (idx < r ? multiplier : 1);
        }
        return result;
    }
};

// Time:  O(n + klogn)
// Space: O(n)
// simulation, heap
class Solution4 {
public:
    vector<int> getFinalState(vector<int>& nums, int k, int multiplier) {
        if (multiplier == 1) {
            return nums;
        }
        using P = pair<int, int>;
        vector<P> vals;
        for (int i = 0; i < size(nums); ++i) {
            vals.emplace_back(nums[i], i);
        }
        priority_queue<P, vector<P>, greater<P>> min_heap(cbegin(vals), cend(vals));
        for (int _ = 0; _ < k; ++_) {
            const auto [x, i] = min_heap.top(); min_heap.pop();
            nums[i] *= multiplier;
            min_heap.emplace(nums[i], i);
        }
        return nums;
    }
};

// Time:  O(k * n)
// Space: O(1)
// simulation
class Solution5 {
public:
    vector<int> getFinalState(vector<int>& nums, int k, int multiplier) {
        if (multiplier == 1) {
            return nums;
        }
        for (int _ = 0; _ < k; ++_) {
            const int i = distance(cbegin(nums), min_element(cbegin(nums), cend(nums)));
            nums[i] *= multiplier;
        }
        return nums;
    }
};

Beginner Explanation

What is Final Array State After K Multiplication Operations I?

Final Array State After K Multiplication Operations I (LeetCode #3264) is a Easy problem that primarily trains two pointers.

How to think about it

  1. Restate the goal in your own words before coding.
  2. Work a tiny example by hand so the invariant becomes obvious.
  3. Identify the pattern — this problem aligns with sort, two pointers, sliding window, heap, and binary search.
  4. Only then translate the idea into code.

Why this problem matters

It builds core muscle memory you will reuse on harder variants. Official solution notes mention: Sort, Two Pointers, Sliding Window, Fast Exponentiation.

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

Interview Walkthrough

Interview approach for Final Array State After K Multiplication Operations I

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 sort, two pointers, sliding window, heap, and binary search.

Core solution narrative

  1. Define the state you track (pointers, DP cell, set membership, stack top, etc.).
  2. Explain the transition when you process the next element.
  3. Call out time (O(nlogn)) and space (O(n)) before coding.
  4. 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(nlogn) time and O(n) space.

Pattern focus: sort, two pointers, sliding window, heap, and binary search

Use the pattern as a checklist:

  • sort — confirm the invariant holds after each step
  • two pointers — confirm the invariant holds after each step
  • sliding window — confirm the invariant holds after each step
  • heap — confirm the invariant holds after each step
  • binary search — 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(nlogn)
Space O(n)

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 Final Array State After K Multiplication Operations I

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

Alternative Approaches

Alternatives

The source file includes more than one method. Compare:

  1. Primary optimized path — best complexity for typical interviews.
  2. 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 sort, two pointers, sliding window, heap, and binary search:

  • 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: two pointers.

Follow-up Interview Questions

Follow-ups

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

Practice Recommendations

What to practice next

  1. Re-solve Final Array State After K Multiplication Operations I in a second language (cpp, python).
  2. Drill 3–5 more problems tagged two pointers.
  3. Teach the solution out loud in under 5 minutes.
  4. Add this problem to your revision calendar in 3 days and 14 days.

Visualization

Conceptual diagram for Final Array State After K Multiplication Operations I: show input structure (two pointers), highlight the moving parts of the sort, two pointers, sliding window, heap, and binary search 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 sort, two pointers, sliding window, heap, and binary search 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

Final Array State After K Multiplication Operations I (#3264) — Easy. Pattern: sort, two pointers, sliding window, heap, and binary search. Complexity: O(nlogn) time / O(n) space. Re-derive the invariant before coding.

FAQs

What is the time complexity of Final Array State After K Multiplication Operations I?+

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

What pattern does Final Array State After K Multiplication Operations I use?+

It primarily maps to sort, two pointers, sliding window, heap, and binary search, within the broader topic of two pointers.

Is Final Array State After K Multiplication Operations I good for interviews?+

Yes — as a Easy 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/final-array-state-after-k-multiplication-operations-i/