What is the time complexity of Find X Value of Array II?

The optimal solution has a time complexity of O(q*n*k).

What data structures are used to solve Find X Value of Array II?

This problem uses Array, Math, Segment Tree. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Find X Value of Array II?

Insight 1: The constraint that `k` is small (<= 5) is crucial. It allows us to maintain counts of the remainders modulo `k` efficiently. Insight 2: A segment tree is an appropriate data structure to handle the update and query operations efficiently. We can precompute and store the product and remainder counts for each subarray.

Which companies ask LeetCode 3525. Find X Value of Array II in interviews?

This problem is frequently asked at Rubrik.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Find X Value of Array II - LeetCode 3525 Solution

Q: How to solve Find X Value of Array II?

The provided code uses a segment tree to efficiently handle updates to the array and queries for the x-value. The segment tree stores the product of the elements in a range modulo `k` and the count of subarrays resulting in different remainders modulo `k`. For each query, the code updates the element at the specified index, then queries the segment tree for the range starting at `start` to the end of the array. The segment tree efficiently returns a `Node` representing the product modulo k and a list of counts, which are then used to find the final result for the x-value.

Q: What is the space complexity of Find X Value of Array II?

The space complexity is O(n*k).

Q: Which companies ask LeetCode 3525. Find X Value of Array II in interviews?

This problem is frequently asked at Rubrik.

Find X Value of Array II - Complete Solution Guide

Find X Value of Array II is LeetCode problem 3525, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an array of positive integers nums and a positive integer k . You are also given a 2D array queries , where queries[i] = [index i , value i , start i , x i ] . You are allowed to perform an operation once on nums , where you can remove any suffix from nums such that nums remains non-empty . The x-value of nums for a given x is defined as the number of ways to perform this operation so that the product of the remaining elements leaves a remainder of x modulo k . For each query in qu

Detailed Explanation

The problem asks us to process a series of queries on an array of positive integers `nums`. For each query, we need to update a specific element in `nums`, remove a prefix of `nums`, and then calculate the "x-value" of the remaining suffix. The x-value is defined as the number of ways to remove a suffix (including an empty suffix) from the remaining array such that the product of the remaining elements has a remainder of `x` when divided by `k`. `nums`, `k`, and a list of `queries` are provided. Each query consists of `index`, `value`, `start`, and `x`. We update `nums[index]` to `value`, remove the prefix `nums[0..(start-1)]`, and compute the x-value (count of suffixes with product % k == x). The output is an array containing the x-values for each query.

Solution Approach

The provided code uses a segment tree to efficiently handle updates to the array and queries for the x-value. The segment tree stores the product of the elements in a range modulo `k` and the count of subarrays resulting in different remainders modulo `k`. For each query, the code updates the element at the specified index, then queries the segment tree for the range starting at `start` to the end of the array. The segment tree efficiently returns a `Node` representing the product modulo k and a list of counts, which are then used to find the final result for the x-value.

Step-by-Step Algorithm

Step 1: Initialize a segment tree with the input array `nums` and the value `k`. The segment tree stores the product of elements modulo k for each node and the counts of subarrays with each remainder modulo k.
Step 2: Iterate through each query in the `queries` array.
Step 3: For each query `[index, value, start, x]`, update the element at index `index` in the `nums` array to `value` using the segment tree's `update_val` method. This propagates the change through the tree.
Step 4: Query the segment tree for the range from `start` to the end of the array (`n-1`) using the `query_range` method. This returns a `Node` containing the product of elements modulo k for that range and the counts of subarray remainders.
Step 5: Extract the count at index `x` from the `counts` array of the returned `Node`. This count represents the x-value for the query.
Step 6: Add the x-value to the `result` array.
Step 7: After processing all queries, return the `result` array.

Key Insights

Insight 1: The constraint that `k` is small (<= 5) is crucial. It allows us to maintain counts of the remainders modulo `k` efficiently.
Insight 2: A segment tree is an appropriate data structure to handle the update and query operations efficiently. We can precompute and store the product and remainder counts for each subarray.
Insight 3: The x-value calculation requires considering all possible suffixes after the prefix removal and update. The segment tree helps in quickly calculating the product modulo k for any range.
Insight 4: Modular arithmetic is key to avoid integer overflow, particularly when multiplying array elements. The product is taken modulo k at each stage.

Complexity Analysis

Time Complexity: O(q*n*k)

Space Complexity: O(n*k)

Topics

This problem involves: Array, Math, Segment Tree.

Companies

Asked at: Rubrik.

Find X Value of Array II | LeetCode 3525

Hard

Array Math Segment Tree

Time: O(q*n*k)Space: O(n*k)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int[] resultArray(int[] nums, int k, int[][] queries) {
        int n = nums.length;
        SegTree segTree = new SegTree(nums, k);

        int[] result = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int index = queries[i][0];
            int value = queries[i][1];
            int start = queries[i][2];
            int x = queries[i][3];

            segTree.update_val(index, value);

            Node queryNode = segTree.query_range(start, n - 1);
            result[i] = queryNode.counts[x];
        }

        return result;
    }
}

class Node {
    public int total_prod;
    public int[] counts;

    public Node(int k_val) {
        this.total_prod = 1;
        this.counts = new int[k_val];
    }
}

class SegTree {
    private int n;
    private int k;
    private Node[] tree;
    private Node identity;

    public SegTree(int[] arr, int k_val) {
        this.n = arr.length;
        this.k = k_val;
        this.tree = new Node[4 * this.n];
        for (int i = 0; i < 4 * this.n; i++) {
            this.tree[i] = new Node(this.k);
        }
        this.identity = new Node(this.k);
        if (this.n > 0) {
            this._build(arr, 0, 0, this.n - 1);
        }
    }

    private Node _merge(Node left, Node right) {
        Node new_node = new Node(this.k);
        new_node.total_prod = (int)(((long)left.total_prod * right.total_prod) % this.k);
        new_node.counts = Arrays.copyOf(left.counts, left.counts.length);

        int left_prod = left.total_prod;
        for (int rem = 0; rem < this.k; rem++) {
            if (right.counts[rem] > 0) {
                int new_rem = (int)(((long)left_prod * rem) % this.k);
                new_node.counts[new_rem] += right.counts[rem];
            }
        }
        return new_node;
    }

    private void _build(int[] arr, int idx, int l, int r) {
        if (l == r) {
            int val = arr[l] % this.k;
            this.tree[idx].total_prod = val;
            this.tree[idx].counts[val] = 1;
            return;
        }
        int mid = (l + r) / 2;
        this._build(arr, 2 * idx + 1, l, mid);
        this._build(arr, 2 * idx + 2, mid + 1, r);
        this.tree[idx] = this._merge(this.tree[2 * idx + 1], this.tree[2 * idx + 2]);
    }

    public void update_val(int pos, int val) {
        this._update(0, 0, this.n - 1, pos, val);
    }

    private void _update(int idx, int l, int r, int pos, int val) {
        if (l == r) {
            int v_mod = val % this.k;
            this.tree[idx].total_prod = v_mod;
            this.tree[idx].counts = new int[this.k];
            this.tree[idx].counts[v_mod] = 1;
            return;
        }
        int mid = (l + r) / 2;
        if (pos <= mid) {
            this._update(2 * idx + 1, l, mid, pos, val);
        } else {
            this._update(2 * idx + 2, mid + 1, r, pos, val);
        }
        this.tree[idx] = this._merge(this.tree[2 * idx + 1], this.tree[2 * idx + 2]);
    }

    public Node query_range(int ql, int qr) {
        if (ql > qr) {
            return this.identity;
        }
        return this._query(0, 0, this.n - 1, ql, qr);
    }

    private Node _query(int idx, int l, int r, int ql, int qr) {
        if (ql > r || qr < l) {
            return this.identity;
        }
        if (ql <= l && r <= qr) {
            return this.tree[idx];
        }
        int mid = (l + r) / 2;
        Node left_res = this._query(2 * idx + 1, l, mid, ql, qr);
        Node right_res = this._query(2 * idx + 2, mid + 1, r, ql, qr);
        return this._merge(left_res, right_res);
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 3525: Find X Value of Array II - Detailed explanation and solution approaches.

Topics

Array Math Segment Tree

Companies

Rubrik

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3525

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize a segment tree with the input array `nums` and the value `k`. The segment tree stores the product of elements modulo k for each node and the counts of subarrays with each remainder modulo k.
2
Step 2: Iterate through each query in the `queries` array.
3
Step 3: For each query `[index, value, start, x]`, update the element at index `index` in the `nums` array to `value` using the segment tree's `update_val` method. This propagates the change through the tree.
4
Step 4: Query the segment tree for the range from `start` to the end of the array (`n-1`) using the `query_range` method. This returns a `Node` containing the product of elements modulo k for that range and the counts of subarray remainders.
5
Step 5: Extract the count at index `x` from the `counts` array of the returned `Node`. This count represents the x-value for the query.
6
Step 6: Add the x-value to the `result` array.
7
Step 7: After processing all queries, return the `result` array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(q*n*k)

💾

Space Complexity

O(n*k)

📝

Why?

The segment tree has a height of O(log n), and each level has at most 2^i nodes. Therefore, there are O(4n) nodes, where each node holds an array `counts` of size `k`. The merge operation takes O(k) since it has to consider all possible remainders. Thus the time complexity is O(q*log(n)*k), where q is number of queries, and the space complexity is O(n*k).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

For each query, update the array, remove the prefix, and then iterate through all possible suffixes, calculating the product modulo `k` for each and counting the ones that equal `x`.

Time: O(q * n^2), where q is the number of queries and n is the length of the array. For each query, iterating through all possible suffixes takes O(n^2) time.Space: O(1). Constant extra space is used.

Trade-offs: The brute force approach is simpler to implement but is much less efficient than the segment tree approach. It would be suitable for small input sizes but will time out for larger inputs. The segment tree approach has a higher initial overhead for building the tree but provides faster query processing.

✨

Key Insights

1Insight 1: The constraint that `k` is small (<= 5) is crucial. It allows us to maintain counts of the remainders modulo `k` efficiently.
2Insight 2: A segment tree is an appropriate data structure to handle the update and query operations efficiently. We can precompute and store the product and remainder counts for each subarray.
3Insight 3: The x-value calculation requires considering all possible suffixes after the prefix removal and update. The segment tree helps in quickly calculating the product modulo k for any range.
4Insight 4: Modular arithmetic is key to avoid integer overflow, particularly when multiplying array elements. The product is taken modulo k at each stage.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: `start` equals `n`. This means an empty array after the prefix removal, the x-value will be 1 if x equals 1 % k, and 0 otherwise. The code handles this correctly by considering the counts from the segment tree.
Edge case 2: `k = 1`. In this case, all numbers modulo `k` are 0. So the number of possible subarrays is simply length of considered segment of the array. Segment tree correctly calculates and maintains required information for this case as well.
Edge case 3: `nums` is empty initially. Code correctly handles this by considering base case of empty array when building a segment tree

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer Overflow: Failing to take the modulo `k` at each step of the multiplication can lead to integer overflow, especially when the values in `nums` are large.

✗Common mistake 2: Incorrect Segment Tree Implementation: Implementing the segment tree incorrectly, such as failing to properly merge the nodes or update the values, will lead to wrong results.

✗Common mistake 3: Ignoring Prefix Removal: Forgetting to consider the `start` index in the query will lead to calculating the x-value for the entire array instead of the required suffix.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data Stream Processing: The problem demonstrates a pattern that can be applied in data stream processing where real-time updates and range queries are required, such as calculating statistics on a rolling window of data.

🏦Real-world scenario 2: Financial Analysis: Analyzing stock prices or trading volumes where you need to calculate rolling averages or other statistical measures over specific time periods, while also incorporating real-time updates.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find X Value of Array II"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?