What is the time complexity of Find Products of Elements of Big Array?

The optimal solution has a time complexity of O(Q * logN).

What data structures are used to solve Find Products of Elements of Big Array?

This problem uses Array, Binary Search, Bit Manipulation. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Find Products of Elements of Big Array?

Insight 1: The powerful array corresponds to the set bits in the binary representation of a number. For example, 13 (1101 in binary) has a powerful array of [1, 4, 8], which are powers of 2 corresponding to the set bits. Insight 2: Calculating the prefix sum of exponents (powers of 2) is crucial for efficiently calculating the product within a given range. Instead of actually constructing the `big_nums` array, we need to compute exponents quickly.

Which companies ask LeetCode 3145. Find Products of Elements of Big Array in interviews?

This problem is frequently asked at IBM.

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Find Products of Elements of Big Array - LeetCode 3145 Solution

Find Products of Elements of Big Array - Complete Solution Guide

Find Products of Elements of Big Array is LeetCode problem 3145, 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

The powerful array of a non-negative integer x is defined as the shortest sorted array of powers of two that sum up to x . The table below illustrates examples of how the powerful array is determined. It can be proven that the powerful array of x is unique. num Binary Representation powerful array 1 0000 1 [1] 8 0 1 000 [8] 10 0 1 0 1 0 [2, 8] 13 0 11 0 1 [1, 4, 8] 23 1 0 111 [1, 2, 4, 16] The array big_nums is created by concatenating the powerful arrays for every positive integer i in ascendin

Detailed Explanation

The problem defines a 'powerful array' for a non-negative integer x as the shortest sorted array of powers of two that sum up to x. The `big_nums` array is created by concatenating the powerful arrays for every positive integer in ascending order (1, 2, 3...). The task is to calculate the product of elements within given ranges [from_i, to_i] of the `big_nums` array, modulo a given value mod_i, for a set of queries. The queries are given in a 2D array, where each query consists of a 'from' index, a 'to' index, and a modulus.

Solution Approach

The solution avoids explicitly constructing the very large `big_nums` array. Instead, it efficiently computes the product of elements within a range by calculating the prefix sum of the exponents of 2 (corresponding to the elements of the powerful arrays). The `_get_prefix_sum_of_exponents` function is used to calculate the sum of exponents up to a certain number. Binary search is used in `_get_prefix_sum_of_exponents` function to determine the value n so as to get the total length. Memoization is used in `_count_total_set_bits` and `_sum_total_exponents` for performance optimization.

Step-by-Step Algorithm

Step 1: Iterate through each query [from_i, to_i, mod_i] in the `queries` array.
Step 2: Calculate the sum of exponents up to `to_i + 1` using the `_get_prefix_sum_of_exponents` function.
Step 3: Calculate the sum of exponents up to `from_i` using the `_get_prefix_sum_of_exponents` function.
Step 4: Calculate the total exponent by subtracting the sum of exponents up to `from_i` from the sum of exponents up to `to_i + 1`.
Step 5: Calculate the result by raising 2 to the power of the total exponent, modulo `mod_i`. This is done using the power function to handle large exponents efficiently.
Step 6: Append the result to the `ans` array.
Step 7: Return the `ans` array containing the results for all queries.
Step 8: `_get_prefix_sum_of_exponents(count)` function uses binary search to find n, where count is equal to number of bits in integers from 1 to n.
Step 9: Then it calculates _sum_total_exponents(n-1), counts remaining numbers of bits from number n and add exponents till total is achieved.

Key Insights

Insight 1: The powerful array corresponds to the set bits in the binary representation of a number. For example, 13 (1101 in binary) has a powerful array of [1, 4, 8], which are powers of 2 corresponding to the set bits.
Insight 2: Calculating the prefix sum of exponents (powers of 2) is crucial for efficiently calculating the product within a given range. Instead of actually constructing the `big_nums` array, we need to compute exponents quickly.
Insight 3: Memoization is used to optimize the calculation of the total number of set bits and the sum of total exponents, preventing redundant calculations since the same numbers can appear multiple times in different queries.

Complexity Analysis

Time Complexity: O(Q * logN)

Space Complexity: O(N)

Topics

This problem involves: Array, Binary Search, Bit Manipulation.

Companies

Asked at: IBM.

Find Products of Elements of Big Array | LeetCode 3145

Hard

Array Binary Search Bit Manipulation

Time: O(Q * logN)Space: O(N)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    private Map<Long, Long> memo_L = new HashMap<>();
    private Map<Long, Long> memo_E = new HashMap<>();

    public int[] findProductsOfElements(long[][] queries) {
        List<Integer> ans = new ArrayList<>();
        for (long[] query : queries) {
            long fromi = query[0];
            long toi = query[1];
            long modi = query[2];

            long count_to = toi + 1;
            long count_from = fromi;

            long exp_to = _get_prefix_sum_of_exponents(count_to);
            long exp_from = _get_prefix_sum_of_exponents(count_from);

            long total_exp = exp_to - exp_from;

            long result = power(2, total_exp, (int)modi);
            ans.add((int)result);
        }

        return ans.stream().mapToInt(Integer::intValue).toArray();
    }

    private long _count_total_set_bits(long n) {
        if (n <= 0) {
            return 0;
        }
        if (memo_L.containsKey(n)) {
            return memo_L.get(n);
        }

        int k = Long.toBinaryString(n).length() - 1;
        long rem = n - (1L << k);

        long res = k * (1L << (k - 1)) ;
        if(k <= 0) res = 0;
        res += (rem + 1);
        res += _count_total_set_bits(rem);

        memo_L.put(n, res);
        return res;
    }

    private long _sum_total_exponents(long n) {
        if (n <= 0) {
            return 0;
        }
        if (memo_E.containsKey(n)) {
            return memo_E.get(n);
        }

        int k = Long.toBinaryString(n).length() - 1;
        long rem = n - (1L << k);

        long res = k * (k - 1) / 2 * (1L << (k - 1));
        if(k <= 0) res = 0;
        res += (rem + 1) * k;
        res += _sum_total_exponents(rem);

        memo_E.put(n, res);
        return res;
    }

    private long _get_prefix_sum_of_exponents(long count) {
        if (count == 0) {
            return 0;
        }

        long low = 1, high = count;
        long n = 0;
        while (low <= high) {
            long mid = low + (high - low) / 2;

            long len_up_to_mid = _count_total_set_bits(mid);
            if (len_up_to_mid >= count) {
                n = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }

        long total_sum = _sum_total_exponents(n - 1);

        long len_up_to_prev = _count_total_set_bits(n - 1);
        long rem_count = count - len_up_to_prev;

        int j = 0;
        while (rem_count > 0) {
            if ((n >> j) % 2 == 1) {
                total_sum += j;
                rem_count -= 1;
            }
            j += 1;
        }

        return total_sum;
    }
    
    private long power(long base, long exp, int mod) {
        long res = 1;
        base = base % mod;
        while (exp > 0) {
            if (exp % 2 == 1) res = (res * base) % mod;
            exp = exp >> 1;
            base = (base * base) % mod;
        }
        return res;
    }
}

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Problem Info

Description

Problem 3145: Find Products of Elements of Big Array - Detailed explanation and solution approaches.

Topics

Array Binary Search Bit Manipulation

Companies

IBM

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3145

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

9 steps

1
Step 1: Iterate through each query [from_i, to_i, mod_i] in the `queries` array.
2
Step 2: Calculate the sum of exponents up to `to_i + 1` using the `_get_prefix_sum_of_exponents` function.
3
Step 3: Calculate the sum of exponents up to `from_i` using the `_get_prefix_sum_of_exponents` function.
4
Step 4: Calculate the total exponent by subtracting the sum of exponents up to `from_i` from the sum of exponents up to `to_i + 1`.
5
Step 5: Calculate the result by raising 2 to the power of the total exponent, modulo `mod_i`. This is done using the power function to handle large exponents efficiently.
6
Step 6: Append the result to the `ans` array.
7
Step 7: Return the `ans` array containing the results for all queries.
8
Step 8: `_get_prefix_sum_of_exponents(count)` function uses binary search to find n, where count is equal to number of bits in integers from 1 to n.
9
Step 9: Then it calculates _sum_total_exponents(n-1), counts remaining numbers of bits from number n and add exponents till total is achieved.

💡

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 * logN)

💾

Space Complexity

O(N)

📝

Why?

The overall time complexity is determined by the number of queries multiplied by the time complexity of processing each query. The space complexity depends on the size of the memoization tables, which can grow up to the maximum input number.

📖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

Precompute big_nums array

One alternative would be to precompute a sufficiently large portion of the `big_nums` array and store it. Then, for each query, you can directly access the elements within the specified range and calculate the product modulo the given value.

Time: O(M + Q*K), where M is the length of the precomputed big_nums, Q is the number of queries, and K is the length of the range in the query.Space: O(M), where M is the length of the precomputed big_nums array.

Trade-offs: This approach has a significant memory overhead because it stores the `big_nums` array. It can be faster for a large number of queries with small ranges if `big_nums` array fits in memory. But the approach is inefficient if ranges are large since it iterates through range, the primary approach of calculating exponents are better.

✨

Key Insights

1Insight 1: The powerful array corresponds to the set bits in the binary representation of a number. For example, 13 (1101 in binary) has a powerful array of [1, 4, 8], which are powers of 2 corresponding to the set bits.
2Insight 2: Calculating the prefix sum of exponents (powers of 2) is crucial for efficiently calculating the product within a given range. Instead of actually constructing the `big_nums` array, we need to compute exponents quickly.
3Insight 3: Memoization is used to optimize the calculation of the total number of set bits and the sum of total exponents, preventing redundant calculations since the same numbers can appear multiple times in different queries.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: When from_i equals to_i, the product is just the big_nums[from_i]. The code handles this correctly as the exponent is calculated accurately using prefix sums.
Edge case 2: Large values of from_i and to_i require handling large numbers and preventing integer overflow during calculations. The use of long long (or equivalent) is essential to prevent integer overflow in all languages.
Edge case 3: `modi` equal to 1. When `modi` is 1, any number modulo 1 will be 0. The power function handles this correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling large values of from_i and to_i and causing integer overflow. Using appropriate data types (long long in C++, Java and int64 in python) is crucial.

✗Common mistake 2: Not using memoization and recalculating the same values repeatedly, leading to Time Limit Exceeded (TLE). Memoization significantly improves the performance.

✗Common mistake 3: Incorrectly calculating the powers of 2 modulo a number. This can be avoided by using the optimized power function (modular exponentiation).

✗Common mistake 4: Misunderstanding the prefix sum calculation, leading to incorrect exponent values and wrong results.

✅

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 compression algorithms can utilize bit manipulation techniques similar to those used in this problem to efficiently encode and decode data.

🏦Real-world scenario 2: Cryptography algorithms often rely on modular arithmetic and exponentiation, which are key components of this problem.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: August 25, 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.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

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 Products of Elements of Big Array"?

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 binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?