What is the time complexity of Construct Target Array With Multiple Sums?

The optimal solution has a time complexity of O(n + k log n).

What data structures are used to solve Construct Target Array With Multiple Sums?

This problem uses Array, Heap (Priority Queue). Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Construct Target Array With Multiple Sums?

Insight 1: Working backward is crucial. Instead of trying to build the `target` array from the initial array, it's more efficient to reduce the `target` array back to the initial array of 1s. Insight 2: Using a max heap (priority queue) allows us to efficiently find the largest element in the `target` array at each step. This largest element was the last element to be modified in the construction process.

Which companies ask LeetCode 1354. Construct Target Array With Multiple Sums in interviews?

This problem is frequently asked at Quora.

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Construct Target Array With Multiple Sums - LeetCode 1354 Solution

Q: How to solve Construct Target Array With Multiple Sums?

The solution uses a max heap to keep track of the largest element in the `target` array. In each iteration, it extracts the largest element and subtracts the sum of the remaining elements from it. This simulates reversing the construction process. If at any point, an invalid state is encountered (e.g., the largest element becomes less than or equal to the sum of other elements, or the remainder is 0), it means the `target` array cannot be constructed. The process continues until the largest element becomes 1, at which point we check if all other elements are also 1.

Construct Target Array With Multiple Sums - Complete Solution Guide

Construct Target Array With Multiple Sums is LeetCode problem 1354, 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 target of n integers. From a starting array arr consisting of n 1's, you may perform the following procedure : let x be the sum of all elements currently in your array. choose index i , such that 0 <= i < n and set the value of arr at index i to x . You may repeat this procedure as many times as needed. Return true if it is possible to construct the target array from arr , otherwise, return false . Example 1: Input: target = [9,3,5] Output: true Explanation: Start with arr

Detailed Explanation

The problem asks whether a given array `target` of positive integers can be constructed from an initial array `arr` of the same size containing only 1s. The construction process involves repeatedly: (1) calculating the sum `x` of all elements in `arr`, (2) choosing an index `i` and setting `arr[i]` to `x`. We need to determine if the `target` array can be reached through this process.

Solution Approach

The solution uses a max heap to keep track of the largest element in the `target` array. In each iteration, it extracts the largest element and subtracts the sum of the remaining elements from it. This simulates reversing the construction process. If at any point, an invalid state is encountered (e.g., the largest element becomes less than or equal to the sum of other elements, or the remainder is 0), it means the `target` array cannot be constructed. The process continues until the largest element becomes 1, at which point we check if all other elements are also 1.

Step-by-Step Algorithm

Step 1: Calculate the sum of all elements in the `target` array.
Step 2: Build a max heap from the `target` array.
Step 3: While the largest element in the heap is greater than 1, perform the following steps:
Step 4: Extract the largest element from the heap.
Step 5: Calculate the sum of the remaining elements (total_sum - largest).
Step 6: Check for invalid states. If rest_sum is 0 or the largest element is less than or equal to rest_sum, return false. If rest_sum is 1, return true because all other elements must be 1.
Step 7: Calculate the previous value of the largest element by taking the modulo of the largest element with the sum of the remaining elements (largest % rest_sum).
Step 8: If the remainder is 0, return false because this would imply the element before the last replacement was 0, and all numbers are positive.
Step 9: Update the total sum and re-insert the previous value into the max heap.
Step 10: If the loop completes without returning false, it means the `target` array can be constructed, so return true.

Key Insights

Insight 1: Working backward is crucial. Instead of trying to build the `target` array from the initial array, it's more efficient to reduce the `target` array back to the initial array of 1s.
Insight 2: Using a max heap (priority queue) allows us to efficiently find the largest element in the `target` array at each step. This largest element was the last element to be modified in the construction process.
Insight 3: The largest element in the current `target` array must have been obtained by replacing some element with the sum of all elements in the array at the previous step. We can reverse this operation by subtracting the sum of all other elements from the largest element.

Complexity Analysis

Time Complexity: O(n + k log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Heap (Priority Queue).

Companies

Asked at: Quora.

Construct Target Array With Multiple Sums | LeetCode 1354

Hard

Array Heap (Priority Queue)

Time: O(n + k log n)Space: O(n)

Solution Code

Available in:

import java.util.PriorityQueue;

class Solution {
    public boolean isPossible(int[] target) {
        long total_sum = 0;
        int n = target.length;

        if (n == 1) {
            return target[0] == 1;
        }

        PriorityQueue<Integer> max_heap = new PriorityQueue<>((a, b) -> b - a);
        for (int num : target) {
            total_sum += num;
            max_heap.add(num);
        }

        while (max_heap.peek() > 1) {
            int largest = max_heap.poll();
            long rest_sum = total_sum - largest;

            if (rest_sum == 1) {
                return true;
            }

            if (rest_sum == 0 || largest <= rest_sum) {
                return false;
            }

            int prev_val = (int) (largest % rest_sum);

            if (prev_val == 0) {
                return false;
            }

            total_sum = total_sum - largest + prev_val;
            max_heap.add(prev_val);
        }

        return true;
    }
}

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Problem Info

Description

Problem 1354: Construct Target Array With Multiple Sums - Detailed explanation and solution approaches.

Topics

Array Heap (Priority Queue)

Companies

Quora

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1354

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

10 steps

1
Step 1: Calculate the sum of all elements in the `target` array.
2
Step 2: Build a max heap from the `target` array.
3
Step 3: While the largest element in the heap is greater than 1, perform the following steps:
4
Step 4: Extract the largest element from the heap.
5
Step 5: Calculate the sum of the remaining elements (total_sum - largest).
6
Step 6: Check for invalid states. If rest_sum is 0 or the largest element is less than or equal to rest_sum, return false. If rest_sum is 1, return true because all other elements must be 1.
7
Step 7: Calculate the previous value of the largest element by taking the modulo of the largest element with the sum of the remaining elements (largest % rest_sum).
8
Step 8: If the remainder is 0, return false because this would imply the element before the last replacement was 0, and all numbers are positive.
9
Step 9: Update the total sum and re-insert the previous value into the max heap.
10
Step 10: If the loop completes without returning false, it means the `target` array can be constructed, so return true.

💡

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(n + k log n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity comes from building the heap (O(n)) and then repeatedly extracting the maximum element and inserting a new element (O(log n) each). The space complexity is dominated by the size of the heap, which stores all elements of the `target` array.

📖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

Try all possible combinations of replacements to see if the target can be created. This is highly inefficient.

Time: ExponentialSpace: O(n)

Trade-offs: Not practical due to extremely high time complexity.

✨

Key Insights

1Insight 1: Working backward is crucial. Instead of trying to build the `target` array from the initial array, it's more efficient to reduce the `target` array back to the initial array of 1s.
2Insight 2: Using a max heap (priority queue) allows us to efficiently find the largest element in the `target` array at each step. This largest element was the last element to be modified in the construction process.
3Insight 3: The largest element in the current `target` array must have been obtained by replacing some element with the sum of all elements in the array at the previous step. We can reverse this operation by subtracting the sum of all other elements from the largest element.

📚

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: `target` array of size 1. The solution directly checks if the single element is equal to 1.
Edge case 2: `target` array contains elements that can be reduced to an invalid state (e.g., rest_sum becomes 0 or largest <= rest_sum or prev_val is 0). The solution explicitly checks for these invalid states within the while loop and returns `false` if encountered.
Edge case 3: large values in target can cause integer overflow if the sum is not stored in a long. The solution uses long for total_sum in java and cpp, and python handles large integers without overflow.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a max heap. Using a regular array or list would result in an O(n) operation to find the maximum element in each iteration, leading to a higher overall time complexity.

✗Common mistake 2: Incorrectly calculating the previous value. It's crucial to use the modulo operator (%) to ensure the previous value is within the correct range. Failing to account for situations where largest <= rest_sum will lead to an infinite loop or incorrect result.

✗Common mistake 3: Integer overflow. Using 'int' instead of 'long' to store the sum of target can cause overflow, leading to incorrect results, especially if the target array elements are large.

✅

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: Resource allocation. Imagine assigning resources to tasks where the initial allocation is minimal (1 unit each). The 'target' array represents the desired resource allocation. This algorithm can help determine if the desired allocation is achievable through a series of resource transfers.

🏦Real-world scenario 2: Dependency resolution. The problem can be viewed as resolving dependencies where each element represents a dependency and its value. Can a complex dependency graph be created from a base state of minimal dependencies (all 1s)? This algorithm can help analyze the feasibility.

💡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.

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 3, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

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

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Construct Target Array With Multiple Sums"?

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 heap (priority queue) techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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