What is the time complexity of Partition Equal Subset Sum?

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

What data structures are used to solve Partition Equal Subset Sum?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Partition Equal Subset Sum?

Insight 1: The problem can be reduced to finding a subset that sums up to half of the total sum of the array. If the total sum is odd, a partition into two equal subsets is impossible. Insight 2: Dynamic programming provides an efficient way to determine if a subset with a specific sum exists.

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Partition Equal Subset Sum - LeetCode 416 Solution

Q: How to solve Partition Equal Subset Sum?

The provided solution utilizes dynamic programming to determine if a subset with a sum equal to half the total sum of the array exists. First, the total sum of the array is computed. If the sum is odd, the function immediately returns `false` because an equal partition is impossible. If the sum is even, the target sum is calculated as half of the total sum. A boolean array `dp` is initialized, where `dp[i]` represents whether a subset exists that sums to `i`. The base case `dp[0]` is set to `true` because an empty subset always sums to 0. Then, the code iterates through each number in the input array. For each number, it iterates backward through the `dp` array from the target sum down to the current number. If `dp[j - num]` is `true`, it means that a subset already exists that sums to `j - num`. By including the current number, a subset with a sum of `j` can be formed. Therefore, `dp[j]` is set to `true`. Finally, the function returns `dp[target]`, which indicates whether a subset exists that sums to the target sum (half of the total sum).

Q: Which companies ask LeetCode 416. Partition Equal Subset Sum in interviews?

This problem is frequently asked at Dream11, EPAM Systems, Flipkart, IBM, ServiceNow and 4 more companies.

Partition Equal Subset Sum - Complete Solution Guide

Partition Equal Subset Sum is LeetCode problem 416, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an integer array nums , return true if you can partition the array into two subsets such that the sum of the elements in both subsets is equal or false otherwise . Example 1: Input: nums = [1,5,11,5] Output: true Explanation: The array can be partitioned as [1, 5, 5] and [11]. Example 2: Input: nums = [1,2,3,5] Output: false Explanation: The array cannot be partitioned into equal sum subsets. Constraints: 1 <= nums.length <= 200 1 <= nums[i] <= 100

Detailed Explanation

The problem asks whether a given array of integers can be partitioned into two subsets such that the sum of the elements in each subset is equal. In other words, can we divide the array into two groups where the sum of the numbers in group A is the same as the sum of the numbers in group B? The input is an integer array `nums`. The output is a boolean value, `true` if such a partition is possible, and `false` otherwise. The constraints state that the array length is between 1 and 200, and each element in the array is between 1 and 100.

Solution Approach

The provided solution utilizes dynamic programming to determine if a subset with a sum equal to half the total sum of the array exists. First, the total sum of the array is computed. If the sum is odd, the function immediately returns `false` because an equal partition is impossible. If the sum is even, the target sum is calculated as half of the total sum. A boolean array `dp` is initialized, where `dp[i]` represents whether a subset exists that sums to `i`. The base case `dp[0]` is set to `true` because an empty subset always sums to 0. Then, the code iterates through each number in the input array. For each number, it iterates backward through the `dp` array from the target sum down to the current number. If `dp[j - num]` is `true`, it means that a subset already exists that sums to `j - num`. By including the current number, a subset with a sum of `j` can be formed. Therefore, `dp[j]` is set to `true`. Finally, the function returns `dp[target]`, which indicates whether a subset exists that sums to the target sum (half of the total sum).

Step-by-Step Algorithm

Step 1: Calculate the total sum of the elements in the `nums` array.
Step 2: If the total sum is odd, return `false` because the array cannot be partitioned into two equal subsets.
Step 3: Calculate the target sum by dividing the total sum by 2.
Step 4: Initialize a boolean array `dp` of size `target + 1`, where `dp[i]` represents whether a subset exists with a sum of `i`. Initialize all elements to `false` except `dp[0]`, which is set to `true`.
Step 5: Iterate through each number `num` in the `nums` array.
Step 6: For each `num`, iterate backward through the `dp` array from `target` down to `num` (inclusive).
Step 7: If `dp[j - num]` is `true`, it means a subset with a sum of `j - num` already exists. Set `dp[j]` to `true` to indicate that a subset with a sum of `j` can be formed by including the current `num`.
Step 8: After iterating through all the numbers, return `dp[target]`. This value indicates whether there exists a subset with a sum equal to the target (half of the total sum).

Key Insights

Insight 1: The problem can be reduced to finding a subset that sums up to half of the total sum of the array. If the total sum is odd, a partition into two equal subsets is impossible.
Insight 2: Dynamic programming provides an efficient way to determine if a subset with a specific sum exists.
Insight 3: The DP table can be optimized to use only a one-dimensional array, as we only need the previous row to compute the current row.

Complexity Analysis

Time Complexity: O(n*t)

Space Complexity: O(t)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Dream11, EPAM Systems, Flipkart, IBM, ServiceNow, Visa, Walmart Labs, Zoho, eBay.

Partition Equal Subset Sum | LeetCode 416

Medium

Array Dynamic Programming

Time: O(n*t)Space: O(t)

Solution Code

Available in:

class Solution {
    public boolean canPartition(int[] nums) {
        int total_sum = 0;
        for (int num : nums) {
            total_sum += num;
        }

        if (total_sum % 2 != 0) {
            return false;
        }

        int target = total_sum / 2;

        boolean[] dp = new boolean[target + 1];
        dp[0] = true;

        for (int num : nums) {
            for (int j = target; j >= num; j--) {
                dp[j] = dp[j] || dp[j - num];
            }
        }

        return dp[target];
    }
}

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Problem Info

Description

Problem 416: Partition Equal Subset Sum - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Dream11EPAM SystemsFlipkartIBMServiceNowVisaWalmart LabsZoho

Stats

Difficulty:Medium

Languages:4

Companies:9

Problem #:416

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

8 steps

1
Step 1: Calculate the total sum of the elements in the `nums` array.
2
Step 2: If the total sum is odd, return `false` because the array cannot be partitioned into two equal subsets.
3
Step 3: Calculate the target sum by dividing the total sum by 2.
4
Step 4: Initialize a boolean array `dp` of size `target + 1`, where `dp[i]` represents whether a subset exists with a sum of `i`. Initialize all elements to `false` except `dp[0]`, which is set to `true`.
5
Step 5: Iterate through each number `num` in the `nums` array.
6
Step 6: For each `num`, iterate backward through the `dp` array from `target` down to `num` (inclusive).
7
Step 7: If `dp[j - num]` is `true`, it means a subset with a sum of `j - num` already exists. Set `dp[j]` to `true` to indicate that a subset with a sum of `j` can be formed by including the current `num`.
8
Step 8: After iterating through all the numbers, return `dp[target]`. This value indicates whether there exists a subset with a sum equal to the target (half of the total sum).

💡

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*t)

💾

Space Complexity

O(t)

📝

Why?

The time complexity is determined by the nested loops used in the dynamic programming approach. The space complexity is determined by the size of the `dp` array, which stores the results of subproblems. The space complexity could be improved to O(1) if we modify the input array in place but this is usually not allowed.

📖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

Recursive Approach (with Memoization)

A recursive function can be used to explore all possible subsets. Memoization can be applied to store the results of subproblems and avoid redundant computations.

Time: O(n*t), where n is the number of elements in the input array and t is the target sum.Space: O(n*t), due to the memoization table and recursion stack. While the runtime is the same, the constant factor is generally worse than the iterative approach.

Trade-offs: The recursive approach with memoization can be more intuitive for some, but it often has a higher overhead due to function calls and recursion stack management compared to the iterative DP solution. The iterative DP solution is preferred for its efficiency.

✨

Key Insights

1Insight 1: The problem can be reduced to finding a subset that sums up to half of the total sum of the array. If the total sum is odd, a partition into two equal subsets is impossible.
2Insight 2: Dynamic programming provides an efficient way to determine if a subset with a specific sum exists.
3Insight 3: The DP table can be optimized to use only a one-dimensional array, as we only need the previous row to compute the current row.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `nums` is an empty array. The problem statement specifies `1 <= nums.length <= 200`, so this edge case isn't relevant according to constraints. However, if the constraint were different, you'd have to consider this and might return true because two empty arrays sum to 0.
Edge case 2: `nums` contains only one element. In this case, it is impossible to create two subsets with an equal sum, unless the sum of that one element is zero (which is not allowed according to the problem's constraints 1 <= nums[i] <= 100). The algorithm returns `false` correctly.
Edge case 3: The sum of `nums` is odd. The algorithm correctly identifies this case and immediately returns `false`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if the total sum is odd before proceeding with the dynamic programming approach. This leads to incorrect results and unnecessary computations.

✗Common mistake 2: Incorrectly initializing the `dp` array. `dp[0]` must be initialized to `true`, as an empty subset always sums to 0.

✗Common mistake 3: Iterating in the wrong order in the inner loop. The inner loop must iterate backwards to avoid overwriting previously computed values. If iterating forward, the same number could be used multiple times, leading to incorrect results.

✗Common mistake 4: Getting confused with indices and sum values while calculating the dp table.

✅

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 - Determining if a set of tasks can be divided into two equally balanced groups based on their resource requirements.

🏦Real-world scenario 2: Knapsack problem - This problem is closely related to the 0/1 knapsack problem, which has applications in resource allocation, investment decisions, and cryptography.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: July 27, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Partition Equal Subset Sum"?

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

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