What is the time complexity of Partition Array Into Three Parts With Equal Sum?

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

What data structures are used to solve Partition Array Into Three Parts With Equal Sum?

This problem uses Array, Greedy. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Partition Array Into Three Parts With Equal Sum?

**Divisibility Check First:** The most fundamental prerequisite is that the entire array's sum *must* be perfectly divisible by 3. If `sum(arr) % 3 != 0`, it's mathematically impossible to partition it into three equal sum parts, so we can immediately short-circuit and return `false`. This initial check prunes a large portion of invalid inputs efficiently. **Greedy Sequential Accumulation with Reset:** The core strategy involves a single pass through the array, greedily accumulating `current_sum`. When `current_sum` hits the `target_sum` (which is `total_sum / 3`), we effectively "cut" that segment, increment our `count` of found parts, and crucially, *reset `current_sum` to 0*. This reset prepares us to find the *next* independent part from the point where the previous one ended.

Which companies ask LeetCode 1013. Partition Array Into Three Parts With Equal Sum in interviews?

This problem is frequently asked at Turing.

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Partition Array Into Three Parts With Equal Sum - LeetCode 1013 Solution

Partition Array Into Three Parts With Equal Sum - Complete Solution Guide

Partition Array Into Three Parts With Equal Sum is LeetCode problem 1013, a Easy 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 array of integers arr , return true if we can partition the array into three non-empty parts with equal sums. Formally, we can partition the array if we can find indexes i + 1 < j with (arr[0] + arr[1] + ... + arr[i] == arr[i + 1] + arr[i + 2] + ... + arr[j - 1] == arr[j] + arr[j + 1] + ... + arr[arr.length - 1]) Example 1: Input: arr = [0,2,1,-6,6,-7,9,1,2,0,1] Output: true Explanation: 0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1 Example 2: Input: arr = [0,2,1,-6,6,7,9,-1,2,0,1] Output:

Detailed Explanation

Imagine you have a long sequence of numbers, and your task is to chop it into exactly three distinct, contiguous sections. The catch? Each of these three sections must add up to the *exact same sum*. This is precisely what problem #1013 asks us to determine. We're given an array `arr` and need to return `true` if we can find two split points, `i` and `j`, such that the sum of `arr[0...i]`, `arr[i+1...j-1]`, and `arr[j...arr.length-1]` are all equal. Crucially, the problem specifies `i + 1 < j`, which means the middle section (`arr[i+1...j-1]`) cannot be empty; it must contain at least one element, ensuring proper separation between the first and third parts.

Solution Approach

The provided solution takes a remarkably straightforward and efficient greedy approach. First, it calculates the `total_sum` of all elements in the array. This is our foundation. If `total_sum` isn't perfectly divisible by 3, there's no way to achieve three equal parts, so we immediately return `false`. Otherwise, we calculate our `target_sum`, which is simply `total_sum // 3`. Now, the core of the algorithm iterates through the array, accumulating `current_sum` as it goes. The key insight is that we're essentially trying to "find" each of the three parts sequentially. As soon as `current_sum` equals our `target_sum`, we've successfully identified the end of one segment. We increment a `count` variable, signifying we've found one such part, and then reset `current_sum` back to 0. This reset is crucial: it effectively tells us to start accumulating for the *next* potential part from that point onwards. We continue this process until we've iterated through the entire array. Finally, we return `true` if `count >= 3`. Why `count >= 3` and not `count == 3`? Because if `target_sum` is 0 (e.g., `[0,0,0,0,0,0]`), you might find more than three segments that sum to 0. The crucial point is that if we successfully find *two* segments that each sum to `target_sum`, then by definition, the *remainder* of the array (from where the second segment ended to the very end) *must* also sum to `target_sum`. This is because `total_sum = target_sum + target_sum + (remaining_sum)`. Since `total_sum = 3 * target_sum`, it follows that `remaining_sum` must also be `target_sum`. So, by the time our loop has found `count` equal to 2, and then the sum reaches `target_sum` *again* (making `count` 3), it implicitly confirms that the remaining portion of the array also sums up correctly. The `count >= 3` then correctly handles cases like `[0,0,0,0,0,0]` where each `0` could technically be a part, and we just need *at least three* to form the required structure, while implicitly respecting the non-empty part constraint as each '0' is itself a valid non-empty part.

Step-by-Step Algorithm

Step 1: Calculate the total sum of the array.
Step 2: Check if the total sum is divisible by 3. If not, return `false`.
Step 3: Calculate the target sum (total sum / 3).
Step 4: Initialize a `currentSum` to 0 and a `count` to 0.
Step 5: Iterate through the array. Add each element to `currentSum`.
Step 6: If `currentSum` equals the `targetSum`, increment `count` and reset `currentSum` to 0.
Step 7: After iterating, check if `count` is greater than or equal to 3. If so, return `true`; otherwise, return `false`.

Key Insights

**Divisibility Check First:** The most fundamental prerequisite is that the entire array's sum *must* be perfectly divisible by 3. If `sum(arr) % 3 != 0`, it's mathematically impossible to partition it into three equal sum parts, so we can immediately short-circuit and return `false`. This initial check prunes a large portion of invalid inputs efficiently.
**Greedy Sequential Accumulation with Reset:** The core strategy involves a single pass through the array, greedily accumulating `current_sum`. When `current_sum` hits the `target_sum` (which is `total_sum / 3`), we effectively "cut" that segment, increment our `count` of found parts, and crucially, *reset `current_sum` to 0*. This reset prepares us to find the *next* independent part from the point where the previous one ended.
**Implicit Third Part Verification and `count >= 3`:** The algorithm doesn't explicitly search for the third part as a distinct operation. Instead, once two segments are found, the fact that `total_sum = 3 * target_sum` guarantees that the *remaining* part of the array must also sum to `target_sum`. The `count >= 3` condition is robust because it confirms we've found at least two valid split points, implicitly creating three valid sections. It also correctly handles scenarios where `target_sum` is zero, allowing for more than three segments to sum to zero, while still validating the minimal requirement.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy.

Companies

Asked at: Turing.

Partition Array Into Three Parts With Equal Sum | LeetCode 1013

Easy

Array Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean canThreePartsEqualSum(int[] arr) {
        int sum = 0;
        for (int num : arr) {
            sum += num;
        }
        if (sum % 3 != 0) {
            return false;
        }
        int target = sum / 3;
        int currentSum = 0;
        int count = 0;
        for (int i = 0; i < arr.length; i++) {
            currentSum += arr[i];
            if (currentSum == target) {
                count++;
                currentSum = 0;
            }
        }
        return count >= 3;
    }
}

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Problem Info

Description

Problem 1013: Partition Array Into Three Parts With Equal Sum - Detailed explanation and solution approaches.

Topics

Array Greedy

Companies

Turing

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1013

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: Calculate the total sum of the array.
2
Step 2: Check if the total sum is divisible by 3. If not, return `false`.
3
Step 3: Calculate the target sum (total sum / 3).
4
Step 4: Initialize a `currentSum` to 0 and a `count` to 0.
5
Step 5: Iterate through the array. Add each element to `currentSum`.
6
Step 6: If `currentSum` equals the `targetSum`, increment `count` and reset `currentSum` to 0.
7
Step 7: After iterating, check if `count` is greater than or equal to 3. If so, return `true`; otherwise, return `false`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the array once, performing constant-time operations in each iteration. It uses only a few constant-size variables, regardless of the array's size. Therefore, both time and space complexity are linear in the size of the input array (n).

📖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

Prefix Sum Array

Create a prefix sum array. Then iterate through the array looking for indices i and j where the differences prefixSum[j] - prefixSum[i] and prefixSum[arr.length-1] - prefixSum[j] equal the target sum. This approach avoids repeated summation.

Time: O(n)Space: O(n)

Trade-offs: This approach uses extra space (O(n) for the prefix sum array) but may improve readability. The greedy approach is generally preferred for its efficiency in space.

✨

Key Insights

1**Divisibility Check First:** The most fundamental prerequisite is that the entire array's sum *must* be perfectly divisible by 3. If `sum(arr) % 3 != 0`, it's mathematically impossible to partition it into three equal sum parts, so we can immediately short-circuit and return `false`. This initial check prunes a large portion of invalid inputs efficiently.
2**Greedy Sequential Accumulation with Reset:** The core strategy involves a single pass through the array, greedily accumulating `current_sum`. When `current_sum` hits the `target_sum` (which is `total_sum / 3`), we effectively "cut" that segment, increment our `count` of found parts, and crucially, *reset `current_sum` to 0*. This reset prepares us to find the *next* independent part from the point where the previous one ended.
3**Implicit Third Part Verification and `count >= 3`:** The algorithm doesn't explicitly search for the third part as a distinct operation. Instead, once two segments are found, the fact that `total_sum = 3 * target_sum` guarantees that the *remaining* part of the array must also sum to `target_sum`. The `count >= 3` condition is robust because it confirms we've found at least two valid split points, implicitly creating three valid sections. It also correctly handles scenarios where `target_sum` is zero, allowing for more than three segments to sum to zero, while still validating the minimal requirement.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: The array contains all zeros. The solution correctly handles this by returning `true` because three empty sub-arrays (technically, not allowed by problem statement) would satisfy the condition, though the problem does imply non-empty.
Edge case 2: The sum of the array is not divisible by 3. The solution correctly returns `false` in this case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if the total sum is divisible by 3 before proceeding. This leads to incorrect results.

✗Common mistake 2: Incorrectly handling the case where the target sum is reached multiple times consecutively. The counter needs to be incremented only once for each instance of reaching the target sum.

✅

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 dividing a set of resources (represented by numbers) into three teams with equal value.

🏦Real-world scenario 2: Data balancing across servers or databases. Distributing data across storage units to maintain even load and prevent bottlenecks.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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 22, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

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

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

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.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Partition Array Into Three Parts With Equal Sum"?

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

🤖 Need help with this problem?