What is the time complexity of Ways to Split Array Into Three Subarrays?

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

What data structures are used to solve Ways to Split Array Into Three Subarrays?

This problem uses Array, Two Pointers, Binary Search, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Ways to Split Array Into Three Subarrays?

Insight 1: Prefix sums are crucial for efficiently calculating the sums of subarrays in O(1) time. This avoids repeatedly summing array elements. Insight 2: The two-pointer technique (or sliding window concept) can be used to efficiently find the valid range of split points for the 'mid' subarray. This avoids nested loops and optimizes the time complexity.

Which companies ask LeetCode 1712. Ways to Split Array Into Three Subarrays in interviews?

This problem is frequently asked at Robinhood, Tekion.

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Ways to Split Array Into Three Subarrays - LeetCode 1712 Solution

Ways to Split Array Into Three Subarrays - Complete Solution Guide

Ways to Split Array Into Three Subarrays is LeetCode problem 1712, 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

A split of an integer array is good if: The array is split into three non-empty contiguous subarrays - named left , mid , right respectively from left to right. The sum of the elements in left is less than or equal to the sum of the elements in mid , and the sum of the elements in mid is less than or equal to the sum of the elements in right . Given nums , an array of non-negative integers, return the number of good ways to split nums . As the number may be too large, return it modulo 10 9 + 7 .

Detailed Explanation

The problem asks us to count the number of 'good' ways to split a given array `nums` into three non-empty contiguous subarrays: `left`, `mid`, and `right`. A split is considered 'good' if the sum of elements in `left` is less than or equal to the sum of elements in `mid`, and the sum of elements in `mid` is less than or equal to the sum of elements in `right`. The input is an array `nums` of non-negative integers. The output is the number of good splits, modulo 10^9 + 7. The constraints are that the array has a length between 3 and 10^5, and each element is between 0 and 10^4.

Solution Approach

The solution uses a prefix sum array to efficiently calculate the sum of any subarray. It iterates through all possible starting positions of the `left` subarray. For each `left` subarray, it uses two pointers, `left_j` and `right_j`, to find the valid range of starting positions for the `mid` subarray. `left_j` represents the earliest possible start of the `mid` subarray such that sum(left) <= sum(mid). `right_j` represents the latest possible start of the `mid` subarray such that sum(mid) <= sum(right). The number of valid splits for the current `left` subarray is then `right_j - left_j`. The final answer is the sum of these counts, modulo 10^9 + 7.

Step-by-Step Algorithm

Step 1: Calculate the prefix sum array `prefix`. `prefix[i]` stores the sum of the first `i` elements of `nums`.
Step 2: Initialize `ans` to 0. This variable will store the number of good splits.
Step 3: Initialize `left_j` and `right_j` to 0. These are the pointers for the start of the mid subarray.
Step 4: Iterate through all possible starting positions `i` of the `left` subarray (from 0 to n-3).
Step 5: Calculate the sum of the `left` subarray: `s_left = prefix[i + 1]`.
Step 6: Find the lower bound `left_j` such that `sum(mid) >= sum(left)`. Increment `left_j` as long as `prefix[left_j + 1] < 2 * s_left` and `left_j < n - 1`. Ensure that `left_j` is at least `i + 1` to guarantee `mid` is non-empty.
Step 7: Find the upper bound `right_j` such that `sum(mid) <= sum(right)`. Increment `right_j` as long as `2 * prefix[right_j + 1] <= prefix[n] + s_left` and `right_j < n - 1`. Ensure that `right_j` is at least `left_j`.
Step 8: The number of valid `mid` subarrays for the current `left` is `right_j - left_j`. Add this count to `ans` modulo 10^9 + 7.
Step 9: Return `ans`.

Key Insights

Insight 1: Prefix sums are crucial for efficiently calculating the sums of subarrays in O(1) time. This avoids repeatedly summing array elements.
Insight 2: The two-pointer technique (or sliding window concept) can be used to efficiently find the valid range of split points for the 'mid' subarray. This avoids nested loops and optimizes the time complexity.
Insight 3: The problem involves counting combinations, and given the constraints, the result may exceed the maximum integer value. Therefore, the modulo operator is essential to prevent overflow.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Two Pointers, Binary Search, Prefix Sum.

Companies

Asked at: Robinhood, Tekion.

Ways to Split Array Into Three Subarrays | LeetCode 1712

Medium

Array Two Pointers Binary Search Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int waysToSplit(int[] nums) {
        int n = nums.length;
        int MOD = (int)1e9 + 7;
        
        int[] prefix = new int[n + 1];
        for (int i = 0; i < n; i++) {
            prefix[i+1] = prefix[i] + nums[i];
        }
        
        int ans = 0;
        int left_j = 0, right_j = 0;
        
        for (int i = 0; i < n - 2; i++) {
            int s_left = prefix[i + 1];
            
            // Find the lower bound for the second split point j (left_j)
            // This is the first j where sum(mid) >= sum(left)
            // j must be at least i + 1
            left_j = Math.max(left_j, i + 1);
            while (left_j < n - 1 && (long)prefix[left_j + 1] < 2 * (long)s_left) {
                left_j += 1;
            }
            
            // Find the upper bound for the second split point j (right_j is exclusive)
            // This is the first j where sum(mid) > sum(right)
            // j must be at least left_j
            right_j = Math.max(right_j, left_j);
            while (right_j < n - 1 && 2 * (long)prefix[right_j + 1] <= (long)prefix[n] + (long)s_left) {
                right_j += 1;
            }
            
            // Valid j's for the current i are in the range [left_j, right_j - 1]
            // j must also be <= n - 2, which is implicitly handled by the loop bounds.
            int count = Math.max(0, right_j - left_j);
            ans = (ans + count) % MOD;
        }
        
        return ans;
    }
}

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Problem Info

Description

Problem 1712: Ways to Split Array Into Three Subarrays - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Prefix Sum

Companies

RobinhoodTekion

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1712

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: Calculate the prefix sum array `prefix`. `prefix[i]` stores the sum of the first `i` elements of `nums`.
2
Step 2: Initialize `ans` to 0. This variable will store the number of good splits.
3
Step 3: Initialize `left_j` and `right_j` to 0. These are the pointers for the start of the mid subarray.
4
Step 4: Iterate through all possible starting positions `i` of the `left` subarray (from 0 to n-3).
5
Step 5: Calculate the sum of the `left` subarray: `s_left = prefix[i + 1]`.
6
Step 6: Find the lower bound `left_j` such that `sum(mid) >= sum(left)`. Increment `left_j` as long as `prefix[left_j + 1] < 2 * s_left` and `left_j < n - 1`. Ensure that `left_j` is at least `i + 1` to guarantee `mid` is non-empty.
7
Step 7: Find the upper bound `right_j` such that `sum(mid) <= sum(right)`. Increment `right_j` as long as `2 * prefix[right_j + 1] <= prefix[n] + s_left` and `right_j < n - 1`. Ensure that `right_j` is at least `left_j`.
8
Step 8: The number of valid `mid` subarrays for the current `left` is `right_j - left_j`. Add this count to `ans` modulo 10^9 + 7.
9
Step 9: Return `ans`.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because we iterate through the array once to build the prefix sum, and then iterate through the array again while using two pointers (`left_j`, `right_j`) that move monotonically forward, each potentially traversing the array at most once in total. The space complexity is linear because we store the prefix sums in an array of size n+1.

📖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

Binary Search

Instead of using two pointers to find the lower and upper bounds for the 'mid' subarray, we could use binary search. For each 'i' (left subarray end position), we can use binary search to find the smallest 'j' such that sum(mid) >= sum(left), and the largest 'j' such that sum(mid) <= sum(right).

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

Trade-offs: Binary search would provide a logarithmic search within each iteration of the main loop (n iterations). This is slower than the two-pointer approach with overall linear time complexity. The two-pointer approach exploits the monotonic nature of prefix sums. The space complexity remains O(n) due to the prefix sum array. Binary search might be preferable if the constraints significantly increased the array size, making memory access more of a bottleneck than computational time.

✨

Key Insights

1Insight 1: Prefix sums are crucial for efficiently calculating the sums of subarrays in O(1) time. This avoids repeatedly summing array elements.
2Insight 2: The two-pointer technique (or sliding window concept) can be used to efficiently find the valid range of split points for the 'mid' subarray. This avoids nested loops and optimizes the time complexity.
3Insight 3: The problem involves counting combinations, and given the constraints, the result may exceed the maximum integer value. Therefore, the modulo operator is essential to prevent overflow.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

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

Edge Cases

Edge case 1: `nums` has a length of 3. The algorithm handles this correctly because the outer loop `for i in range(n - 2)` will run once. If there's a good split, it's counted; otherwise, `ans` remains 0.
Edge case 2: All elements in `nums` are 0. In this case, the prefix sum will also be all 0s. The conditions `sum(left) <= sum(mid) <= sum(right)` will always be true. The loops will correctly count the valid splits because left_j and right_j will correctly converge to valid values. No special handling is needed.
Edge case 3: An array that cannot be split in a valid way. Example: [3, 2, 1]. The algorithm handles this by incrementing `left_j` and `right_j` until they violate the loop condition, resulting in a count of 0 for that specific 'i' which correctly propagates to the final 'ans'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. The problem requires returning the result modulo 10^9 + 7. Forgetting to apply the modulo operator at each step (especially when adding `count` to `ans`) can lead to incorrect results due to integer overflow.

✗Common mistake 2: Incorrectly calculating subarray sums. Using the prefix sum array incorrectly (e.g., off-by-one errors when calculating the sum of a subarray) is a common mistake. Double-checking the indices used to access the prefix sum array is crucial.

✗Common mistake 3: Not considering the non-empty subarray constraint. The code needs to guarantee that `left`, `mid`, and `right` are non-empty. Failing to appropriately update `left_j` to at least `i+1` may violate this condition.

✅

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: Load balancing. Imagine dividing a server's workload into three parts such that each part is roughly equally loaded. This algorithm can help determine the number of ways to distribute the load such that no part is overloaded compared to the others.

🏦Real-world scenario 2: Resource allocation. Consider allocating resources (e.g., budget, manpower) to three different projects. The algorithm can help in determining the number of different allocations that satisfy certain fairness criteria (e.g., no project receives significantly less resource than the others).

💡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: September 1, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 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.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

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 "Ways to Split Array Into Three Subarrays"?

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

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