What is the time complexity of Split Array Largest Sum?

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

What data structures are used to solve Split Array Largest Sum?

This problem uses Array, Binary Search, Dynamic Programming, Greedy, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Split Array Largest Sum?

Insight 1: The problem can be solved using binary search. The search space is the range of possible values for the maximum sum of a subarray, which lies between the largest element in `nums` and the sum of all elements in `nums`. Insight 2: A helper function `can_split` is crucial. This function checks if we can split the array into at most `k` subarrays given a certain maximum sum constraint. This allows us to efficiently evaluate whether a given `mid` value (from the binary search) is a feasible maximum sum.

Which companies ask LeetCode 410. Split Array Largest Sum in interviews?

This problem is frequently asked at Baidu, DE Shaw, MathWorks, PhonePe, Pinterest and 4 more companies.

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Split Array Largest Sum - LeetCode 410 Solution

Split Array Largest Sum - Complete Solution Guide

Split Array Largest Sum is LeetCode problem 410, 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

Given an integer array nums and an integer k , split nums into k non-empty subarrays such that the largest sum of any subarray is minimized . Return the minimized largest sum of the split . A subarray is a contiguous part of the array. Example 1: Input: nums = [7,2,5,10,8], k = 2 Output: 18 Explanation: There are four ways to split nums into two subarrays. The best way is to split it into [7,2,5] and [10,8], where the largest sum among the two subarrays is only 18. Example 2: Input: nums = [1,2,

Detailed Explanation

The problem asks us to split an integer array `nums` into `k` non-empty subarrays such that the largest sum among all the subarrays is minimized. In other words, we want to find a split where the maximum subarray sum is as small as possible. The input consists of the array `nums` and the integer `k`, representing the number of subarrays. The output should be the minimized largest sum among the `k` subarrays after splitting the array.

Solution Approach

The solution utilizes binary search to find the minimized largest sum of the split array. The binary search is performed on the range of possible sums, from the maximum individual element in `nums` to the total sum of `nums`. For each mid-point in the search, we use a helper function, `can_split`, to check if we can split the array into at most `k` subarrays, where the maximum sum of each subarray does not exceed `mid`. If we can split the array into at most k subarrays, it means `mid` is a possible largest sum, so we try a smaller largest sum by updating `high = mid - 1`. Otherwise, we need to try a larger largest sum, so we update `low = mid + 1`.

Step-by-Step Algorithm

Step 1: Initialize `low` to the maximum element in `nums` and `high` to the sum of all elements in `nums`. These define the range for the binary search.
Step 2: Initialize `ans` to `high`. This represents the current best possible answer. If the loop doesn't refine ans, this covers the scenario where only one group fulfills the requirements. This acts as an initial upper bound for the minimized largest sum.
Step 3: Perform binary search while `low <= high`.
Step 4: Calculate the `mid` value as `low + (high - low) // 2` to prevent potential integer overflow.
Step 5: Call the `can_split` function with `mid` to determine if the array can be split into at most `k` subarrays with the maximum sum of any subarray being at most `mid`.
Step 6: If `can_split(mid)` returns true, it means `mid` is a feasible maximum sum. Update `ans` to `mid` and search for a smaller possible sum by setting `high = mid - 1`.
Step 7: If `can_split(mid)` returns false, it means `mid` is too small, and we need to increase the possible sum. Update `low = mid + 1`.
Step 8: Repeat steps 3-7 until `low > high`.
Step 9: Return `ans`, which will hold the minimized largest sum.

Key Insights

Insight 1: The problem can be solved using binary search. The search space is the range of possible values for the maximum sum of a subarray, which lies between the largest element in `nums` and the sum of all elements in `nums`.
Insight 2: A helper function `can_split` is crucial. This function checks if we can split the array into at most `k` subarrays given a certain maximum sum constraint. This allows us to efficiently evaluate whether a given `mid` value (from the binary search) is a feasible maximum sum.
Insight 3: The core idea is to find the minimum possible largest sum. Binary search efficiently narrows down the possible largest sum by repeatedly dividing the search interval in half, until the smallest possible largest sum is found.

Complexity Analysis

Time Complexity: O(n*log(sum))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Dynamic Programming, Greedy, Prefix Sum.

Companies

Asked at: Baidu, DE Shaw, MathWorks, PhonePe, Pinterest, PornHub, Salesforce, Tekion, Zeta.

Split Array Largest Sum | LeetCode 410

Hard

Array Binary Search Dynamic Programming Greedy Prefix Sum

Time: O(n*log(sum))Space: O(1)

Solution Code

Available in:

class Solution {
    public int splitArray(int[] nums, int k) {
        
        // Helper function to check if we can split the array into at most k subarrays
        // such that the maximum sum of any subarray is at most max_sum.
        
        
        
        int low = 0;
        int high = 0;
        for(int num : nums){
            high += num;
            low = Math.max(low, num);
        }
        int ans = high;
        
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (can_split(nums, mid, k)) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        
        return ans;
    }
    
    private boolean can_split(int[] nums, int max_sum, int k) {
            int num_subarrays = 1;
            int current_sum = 0;
            for (int num : nums) {
                if (current_sum + num > max_sum) {
                    num_subarrays += 1;
                    current_sum = num;
                } else {
                    current_sum += num;
                }
            }
            return num_subarrays <= k;
        }
}

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Problem Info

Description

Problem 410: Split Array Largest Sum - Detailed explanation and solution approaches.

Topics

Array Binary Search Dynamic Programming Greedy Prefix Sum

Companies

BaiduDE ShawMathWorksPhonePePinterestPornHubSalesforceTekion

Stats

Difficulty:Hard

Languages:4

Companies:9

Problem #:410

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: Initialize `low` to the maximum element in `nums` and `high` to the sum of all elements in `nums`. These define the range for the binary search.
2
Step 2: Initialize `ans` to `high`. This represents the current best possible answer. If the loop doesn't refine ans, this covers the scenario where only one group fulfills the requirements. This acts as an initial upper bound for the minimized largest sum.
3
Step 3: Perform binary search while `low <= high`.
4
Step 4: Calculate the `mid` value as `low + (high - low) // 2` to prevent potential integer overflow.
5
Step 5: Call the `can_split` function with `mid` to determine if the array can be split into at most `k` subarrays with the maximum sum of any subarray being at most `mid`.
6
Step 6: If `can_split(mid)` returns true, it means `mid` is a feasible maximum sum. Update `ans` to `mid` and search for a smaller possible sum by setting `high = mid - 1`.
7
Step 7: If `can_split(mid)` returns false, it means `mid` is too small, and we need to increase the possible sum. Update `low = mid + 1`.
8
Step 8: Repeat steps 3-7 until `low > high`.
9
Step 9: Return `ans`, which will hold the minimized largest 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*log(sum))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The binary search takes O(log(sum)) time, where 'sum' is the difference between the maximum and minimum possible values for the minimized largest sum. For each 'mid' value during the binary search, we call the 'can_split' function, which iterates through the 'n' elements of the array once, taking O(n) time. Thus the overall time complexity is O(n * log(sum)). The algorithm uses a constant amount of extra space for variables such as low, high, mid, and the variables inside can_split.

📖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

Dynamic Programming

Dynamic programming can be used to solve this problem. We can define `dp[i][j]` as the minimum possible largest sum when splitting the first `i` elements into `j` subarrays. The transition is `dp[i][j] = min(dp[k][j-1], sum(nums[k+1...i])) for k in range(j-1, i)`. In other words, for each element i and number of splits j, iterate through k previous points to find best split point.

Time: O(n^2 * k)Space: O(n * k)

Trade-offs: Dynamic programming has a higher time complexity O(n^2 * k) than the binary search approach O(n * log(sum)). It also has a higher space complexity. DP may be preferred if constraints on n and k are small but should otherwise be avoided.

✨

Key Insights

1Insight 1: The problem can be solved using binary search. The search space is the range of possible values for the maximum sum of a subarray, which lies between the largest element in `nums` and the sum of all elements in `nums`.
2Insight 2: A helper function `can_split` is crucial. This function checks if we can split the array into at most `k` subarrays given a certain maximum sum constraint. This allows us to efficiently evaluate whether a given `mid` value (from the binary search) is a feasible maximum sum.
3Insight 3: The core idea is to find the minimum possible largest sum. Binary search efficiently narrows down the possible largest sum by repeatedly dividing the search interval in half, until the smallest possible largest sum is found.

📚

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!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `k = 1`. In this case, the entire array is one subarray, and the minimized largest sum is the sum of all elements in `nums`. The algorithm correctly handles this because the binary search eventually converges to this sum if no smaller value satisfies the constraint.
Edge case 2: `k = nums.length`. In this case, each element is a subarray, and the minimized largest sum is the maximum element in `nums`. The binary search algorithm correctly handles this as the low value is initialized as the max element. The binary search will then converge to a solution at or close to the max element.
Edge case 3: All elements in `nums` are zero. The algorithm still functions correctly. The low value will be zero, and the high value will be zero, so the binary search executes once at mid = 0 and returns the value zero.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly defining the search space for the binary search. The search space should range from the largest element in the array to the sum of all elements.

✗Common mistake 2: Not handling the edge cases where `k = 1` or `k = nums.length` correctly. These cases need to be accounted for in the logic or initialization of the binary search.

✗Common mistake 3: Integer overflow when calculating the mid value. Use `low + (high - low) // 2` instead of `(low + high) // 2` to prevent overflow.

✅

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: Task scheduling: Dividing a set of tasks with varying durations among a team of workers such that the maximum workload assigned to any worker is minimized.

🏦Real-world scenario 2: Load balancing: Distributing data across multiple servers to minimize the maximum load on any single server.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: July 21, 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 14

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

@AlgoMasterJul 7

❓

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 avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

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 "Split Array Largest Sum"?

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?