What is the time complexity of Partition to K Equal Sum Subsets?

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

What data structures are used to solve Partition to K Equal Sum Subsets?

This problem uses Array, Dynamic Programming, Backtracking, Bit Manipulation, Memoization, Bitmask. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Partition to K Equal Sum Subsets?

Insight 1: The sum of all elements in `nums` must be divisible by `k` for a valid partition to exist. The target sum for each subset is then the total sum divided by `k`. Insight 2: Backtracking is an effective approach because it explores different combinations of elements to form subsets, allowing us to find a valid partition if one exists.

Which companies ask LeetCode 698. Partition to K Equal Sum Subsets in interviews?

This problem is frequently asked at ByteDance, LinkedIn.

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Partition to K Equal Sum Subsets - LeetCode 698 Solution

Q: How to solve Partition to K Equal Sum Subsets?

The provided code uses a backtracking algorithm to solve the problem. First, it checks if the total sum of the numbers is divisible by `k`. If not, it immediately returns `false`. Otherwise, it calculates the target sum for each subset. The array is then sorted in descending order to prioritize placing larger numbers first, potentially pruning search space early on. A recursive `backtrack` function is used to explore possible subset combinations. For each number in the array, the function attempts to place it into one of the `k` subsets, ensuring that the subset's sum does not exceed the target sum. If a valid partition is found, the function returns `true`. If all possibilities are exhausted without finding a valid partition, it returns `false`.

Partition to K Equal Sum Subsets - Complete Solution Guide

Partition to K Equal Sum Subsets is LeetCode problem 698, 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 and an integer k , return true if it is possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4,3,2,3,5,2,1], k = 4 Output: true Explanation: It is possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Example 2: Input: nums = [1,2,3,4], k = 3 Output: false Constraints: 1 <= k <= nums.length <= 16 1 <= nums[i] <= 10 4 The frequency of each element is in the range [1, 4] .

Detailed Explanation

The problem asks us to determine if a given array of integers `nums` can be partitioned into `k` non-empty subsets, such that the sum of the elements in each subset is equal. The array `nums` and the number of subsets `k` are the inputs. The output should be a boolean value: `true` if such a partition is possible, and `false` otherwise. The constraints specify the ranges for `k`, the length of `nums`, and the values within `nums`.

Solution Approach

The provided code uses a backtracking algorithm to solve the problem. First, it checks if the total sum of the numbers is divisible by `k`. If not, it immediately returns `false`. Otherwise, it calculates the target sum for each subset. The array is then sorted in descending order to prioritize placing larger numbers first, potentially pruning search space early on. A recursive `backtrack` function is used to explore possible subset combinations. For each number in the array, the function attempts to place it into one of the `k` subsets, ensuring that the subset's sum does not exceed the target sum. If a valid partition is found, the function returns `true`. If all possibilities are exhausted without finding a valid partition, it returns `false`.

Step-by-Step Algorithm

Step 1: Calculate the total sum of the elements in `nums`.
Step 2: Check if the total sum is divisible by `k`. If not, return `false`.
Step 3: Calculate the target sum for each subset (total_sum / k).
Step 4: Sort the `nums` array in descending order.
Step 5: Initialize an array `subsets` of size `k` to store the current sum of each subset, initialized to 0.
Step 6: Define a recursive `backtrack` function that takes an index `index` as input, representing the current number being considered.
Step 7: Base case: If `index` is equal to the length of `nums`, it means all numbers have been placed into subsets. Return `true`.
Step 8: In the `backtrack` function, iterate through each subset `i` from 0 to `k-1`.
Step 9: For each subset `i`, check if adding the current number `nums[index]` to the subset's sum (`subsets[i]`) would exceed the `target`. If not, add the number to the subset.
Step 10: Recursively call `backtrack` with `index + 1` to explore the next number.
Step 11: If the recursive call returns `true`, it means a valid partition has been found. Return `true`.
Step 12: If the recursive call returns `false`, backtrack by removing the number from the subset (`subsets[i] -= nums[index]`).
Step 13: Optimize by breaking if the current bucket is empty after backtracking in the current iteration of the for loop
Step 14: If no valid subset is found for the current number, return `false`.

Key Insights

Insight 1: The sum of all elements in `nums` must be divisible by `k` for a valid partition to exist. The target sum for each subset is then the total sum divided by `k`.
Insight 2: Backtracking is an effective approach because it explores different combinations of elements to form subsets, allowing us to find a valid partition if one exists.
Insight 3: Sorting the input array in descending order can improve performance by prioritizing larger numbers, which can lead to faster pruning of invalid branches in the backtracking search.
Insight 4: Optimization to break if a bucket is empty after the subtraction of a number in backtrack. This avoids trying duplicate configurations.

Complexity Analysis

Time Complexity: O(k^(n-k))

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Backtracking, Bit Manipulation, Memoization, Bitmask.

Companies

Asked at: ByteDance, LinkedIn.

Partition to K Equal Sum Subsets | LeetCode 698

Medium

Array Dynamic Programming Backtracking Bit Manipulation Memoization Bitmask

Time: O(k^(n-k))Space: O(n)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public boolean canPartitionKSubsets(int[] nums, int k) {
        int totalSum = 0;
        for (int num : nums) {
            totalSum += num;
        }

        if (totalSum % k != 0) {
            return false;
        }

        int target = totalSum / k;
        int n = nums.length;

        Arrays.sort(nums);
        for (int i = 0; i < n / 2; i++) {
            int temp = nums[i];
            nums[i] = nums[n - 1 - i];
            nums[n - 1 - i] = temp;
        }

        if (nums[0] > target) {
            return false;
        }

        int[] subsets = new int[k];

        return backtrack(nums, k, 0, subsets, target);
    }

    private boolean backtrack(int[] nums, int k, int index, int[] subsets, int target) {
        if (index == nums.length) {
            return true;
        }

        int numToPlace = nums[index];

        for (int i = 0; i < k; i++) {
            if (subsets[i] + numToPlace <= target) {
                subsets[i] += numToPlace;

                if (backtrack(nums, k, index + 1, subsets, target)) {
                    return true;
                }

                subsets[i] -= numToPlace;

            }

            if (subsets[i] == 0) {
                break;
            }
        }

        return false;
    }
}

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Problem Info

Description

Problem 698: Partition to K Equal Sum Subsets - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Backtracking Bit Manipulation Memoization Bitmask

Companies

ByteDanceLinkedIn

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:698

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

14 steps

1
Step 1: Calculate the total sum of the elements in `nums`.
2
Step 2: Check if the total sum is divisible by `k`. If not, return `false`.
3
Step 3: Calculate the target sum for each subset (total_sum / k).
4
Step 4: Sort the `nums` array in descending order.
5
Step 5: Initialize an array `subsets` of size `k` to store the current sum of each subset, initialized to 0.
6
Step 6: Define a recursive `backtrack` function that takes an index `index` as input, representing the current number being considered.
7
Step 7: Base case: If `index` is equal to the length of `nums`, it means all numbers have been placed into subsets. Return `true`.
8
Step 8: In the `backtrack` function, iterate through each subset `i` from 0 to `k-1`.
9
Step 9: For each subset `i`, check if adding the current number `nums[index]` to the subset's sum (`subsets[i]`) would exceed the `target`. If not, add the number to the subset.
10
Step 10: Recursively call `backtrack` with `index + 1` to explore the next number.
11
Step 11: If the recursive call returns `true`, it means a valid partition has been found. Return `true`.
12
Step 12: If the recursive call returns `false`, backtrack by removing the number from the subset (`subsets[i] -= nums[index]`).
13
Step 13: Optimize by breaking if the current bucket is empty after backtracking in the current iteration of the for loop
14
Step 14: If no valid subset is found for the current number, 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(k^(n-k))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived from the branching factor of `k` in the backtracking search, and the maximum depth of the search tree is related to `n`. The space complexity is determined by the call stack depth of the recursive backtracking and the auxiliary space used to store subset sums.

📖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 with Bitmasking

This approach represents the subsets as bitmasks, where each bit corresponds to an element in the array. A DP table stores whether a particular sum can be achieved using a specific bitmask. It's particularly useful when k is not very large.

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

Trade-offs: DP with bitmasking may be faster for smaller 'n' or when memoization is highly effective but will use more space. Backtracking is often easier to implement and may be more efficient for certain input distributions.

✨

Key Insights

1Insight 1: The sum of all elements in `nums` must be divisible by `k` for a valid partition to exist. The target sum for each subset is then the total sum divided by `k`.
2Insight 2: Backtracking is an effective approach because it explores different combinations of elements to form subsets, allowing us to find a valid partition if one exists.
3Insight 3: Sorting the input array in descending order can improve performance by prioritizing larger numbers, which can lead to faster pruning of invalid branches in the backtracking search.
4Insight 4: Optimization to break if a bucket is empty after the subtraction of a number in backtrack. This avoids trying duplicate configurations.

📚

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 empty. The algorithm should handle this case gracefully, likely returning `false` as it is impossible to partition an empty array into non-empty subsets. The code implicitly handles this, where `n` becomes 0, and backtrack will check if index == n immediately return true, which then returns true to the main function. Since total_sum is also zero, target is 0, and nums[0] > target will be false, and it proceeds to backtrack
Edge case 2: `k` is 1. In this case, the entire array forms a single subset. The algorithm correctly returns `true` if the total sum is divisible by `k`, which is always the case when `k` is 1. Also the code does short-circuit so it avoids the whole backtrack section if k is 1
Edge case 3: `k` is equal to the length of `nums`. In this case, each number must form its own subset. The algorithm checks that each number is less than or equal to the target sum. Because the elements are sorted in descending order, if the first number is bigger than the target, the answer must be false.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if the total sum is divisible by `k` before proceeding. This can lead to unnecessary computation and incorrect results.

✗Common mistake 2: Not sorting the array in descending order. Sorting can significantly improve performance by pruning the search space more effectively.

✗Common mistake 3: Not understanding the base case for the recursion, leading to incorrect termination or infinite loops.

✗Common mistake 4: Failing to properly reset the state of the subsets array during backtracking (i.e., not subtracting the number after an unsuccessful attempt), causing incorrect results.

✅

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 distributing tasks (represented by their effort/time) among team members (k). The algorithm helps determine if it's possible to divide the tasks so each team member gets an equal amount of work.

🏦Real-world scenario 2: Load balancing: Distributing workloads across servers to ensure even utilization. `nums` would represent individual workload sizes, and `k` would be the number of servers.

💡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: November 6, 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

❓

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 to K Equal Sum Subsets"?

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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