What is the time complexity of Number of Great Partitions?

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

What data structures are used to solve Number of Great Partitions?

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

What are the key insights for solving Number of Great Partitions?

Insight 1: Realize that simply iterating through all possible partitions is inefficient due to the exponential number of possibilities (2^n). Insight 2: Employ dynamic programming to count the number of partitions where the sum of one group is *less than* k. This allows us to calculate the number of *bad* partitions and subtract it from the total number of partitions.

Which companies ask LeetCode 2518. Number of Great Partitions in interviews?

This problem is frequently asked at Darwinbox, Sprinklr.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Number of Great Partitions - LeetCode 2518 Solution

Number of Great Partitions - Complete Solution Guide

Number of Great Partitions is LeetCode problem 2518, 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

You are given an array nums consisting of positive integers and an integer k . Partition the array into two ordered groups such that each element is in exactly one group. A partition is called great if the sum of elements of each group is greater than or equal to k . Return the number of distinct great partitions . Since the answer may be too large, return it modulo 10 9 + 7 . Two partitions are considered distinct if some element nums[i] is in different groups in the two partitions. Example 1:

Detailed Explanation

The problem asks us to divide an array of positive integers `nums` into two non-empty groups, such that the sum of elements in each group is greater than or equal to a given value `k`. We need to find the number of distinct ways to perform this partitioning and return the result modulo 10^9 + 7. Two partitions are considered distinct if at least one element from `nums` is placed in different groups across the two partitions.

Solution Approach

The solution uses dynamic programming to efficiently count the number of subsets of `nums` that have a sum less than `k`. We create a DP table `dp` where `dp[i]` represents the number of subsets with a sum equal to `i`. We iterate through the `nums` array, and for each number, we update the `dp` table to include the possibility of adding that number to existing subsets. After calculating the number of subsets with a sum less than `k`, we use it to determine the number of 'bad' partitions. We subtract the number of bad partitions from the total number of possible partitions (2^n) to find the number of great partitions.

Step-by-Step Algorithm

Step 1: Calculate the total sum of the elements in the `nums` array.
Step 2: Check if the total sum is less than 2 * k. If it is, return 0 because no valid partitions are possible.
Step 3: Initialize a dynamic programming array `dp` of size `k` with all elements set to 0. Set `dp[0]` to 1, indicating that there is one way to have a sum of 0 (the empty set).
Step 4: Iterate through the `nums` array. For each number `num`, iterate through the `dp` array from `k - 1` down to `num`. Update `dp[i]` to be `(dp[i] + dp[i - num]) % MOD`. This step effectively adds `num` to all possible subset sums less than `k` and counts these new sums.
Step 5: Calculate the sum of all elements in the `dp` array. This represents the total number of subsets with a sum less than `k`.
Step 6: Calculate the total number of possible partitions, which is 2^n (where n is the length of the `nums` array), modulo MOD.
Step 7: Calculate the number of 'bad' partitions. Since we counted the number of subsets with sum less than k for only one group, we need to multiply this value by 2. Taking the modulo at each step
Step 8: Subtract the number of bad partitions from the total number of partitions and return the result modulo MOD. Adding MOD before subtracting ensures that the answer remains positive.

Key Insights

Insight 1: Realize that simply iterating through all possible partitions is inefficient due to the exponential number of possibilities (2^n).
Insight 2: Employ dynamic programming to count the number of partitions where the sum of one group is *less than* k. This allows us to calculate the number of *bad* partitions and subtract it from the total number of partitions.
Insight 3: The total number of ways to partition an array of size 'n' into two groups is 2^n. Since we want great partitions where both groups are >= k, we need to subtract 'bad' partitions from the total. However, we have counted each 'bad' partition twice (once for each group with sum < k), so we account for that by multiplying the number of 'less than k' sums by 2.
Insight 4: If the sum of all elements is less than 2*k, then there can be no great partitions since at least one of the groups will have a sum < k.

Complexity Analysis

Time Complexity: O(n*k)

Space Complexity: O(k)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Darwinbox, Sprinklr.

Number of Great Partitions | LeetCode 2518

Hard

Array Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int countPartitions(int[] nums, int k) {
        int MOD = 1_000_000_007;
        
        long total_sum = 0;
        for (int num : nums) {
            total_sum += num;
        }
        if (total_sum < 2 * k) {
            return 0;
        }
        
        int n = nums.length;
        
        long[] dp = new long[k];
        dp[0] = 1;
        
        for (int num : nums) {
            for (int i = k - 1; i >= num; i--) {
                dp[i] = (dp[i] + dp[i - num]) % MOD;
            }
        }
        
        long count_less_than_k = 0;
        for (long val : dp) {
            count_less_than_k = (count_less_than_k + val) % MOD;
        }
        
        long total_partitions = power(2, n, MOD);
        
        long bad_partitions = (2 * count_less_than_k) % MOD;
        
        long ans = (total_partitions - bad_partitions + MOD) % MOD;
        
        return (int) ans;
    }
    
    private long power(long base, long exp, long mod) {
        long res = 1;
        base %= mod;
        while (exp > 0) {
            if (exp % 2 == 1) res = (res * base) % mod;
            exp >>= 1;
            base = (base * base) % mod;
        }
        return res;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2518: Number of Great Partitions - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

DarwinboxSprinklr

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2518

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: Check if the total sum is less than 2 * k. If it is, return 0 because no valid partitions are possible.
3
Step 3: Initialize a dynamic programming array `dp` of size `k` with all elements set to 0. Set `dp[0]` to 1, indicating that there is one way to have a sum of 0 (the empty set).
4
Step 4: Iterate through the `nums` array. For each number `num`, iterate through the `dp` array from `k - 1` down to `num`. Update `dp[i]` to be `(dp[i] + dp[i - num]) % MOD`. This step effectively adds `num` to all possible subset sums less than `k` and counts these new sums.
5
Step 5: Calculate the sum of all elements in the `dp` array. This represents the total number of subsets with a sum less than `k`.
6
Step 6: Calculate the total number of possible partitions, which is 2^n (where n is the length of the `nums` array), modulo MOD.
7
Step 7: Calculate the number of 'bad' partitions. Since we counted the number of subsets with sum less than k for only one group, we need to multiply this value by 2. Taking the modulo at each step
8
Step 8: Subtract the number of bad partitions from the total number of partitions and return the result modulo MOD. Adding MOD before subtracting ensures that the answer remains positive.

💡

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*k)

💾

Space Complexity

O(k)

📝

Why?

The time complexity arises from iterating through the `nums` array of size 'n' and for each number, iterating through the `dp` array of size 'k'. The space complexity is due to the `dp` array storing the number of ways to achieve a sum less than k.

📖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

Brute-Force

Generate all possible partitions of the array and check if each partition is a great partition. This involves creating all 2^n subsets and evaluating their sums.

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

Trade-offs: This approach is very slow and impractical for larger arrays. Use only for very small inputs (n <= 20).

✨

Key Insights

1Insight 1: Realize that simply iterating through all possible partitions is inefficient due to the exponential number of possibilities (2^n).
2Insight 2: Employ dynamic programming to count the number of partitions where the sum of one group is *less than* k. This allows us to calculate the number of *bad* partitions and subtract it from the total number of partitions.
3Insight 3: The total number of ways to partition an array of size 'n' into two groups is 2^n. Since we want great partitions where both groups are >= k, we need to subtract 'bad' partitions from the total. However, we have counted each 'bad' partition twice (once for each group with sum < k), so we account for that by multiplying the number of 'less than k' sums by 2.
4Insight 4: If the sum of all elements is less than 2*k, then there can be no great partitions since at least one of the groups will have a sum < k.

📚

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: If the total sum of `nums` is less than 2*k, there are no great partitions. The code correctly handles this by returning 0.
Edge case 2: k=1. In this case, the dp table will be small, but since nums[i] >= 1, we only need to see how many ways we can get 0 as the sum. The total number of partitions less than k will thus be 1. The logic is handled correctly.
Edge case 3: nums contains large numbers, but still less than k. The code will handle it because the inner loop iterates from k-1 down to nums[i].

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the total number of partitions (2^n) without taking the modulo at each step, leading to integer overflow issues.

✗Common mistake 2: Failing to account for the fact that 'bad' partitions are counted twice (once for each group with a sum less than k). Need to multiply the 'bad' count by 2.

✗Common mistake 3: Not handling the case where the total sum of `nums` is less than 2*k.

✗Common mistake 4: Not taking the modulo in each step of DP calculations, leading to incorrect results when dealing with large numbers.

✅

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: Deciding how to split tasks/resources between two teams such that each team has a minimum workload/resource allocation.

🏦Real-world scenario 2: Load balancing: Distributing workload across two servers such that each server's load is above a certain threshold.

🎮Real-world scenario 3: Knapsack-like problems with constraints on the minimum value of each knapsack.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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 26, 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 13

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

@TechLeadJul 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."

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 "Number of Great Partitions"?

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

🤖 Need help with this problem?