What is the time complexity of Minimum Elements to Add to Form a Given Sum?

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

What data structures are used to solve Minimum Elements to Add to Form a Given Sum?

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

What are the key insights for solving Minimum Elements to Add to Form a Given Sum?

Insight 1: The core idea is to calculate the difference between the current sum of the array and the target `goal`. This difference tells us how much we need to adjust the sum by. Insight 2: Since we can only add numbers within the range [-limit, limit], we want to use the largest possible absolute value (i.e., either `limit` or `-limit`) to minimize the number of additions.

Which companies ask LeetCode 1785. Minimum Elements to Add to Form a Given Sum in interviews?

This problem is frequently asked at X.

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Minimum Elements to Add to Form a Given Sum - LeetCode 1785 Solution

Q: How to solve Minimum Elements to Add to Form a Given Sum?

The solution calculates the current sum of the array, finds the absolute difference between the current sum and the target `goal`, and then divides the absolute difference by `limit`, rounding up to the nearest integer. This gives us the minimum number of elements to add to reach the goal. Ceiling division is crucial here.

Minimum Elements to Add to Form a Given Sum - Complete Solution Guide

Minimum Elements to Add to Form a Given Sum is LeetCode problem 1785, 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

You are given an integer array nums and two integers limit and goal . The array nums has an interesting property that abs(nums[i]) <= limit . Return the minimum number of elements you need to add to make the sum of the array equal to goal . The array must maintain its property that abs(nums[i]) <= limit . Note that abs(x) equals x if x >= 0 , and -x otherwise. Example 1: Input: nums = [1,-1,1], limit = 3, goal = -4 Output: 2 Explanation: You can add -2 and -3, then the sum of the array will be 1

Detailed Explanation

The problem asks us to find the minimum number of integers we need to add to an existing array `nums` such that the sum of the modified array equals a given `goal`. We are given a `limit`, which constrains the absolute value of any new element we add (i.e., `abs(new_element) <= limit`). Essentially, we need to bring the current sum of `nums` closer to the `goal` by adding integers within the specified limit, and we want to do this using the fewest possible additions.

Solution Approach

The solution calculates the current sum of the array, finds the absolute difference between the current sum and the target `goal`, and then divides the absolute difference by `limit`, rounding up to the nearest integer. This gives us the minimum number of elements to add to reach the goal. Ceiling division is crucial here.

Step-by-Step Algorithm

Step 1: Calculate the `current_sum` of all elements in the input array `nums`.
Step 2: Calculate the `diff` which is the absolute value of the difference between the `goal` and the `current_sum` (i.e., `abs(goal - current_sum)`). This represents the amount the sum needs to change.
Step 3: If `diff` is 0, the array already sums to the `goal`, so return 0.
Step 4: Calculate the minimum number of elements to add: `(diff + limit - 1) // limit`. This is ceiling division, equivalent to dividing `diff` by `limit` and rounding up to the nearest integer.

Key Insights

Insight 1: The core idea is to calculate the difference between the current sum of the array and the target `goal`. This difference tells us how much we need to adjust the sum by.
Insight 2: Since we can only add numbers within the range [-limit, limit], we want to use the largest possible absolute value (i.e., either `limit` or `-limit`) to minimize the number of additions.
Insight 3: The number of elements to add can be determined by dividing the absolute difference by the limit and rounding up to the nearest integer (using ceiling division).

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy.

Companies

Asked at: X.

Minimum Elements to Add to Form a Given Sum | LeetCode 1785

Medium

Array Greedy

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

Solution Code

Available in:

class Solution {
    public int minElements(int[] nums, int limit, int goal) {
        long current_sum = 0;
        for(int num : nums){
            current_sum += num;
        }
        long diff = Math.abs(goal - current_sum);
        
        if (diff == 0){
            return 0;
        }
        
        return (int) ((diff + limit - 1) / limit);
    }
}

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Problem Info

Description

Problem 1785: Minimum Elements to Add to Form a Given Sum - Detailed explanation and solution approaches.

Topics

Array Greedy

Companies

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1785

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

4 steps

1
Step 1: Calculate the `current_sum` of all elements in the input array `nums`.
2
Step 2: Calculate the `diff` which is the absolute value of the difference between the `goal` and the `current_sum` (i.e., `abs(goal - current_sum)`). This represents the amount the sum needs to change.
3
Step 3: If `diff` is 0, the array already sums to the `goal`, so return 0.
4
Step 4: Calculate the minimum number of elements to add: `(diff + limit - 1) // limit`. This is ceiling division, equivalent to dividing `diff` by `limit` and rounding up to the nearest integer.

💡

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 time complexity is determined by the linear traversal of the input array `nums` to calculate the sum. All other operations (absolute difference, division) take constant time. The space complexity is constant because we only use a fixed number of variables, irrespective of the size of the input array.

📖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

Iterative addition

Instead of calculating the number of additions directly, we could iteratively add either `limit` or `-limit` to the current sum until it reaches the `goal`, counting the number of additions. However, this is less efficient.

Time: O(diff / limit) in the best case, O(diff/limit) in the worst case. Equivalent to O(k) where k is the result of the original algorithm.Space: O(1)

Trade-offs: The iterative approach is less efficient than the direct calculation approach because it involves repeated addition. The direct calculation is preferable due to its superior time complexity. However, the iterative approach may be more intuitive for some.

✨

Key Insights

1Insight 1: The core idea is to calculate the difference between the current sum of the array and the target `goal`. This difference tells us how much we need to adjust the sum by.
2Insight 2: Since we can only add numbers within the range [-limit, limit], we want to use the largest possible absolute value (i.e., either `limit` or `-limit`) to minimize the number of additions.
3Insight 3: The number of elements to add can be determined by dividing the absolute difference by the limit and rounding up to the nearest integer (using ceiling division).

📚

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: `diff` is 0, meaning the current sum is equal to the goal. In this case, no elements need to be added, and the function should return 0. The code correctly handles this case.
Edge case 2: `limit` is 1. The code correctly calculates the number of additions required to reach the `goal` in this scenario, which is simply the absolute difference.
Edge case 3: `goal` and `nums` can be very large, necessitating the use of `long` datatype to avoid integer overflow. The Java and C++ solutions use `long`, Python solution uses arbitrary precision integers and C solution casts to `long long`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly using integer division instead of ceiling division. This will underestimate the number of elements required to reach the goal.

✗Common mistake 2: Ignoring potential integer overflow when calculating the `current_sum` or the `diff`. Using a larger data type (e.g., `long` in Java/C++) helps prevent overflow.

✗Common mistake 3: Not handling the edge case where the current sum is already equal to the `goal`.

✅

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: Optimizing resource allocation. Suppose you have a budget goal, and you need to allocate resources to different projects. The `nums` array represents the current allocation of resources, the `limit` represents the maximum amount you can allocate to any one project, and the `goal` represents the budget target. This algorithm finds the minimum number of new projects you need to fund to reach the budget goal.

🏦Real-world scenario 2: Inventory management. Suppose you need to meet a specific demand for a product. The `nums` array represents the current inventory levels, the `limit` represents the maximum production capacity per day, and the `goal` represents the target inventory level. The algorithm helps determine the minimum number of production days required to reach the target inventory.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 13, 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 13

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

@AlgoMasterJul 6

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

@InterviewProJun 29

❓

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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Minimum Elements to Add to Form a Given Sum"?

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

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