What is the time complexity of Apply Operations to Make Sum of Array Greater Than or Equal to k?

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

What data structures are used to solve Apply Operations to Make Sum of Array Greater Than or Equal to k?

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

What are the key insights for solving Apply Operations to Make Sum of Array Greater Than or Equal to k?

Insight 1: The optimal strategy involves increasing the initial value to some number 'val' and then duplicating this value as needed to reach the target sum 'k'. Insight 2: We can iterate through possible values of 'val' (from 1 to k) and calculate the number of increase operations and duplication operations required for each 'val'.

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Apply Operations to Make Sum of Array Greater Than or Equal to k - LeetCode 3091 Solution

Q: Which companies ask LeetCode 3091. Apply Operations to Make Sum of Array Greater Than or Equal to k in interviews?

This problem is frequently asked at Turing, ZScaler.

Apply Operations to Make Sum of Array Greater Than or Equal to k - Complete Solution Guide

Apply Operations to Make Sum of Array Greater Than or Equal to k is LeetCode problem 3091, 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 a positive integer k . Initially, you have an array nums = [1] . You can perform any of the following operations on the array any number of times ( possibly zero ): Choose any element in the array and increase its value by 1 . Duplicate any element in the array and add it to the end of the array. Return the minimum number of operations required to make the sum of elements of the final array greater than or equal to k . Example 1: Input: k = 11 Output: 5 Explanation: We can do the f

Detailed Explanation

The problem asks us to find the minimum number of operations needed to make the sum of the elements in an array `nums` greater than or equal to `k`. We start with `nums = [1]`. We can perform two types of operations: (1) Increase any element by 1, and (2) Duplicate any element and add it to the end of the array. The goal is to minimize the total count of these two types of operations to reach a sum >= k.

Solution Approach

The solution iterates through all possible values 'val' that the initial element can be increased to. For each 'val', it calculates how many times that 'val' needs to be duplicated to make the total sum greater than or equal to 'k'. It then calculates the number of operations needed to increase the initial element to 'val' (which is val - 1) and the number of operations needed to duplicate the element to reach the required sum (which is num_copies - 1). The minimum number of total operations is tracked, and returned.

Step-by-Step Algorithm

Step 1: Initialize `min_ops` to `k - 1`. This is the maximum possible number of operations, achieved by increasing the initial value to `k` (k-1 operations) and not duplicating at all.
Step 2: Iterate through possible values of `val` from 1 to `k`. We are essentially trying out increasing the initial value (1) to val and then duplicating val to meet the required sum k.
Step 3: For each `val`, calculate the number of copies `num_copies` needed to achieve a sum greater than or equal to `k`. This is done using the formula `num_copies = (k + val - 1) // val`.
Step 4: Calculate the total number of operations `current_ops` required for this value of `val`. This is the sum of the operations needed to increase the initial value to `val` (val - 1 operations) and the operations needed to duplicate the element (num_copies - 1 operations). So `current_ops = (val - 1) + (num_copies - 1)`.
Step 5: Update `min_ops` with the minimum of the current `min_ops` and `current_ops`. This ensures we keep track of the best (smallest) number of operations we've found so far.
Step 6: Optimization: If `val - 1` is greater than or equal to `min_ops`, then the current value of `val` will require at least `min_ops` operations even without any duplication. Therefore, we can break out of the loop because all subsequent iterations will only produce worse (larger) results.
Step 7: Return `min_ops` after the loop completes. This is the minimum number of operations found.

Key Insights

Insight 1: The optimal strategy involves increasing the initial value to some number 'val' and then duplicating this value as needed to reach the target sum 'k'.
Insight 2: We can iterate through possible values of 'val' (from 1 to k) and calculate the number of increase operations and duplication operations required for each 'val'.
Insight 3: We can use integer division to calculate the number of duplicates needed: `num_copies = (k + val - 1) // val`. The ceiling operator is achieved by `(k + val - 1) // val` which computes `ceil(k / val)` efficiently.

Complexity Analysis

Time Complexity: O(k)

Space Complexity: O(1)

Topics

This problem involves: Math, Greedy, Enumeration.

Companies

Asked at: Turing, ZScaler.

Apply Operations to Make Sum of Array Greater Than or Equal to k | LeetCode 3091

Medium

Math Greedy Enumeration

Time: O(k)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minOperations(int k) {
        if (k == 1) {
            return 0;
        }

        int min_ops = k - 1;

        for (int val = 1; val <= k; val++) {
            if ((val - 1) >= min_ops) {
                break;
            }
            
            int num_copies = (k + val - 1) / val;
            int current_ops = (val - 1) + (num_copies - 1);
            min_ops = Math.min(min_ops, current_ops);
        }
        
        return min_ops;
    }
}

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Problem Info

Description

Problem 3091: Apply Operations to Make Sum of Array Greater Than or Equal to k - Detailed explanation and solution approaches.

Topics

Math Greedy Enumeration

Companies

TuringZScaler

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:3091

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

7 steps

1
Step 1: Initialize `min_ops` to `k - 1`. This is the maximum possible number of operations, achieved by increasing the initial value to `k` (k-1 operations) and not duplicating at all.
2
Step 2: Iterate through possible values of `val` from 1 to `k`. We are essentially trying out increasing the initial value (1) to val and then duplicating val to meet the required sum k.
3
Step 3: For each `val`, calculate the number of copies `num_copies` needed to achieve a sum greater than or equal to `k`. This is done using the formula `num_copies = (k + val - 1) // val`.
4
Step 4: Calculate the total number of operations `current_ops` required for this value of `val`. This is the sum of the operations needed to increase the initial value to `val` (val - 1 operations) and the operations needed to duplicate the element (num_copies - 1 operations). So `current_ops = (val - 1) + (num_copies - 1)`.
5
Step 5: Update `min_ops` with the minimum of the current `min_ops` and `current_ops`. This ensures we keep track of the best (smallest) number of operations we've found so far.
6
Step 6: Optimization: If `val - 1` is greater than or equal to `min_ops`, then the current value of `val` will require at least `min_ops` operations even without any duplication. Therefore, we can break out of the loop because all subsequent iterations will only produce worse (larger) results.
7
Step 7: Return `min_ops` after the loop completes. This is the minimum number of operations found.

💡

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)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(k) because the loop iterates at most `k` times. Each iteration performs constant time operations. The space complexity is O(1) because only a few constant space variables are used, regardless of the input value `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

Binary Search (Conceptual)

One could attempt to apply binary search to find the optimal 'val', but it's not straightforward because the number of duplications isn't monotonic with respect to 'val'. It is possible but more complicated and likely doesn't improve the time complexity given the constraint `1 <= k <= 10^5`.

Time: O(k log k) if a suitable monotonic function can be identified with properties amenable to effective binary search.Space: O(1)

Trade-offs: Potentially better performance for much larger values of k, but more difficult to implement and may not be faster in practice given the problem constraints. The current solution is very readable and efficient given the constraints.

✨

Key Insights

1Insight 1: The optimal strategy involves increasing the initial value to some number 'val' and then duplicating this value as needed to reach the target sum 'k'.
2Insight 2: We can iterate through possible values of 'val' (from 1 to k) and calculate the number of increase operations and duplication operations required for each 'val'.
3Insight 3: We can use integer division to calculate the number of duplicates needed: `num_copies = (k + val - 1) // val`. The ceiling operator is achieved by `(k + val - 1) // val` which computes `ceil(k / val)` efficiently.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `k = 1`. The initial array `[1]` already has a sum >= `k`, so the solution correctly returns 0.
Edge case 2: Large values of `k`. The code handles this by iterating up to `k`, which might be inefficient for extremely large values of `k` but is acceptable given the constraints. The integer division `(k + val - 1) / val` is safe because `k` and `val` are positive.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the initial value of 1 in the array. The initial element must also be increased.

✗Common mistake 2: Using floating-point division and rounding to calculate the number of copies. Integer division `//` (Python) or `/` (Java, C++) is crucial for correct results and efficiency.

✗Common mistake 3: Failing to minimize `min_ops`. The code must track the minimum number of operations across all possible 'val' values.

✗Common mistake 4: Not optimizing by breaking the loop when `val - 1 >= min_ops`. This optimization can save some unnecessary iterations.

✅

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 you have an initial resource and can either increase its capacity or create duplicates. This problem models finding the minimum number of actions to reach a target resource level.

🏦Real-world scenario 2: Production optimization. A company has a single machine producing items. They can either upgrade the machine to produce more items per unit time or replicate the machine. This problem can help determine the optimal combination of upgrades and replications to meet a production target with minimal investment.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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 2, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

Frequently Asked Questions

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 "Apply Operations to Make Sum of Array Greater Than or Equal to k"?

For this medium-level Math 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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