What is the time complexity of Minimum Operations to Form Subsequence With Target Sum?

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

What is the space complexity of Minimum Operations to Form Subsequence With Target Sum?

The space complexity is O(log(max(nums))).

What data structures are used to solve Minimum Operations to Form Subsequence With Target Sum?

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

What are the key insights for solving Minimum Operations to Form Subsequence With Target Sum?

Insight 1: The problem deals with powers of 2, making bit manipulation a suitable technique. Insight 2: Greedy approach works because splitting a power of 2 (say 2^k) into two 2^(k-1)s always helps in minimizing operations towards forming smaller values in the target.

Which companies ask LeetCode 2835. Minimum Operations to Form Subsequence With Target Sum in interviews?

This problem is frequently asked at Media.net.

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Minimum Operations to Form Subsequence With Target Sum - LeetCode 2835 Solution

Minimum Operations to Form Subsequence With Target Sum - Complete Solution Guide

Minimum Operations to Form Subsequence With Target Sum is LeetCode problem 2835, 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 a 0-indexed array nums consisting of non-negative powers of 2 , and an integer target . In one operation, you must apply the following changes to the array: Choose any element of the array nums[i] such that nums[i] > 1 . Remove nums[i] from the array. Add two occurrences of nums[i] / 2 to the end of nums . Return the minimum number of operations you need to perform so that nums contains a subsequence whose elements sum to target . If it is impossible to obtain such a subsequence, r

Detailed Explanation

The problem asks us to find the minimum number of operations required to form a subsequence within a given array `nums` whose elements sum up to a given `target`. The array `nums` consists of non-negative powers of 2. An operation involves choosing an element `nums[i]` greater than 1, removing it from the array, and adding two occurrences of `nums[i] / 2` to the end of the array. If it's impossible to form such a subsequence, we return -1.

Solution Approach

The solution uses a greedy approach combined with bit manipulation. It first counts the occurrences of each power of 2 present in the input array. Then, it iterates through the bits of the target, from the least significant bit to the most significant bit. If the current bit in the target is set to 1, we try to find a power of 2 in the array equal to 2^i (where i is the bit position). If we find such a power of 2, we use it. If not, we look for a larger power of 2 and repeatedly split it until we obtain the desired power of 2. The number of splits represents the number of operations. If the total sum of numbers in `nums` is less than `target`, we return -1 because no subsequence will sum up to target.

Step-by-Step Algorithm

Step 1: Check if the sum of elements in `nums` is less than `target`. If it is, return -1.
Step 2: Create an array `counts` to store the number of occurrences of each power of 2. Iterate through `nums` and populate the `counts` array. Specifically, `counts[i]` will store how many 2^i are present.
Step 3: Initialize the number of operations `ops` to 0.
Step 4: Iterate through the bits of `target` from the least significant bit (i=0) to the most significant bit.
Step 5: For each bit `i`, check if the `i`-th bit of `target` is set to 1. If it is, then try to find 2^i in the array. If counts[i] > 0, then decrement counts[i]. If not, search for the next available power of 2 (counts[j] > 0, j > i) to split to reach 2^i. Each split operation increments ops. After splitting, the missing power of 2 has been created ( counts[i] is incremented and counts[j] is decremented).
Step 6: After processing each bit, update `counts[i+1]` by adding `counts[i] / 2`. This is because if there are counts[i] powers of 2^i left, after fulfilling target's needs at the i-th bit, the excess becomes 2^i / 2 , which are powers of 2^(i+1).
Step 7: Return the total number of operations `ops`.

Key Insights

Insight 1: The problem deals with powers of 2, making bit manipulation a suitable technique.
Insight 2: Greedy approach works because splitting a power of 2 (say 2^k) into two 2^(k-1)s always helps in minimizing operations towards forming smaller values in the target.
Insight 3: We can keep track of the count of each power of 2 present in the array and iterate through the bits of the target to decide when to split numbers.

Complexity Analysis

Time Complexity: O(n + log(target))

Space Complexity: O(log(max(nums)))

Topics

This problem involves: Array, Greedy, Bit Manipulation.

Companies

Asked at: Media.net.

Minimum Operations to Form Subsequence With Target Sum | LeetCode 2835

Hard

Array Greedy Bit Manipulation

Time: O(n + log(target))Space: O(log(max(nums)))

Solution Code

Available in:

import java.util.List;

class Solution {
    public int minOperations(List<Integer> nums, int target) {
        long sum = 0;
        for (int num : nums) {
            sum += num;
        }
        if (sum < target) {
            return -1;
        }

        int[] counts = new int[45];
        for (int num : nums) {
            if (num > 0) {
                int exponent = 31 - Integer.numberOfLeadingZeros(num);
                counts[exponent] += 1;
            }
        }

        int ops = 0;
        
        for (int i = 0; i < counts.length; i++) {
            if (((target >> i) & 1) == 1) {
                if (counts[i] > 0) {
                    counts[i] -= 1;
                } else {
                    int j = i + 1;
                    while (j < counts.length && counts[j] == 0) {
                        j += 1;
                    }
                    
                    ops += j - i;
                    
                    counts[j] -= 1;
                    for (int k = j - 1; k >= i; k--) {
                        counts[k] += 1;
                    }
                    
                    counts[i] -= 1;
                }
            }

            if (i + 1 < counts.length) {
                counts[i+1] += counts[i] / 2;
            }
        }
        
        return ops;
    }
}

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Problem Info

Description

Problem 2835: Minimum Operations to Form Subsequence With Target Sum - Detailed explanation and solution approaches.

Topics

Array Greedy Bit Manipulation

Companies

Media.net

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2835

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: Check if the sum of elements in `nums` is less than `target`. If it is, return -1.
2
Step 2: Create an array `counts` to store the number of occurrences of each power of 2. Iterate through `nums` and populate the `counts` array. Specifically, `counts[i]` will store how many 2^i are present.
3
Step 3: Initialize the number of operations `ops` to 0.
4
Step 4: Iterate through the bits of `target` from the least significant bit (i=0) to the most significant bit.
5
Step 5: For each bit `i`, check if the `i`-th bit of `target` is set to 1. If it is, then try to find 2^i in the array. If counts[i] > 0, then decrement counts[i]. If not, search for the next available power of 2 (counts[j] > 0, j > i) to split to reach 2^i. Each split operation increments ops. After splitting, the missing power of 2 has been created ( counts[i] is incremented and counts[j] is decremented).
6
Step 6: After processing each bit, update `counts[i+1]` by adding `counts[i] / 2`. This is because if there are counts[i] powers of 2^i left, after fulfilling target's needs at the i-th bit, the excess becomes 2^i / 2 , which are powers of 2^(i+1).
7
Step 7: Return the total number of operations `ops`.

💡

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(target))

💾

Space Complexity

O(log(max(nums)))

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The time complexity is derived from the initial sum calculation (O(n)), the loop that constructs the `counts` array (O(n)), and the loop that iterates through the bits of the `target` (O(log(target))). The space complexity is derived from the `counts` array, which stores the count of each power of 2.

📖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

A dynamic programming approach could be used to determine if a sum equal to the target can be formed from a subset of the processed array at each step. However, it is expected to be less efficient than the greedy approach, due to higher computational cost.

Time: O(n*target) where n is nums.lengthSpace: O(target)

Trade-offs: The DP approach is less efficient due to its higher time complexity O(n*target), making it suitable only for very small target values. The greedy approach performs better in this problem, making DP solution undesirable.

✨

Key Insights

1Insight 1: The problem deals with powers of 2, making bit manipulation a suitable technique.
2Insight 2: Greedy approach works because splitting a power of 2 (say 2^k) into two 2^(k-1)s always helps in minimizing operations towards forming smaller values in the target.
3Insight 3: We can keep track of the count of each power of 2 present in the array and iterate through the bits of the target to decide when to split numbers.

📚

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: `target` is larger than the sum of elements in `nums`. The code handles this by checking `if sum(nums) < target: return -1` at the beginning.
Edge case 2: `nums` is empty. The code will still function correctly and return -1 if the target is larger than the sum of elements in `nums` (which is 0 in this case).
Edge case 3: Some counts[i] can become very large which can potentially cause overflow. However, based on the constraints, it is highly unlikely to occur. The problem constraints limit the length of `nums` to 1000, and each `nums[i]` is at most 2^30. Therefore, the counts array will be at most 1000 at a particular index and therefore, integer overflow isn't a concern.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the sum of elements in `nums` is less than `target`. This results in incorrect output.

✗Common mistake 2: Not updating `counts[i+1]` after processing each bit, leading to incorrect splitting logic and a wrong operation count.

✗Common mistake 3: Integer overflow due to splitting a number a very large number of times. Although in this particular question constraints are designed so it won't happen. Be careful while applying these types of solutions to the real world.

✅

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. Suppose you have a set of resources (powers of 2) and need to allocate them to fulfill a request (target). The minimum operations to split the resources represent the most efficient way to fulfill the request.

🏦Real-world scenario 2: Financial modeling. Representing financial assets or liabilities as powers of 2 and determining the minimum number of transactions (splits) to reach a specific investment target.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: October 14, 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

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

@InterviewProJun 30

❓

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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 Operations to Form Subsequence With Target 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 greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?