What is the time complexity of Create Maximum Number?

The optimal solution has a time complexity of O((m+n)^3).

What is the space complexity of Create Maximum Number?

The space complexity is O(m+n).

What data structures are used to solve Create Maximum Number?

This problem uses Array, Two Pointers, Stack, Greedy, Monotonic Stack. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Create Maximum Number?

Insight 1: The problem can be broken down into three subproblems: finding the maximum subsequence of a given length from each array, merging two subsequences to form a candidate number, and comparing candidate numbers to find the maximum. Insight 2: A monotonic stack is an effective data structure for finding the maximum subsequence of a given length from a single array in O(n) time. This is because it allows us to greedily maintain the largest possible digits in the subsequence.

Which companies ask LeetCode 321. Create Maximum Number in interviews?

This problem is frequently asked at Flipkart.

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Create Maximum Number - LeetCode 321 Solution

Q: How to solve Create Maximum Number?

The solution adopts a greedy approach by first generating the maximum possible subsequences from each of the input arrays and then merging them optimally. It iterates through all possible lengths `i` for the subsequence from `nums1`, where `max(0, k - n) <= i <= min(k, m)`. For each `i`, it calculates the length of the subsequence from `nums2` as `k - i`. Then it merges these subsequences and compares the result with the current maximum number found so far. It finds the maximum subsequence with `_get_max_subsequence` and merges them with `_merge`.

Create Maximum Number - Complete Solution Guide

Create Maximum Number is LeetCode problem 321, 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 two integer arrays nums1 and nums2 of lengths m and n respectively. nums1 and nums2 represent the digits of two numbers. You are also given an integer k . Create the maximum number of length k <= m + n from digits of the two numbers. The relative order of the digits from the same array must be preserved. Return an array of the k digits representing the answer. Example 1: Input: nums1 = [3,4,6,5], nums2 = [9,1,2,5,8,3], k = 5 Output: [9,8,6,5,3] Example 2: Input: nums1 = [6,7], nums

Detailed Explanation

The problem asks us to create the largest possible number of length `k` from the digits of two given integer arrays, `nums1` and `nums2`. We must preserve the relative order of digits within each original array. In other words, we're effectively selecting subsequences from `nums1` and `nums2` and merging them to create the maximum possible `k`-length number. The constraints specify the lengths of the arrays and the range of `k`, ensuring it's a valid length for the merged number. No leading zeros in the input are mentioned but the logic ensures there will not be any in the output.

Solution Approach

The solution adopts a greedy approach by first generating the maximum possible subsequences from each of the input arrays and then merging them optimally. It iterates through all possible lengths `i` for the subsequence from `nums1`, where `max(0, k - n) <= i <= min(k, m)`. For each `i`, it calculates the length of the subsequence from `nums2` as `k - i`. Then it merges these subsequences and compares the result with the current maximum number found so far. It finds the maximum subsequence with `_get_max_subsequence` and merges them with `_merge`.

Step-by-Step Algorithm

Step 1: Iterate through all possible lengths `i` of the subsequence to be taken from `nums1`. This range is `max(0, k - n) <= i <= min(k, m)`, ensuring that the lengths of the subsequences from `nums1` and `nums2` sum up to `k` and that the lengths are valid (non-negative and not exceeding the array lengths).
Step 2: For each `i`, extract the maximum subsequence of length `i` from `nums1` using the `_get_max_subsequence` function, resulting in `sub1`.
Step 3: Extract the maximum subsequence of length `k - i` from `nums2` using the `_get_max_subsequence` function, resulting in `sub2`.
Step 4: Merge `sub1` and `sub2` using the `_merge` function to create a candidate number of length `k`. The merge operation always chooses the larger digit between the two subsequences' current heads, handling ties by comparing the remaining subsequences.
Step 5: Compare the candidate number with the current maximum number found so far. If the candidate is greater, update the maximum number.
Step 6: After iterating through all possible lengths `i`, return the maximum number found.

Key Insights

Insight 1: The problem can be broken down into three subproblems: finding the maximum subsequence of a given length from each array, merging two subsequences to form a candidate number, and comparing candidate numbers to find the maximum.
Insight 2: A monotonic stack is an effective data structure for finding the maximum subsequence of a given length from a single array in O(n) time. This is because it allows us to greedily maintain the largest possible digits in the subsequence.
Insight 3: When merging two subsequences, we need to compare the subsequences lexicographically starting from the current indices. The 'greater than' comparison needs to handle ties appropriately.

Complexity Analysis

Time Complexity: O((m+n)^3)

Space Complexity: O(m+n)

Topics

This problem involves: Array, Two Pointers, Stack, Greedy, Monotonic Stack.

Companies

Asked at: Flipkart.

Create Maximum Number | LeetCode 321

Hard

Array Two Pointers Stack Greedy Monotonic Stack

Time: O((m+n)^3)Space: O(m+n)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int[] maxNumber(int[] nums1, int[] nums2, int k) {
        int m = nums1.length;
        int n = nums2.length;
        int[] max_ans = new int[0];

        for (int i = Math.max(0, k - n); i <= Math.min(k, m); i++) {
            int[] sub1 = _get_max_subsequence(nums1, i);
            int[] sub2 = _get_max_subsequence(nums2, k - i);
            
            int[] candidate = _merge(sub1, sub2);
            
            if (greater(candidate, 0, max_ans, 0)) {
                max_ans = candidate;
            }
        }

        return max_ans;
    }

    private int[] _get_max_subsequence(int[] nums, int length) {
        int[] stack = new int[length];
        int top = -1;
        int to_drop = nums.length - length;
        for (int num : nums) {
            while (to_drop > 0 && top >= 0 && stack[top] < num) {
                top--;
                to_drop--;
            }
            if (top < length - 1) {
                stack[++top] = num;
            }
        }
        return stack;
    }

    private int[] _merge(int[] nums1, int[] nums2) {
        int[] res = new int[nums1.length + nums2.length];
        int p1 = 0, p2 = 0, p = 0;
        while (p1 < nums1.length || p2 < nums2.length) {
            if (greater(nums1, p1, nums2, p2)) {
                res[p++] = nums1[p1++];
            } else {
                res[p++] = nums2[p2++];
            }
        }
        return res;
    }

    private boolean greater(int[] nums1, int i, int[] nums2, int j) {
        while (i < nums1.length && j < nums2.length && nums1[i] == nums2[j]) {
            i++;
            j++;
        }
        return j == nums2.length || (i < nums1.length && nums1[i] > nums2[j]);
    }
}

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Problem Info

Description

Problem 321: Create Maximum Number - Detailed explanation and solution approaches.

Topics

Array Two Pointers Stack Greedy Monotonic Stack

Companies

Flipkart

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:321

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

6 steps

1
Step 1: Iterate through all possible lengths `i` of the subsequence to be taken from `nums1`. This range is `max(0, k - n) <= i <= min(k, m)`, ensuring that the lengths of the subsequences from `nums1` and `nums2` sum up to `k` and that the lengths are valid (non-negative and not exceeding the array lengths).
2
Step 2: For each `i`, extract the maximum subsequence of length `i` from `nums1` using the `_get_max_subsequence` function, resulting in `sub1`.
3
Step 3: Extract the maximum subsequence of length `k - i` from `nums2` using the `_get_max_subsequence` function, resulting in `sub2`.
4
Step 4: Merge `sub1` and `sub2` using the `_merge` function to create a candidate number of length `k`. The merge operation always chooses the larger digit between the two subsequences' current heads, handling ties by comparing the remaining subsequences.
5
Step 5: Compare the candidate number with the current maximum number found so far. If the candidate is greater, update the maximum number.
6
Step 6: After iterating through all possible lengths `i`, return the maximum number 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((m+n)^3)

💾

Space Complexity

O(m+n)

📝

Why?

The time complexity is dominated by the comparison within the merge process that is repeated for various lengths `i` of the subsequences. The space complexity is derived from the storage required for the stacks in `_get_max_subsequence` and the merged array in `_merge`.

📖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

Dynamic programming could be used to find optimal subsequences. However, it might not be as efficient as the greedy approach.

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

Trade-offs: DP would likely lead to larger time and space complexities, the current solution is already at O((m+n)^3) so DP is probably not a great direction.

✨

Key Insights

1Insight 1: The problem can be broken down into three subproblems: finding the maximum subsequence of a given length from each array, merging two subsequences to form a candidate number, and comparing candidate numbers to find the maximum.
2Insight 2: A monotonic stack is an effective data structure for finding the maximum subsequence of a given length from a single array in O(n) time. This is because it allows us to greedily maintain the largest possible digits in the subsequence.
3Insight 3: When merging two subsequences, we need to compare the subsequences lexicographically starting from the current indices. The 'greater than' comparison needs to handle ties appropriately.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: When either `nums1` or `nums2` is empty. The loop conditions `max(0, k - n)` and `min(k, m)` correctly handle this by ensuring the subsequence length from the other array is `k`.
Edge case 2: When `k` is larger than the length of one of the arrays. The loop conditions ensure that it attempts to extract as much as possible from both arrays.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling ties during the merge operation. If the current digits are the same, it's crucial to compare the remaining subsequences lexicographically to determine which subsequence contributes the next digit.

✗Common mistake 2: Failing to consider all possible combinations of subsequence lengths from `nums1` and `nums2`. The loop condition `max(0, k - n) <= i <= min(k, m)` is essential to cover all valid possibilities.

✅

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 when you have two different sources, such as choosing the best parts from two different suppliers to build the highest performing product, given a limited budget/number of parts.

🏦Real-world scenario 2: Combining data from multiple sensors to create the most accurate representation of a system. For example, merging readings from temperature and pressure sensors to predict weather patterns.

💡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.

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 20, 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 13

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

@InterviewProJul 6

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

@CodeNinjaJun 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.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

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 "Create Maximum Number"?

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

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