What is the time complexity of Get the Maximum Score?

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

What data structures are used to solve Get the Maximum Score?

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

What are the key insights for solving Get the Maximum Score?

Insight 1: Since the arrays are sorted, we can use a two-pointer approach to efficiently traverse them. Insight 2: We need to keep track of the sums of the elements encountered in each array before a common element is found.

Which companies ask LeetCode 1537. Get the Maximum Score in interviews?

This problem is frequently asked at Intuit, Mindtickle.

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Get the Maximum Score - LeetCode 1537 Solution

Get the Maximum Score - Complete Solution Guide

Get the Maximum Score is LeetCode problem 1537, 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 sorted arrays of distinct integers nums1 and nums2 . A valid path is defined as follows: Choose array nums1 or nums2 to traverse (from index-0). Traverse the current array from left to right. If you are reading any value that is present in nums1 and nums2 you are allowed to change your path to the other array. (Only one repeated value is considered in the valid path). The score is defined as the sum of unique values in a valid path. Return the maximum score you can obtain of al

Detailed Explanation

The problem requires finding the maximum possible score from traversing two sorted arrays, `nums1` and `nums2`. A valid path starts from either array and progresses from left to right. If a value appears in both arrays, we can switch paths at that element. The score is the sum of the unique values in the chosen path, and we need to return the maximum score modulo 10^9 + 7.

Solution Approach

The solution utilizes a two-pointer approach. Two pointers, `i` and `j`, are used to iterate through `nums1` and `nums2`, respectively. We maintain two variables, `sum1` and `sum2`, to store the sum of elements encountered in `nums1` and `nums2` before reaching a common element. When a common element is found (nums1[i] == nums2[j]), we add the maximum of `sum1` and `sum2` plus the common element to the total `max_score`. Then, we reset `sum1` and `sum2` to zero and increment both pointers. After one of the arrays is exhausted, we add the remaining elements of the other array to their respective `sum`. Finally, we choose the max of the final `sum1` and `sum2` and add it to the `max_score`. The result is returned modulo 10^9 + 7.

Step-by-Step Algorithm

Step 1: Initialize two pointers, `i` and `j`, to 0, representing the current index in `nums1` and `nums2`, respectively.
Step 2: Initialize `sum1` and `sum2` to 0 to store the sum of elements in `nums1` and `nums2` before a common element.
Step 3: Initialize `max_score` to 0 to store the maximum score obtained.
Step 4: While `i` is less than the length of `nums1` and `j` is less than the length of `nums2`:
Step 5: If `nums1[i]` is less than `nums2[j]`, add `nums1[i]` to `sum1` and increment `i`.
Step 6: Else if `nums1[i]` is greater than `nums2[j]`, add `nums2[j]` to `sum2` and increment `j`.
Step 7: Else (if `nums1[i]` equals `nums2[j]`): Add `max(sum1, sum2) + nums1[i]` to `max_score`, reset `sum1` and `sum2` to 0, and increment both `i` and `j`.
Step 8: While `i` is less than the length of `nums1`, add `nums1[i]` to `sum1` and increment `i`.
Step 9: While `j` is less than the length of `nums2`, add `nums2[j]` to `sum2` and increment `j`.
Step 10: Add `max(sum1, sum2)` to `max_score`.
Step 11: Return `max_score` modulo 10^9 + 7.

Key Insights

Insight 1: Since the arrays are sorted, we can use a two-pointer approach to efficiently traverse them.
Insight 2: We need to keep track of the sums of the elements encountered in each array before a common element is found.
Insight 3: At each common element, we choose the path with the larger sum and add the common element's value to the total score. Reset the path sums after each merge point.

Complexity Analysis

Time Complexity: O(m+n)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Dynamic Programming, Greedy.

Companies

Asked at: Intuit, Mindtickle.

Get the Maximum Score | LeetCode 1537

Hard

Array Two Pointers Dynamic Programming Greedy

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

Solution Code

Available in:

class Solution {
    public int maxSum(int[] nums1, int[] nums2) {
        int i = 0, j = 0;
        long sum1 = 0, sum2 = 0, max_score = 0;
        int MOD = (int)1e9 + 7;

        while (i < nums1.length && j < nums2.length) {
            if (nums1[i] < nums2[j]) {
                sum1 += nums1[i];
                i++;
            } else if (nums1[i] > nums2[j]) {
                sum2 += nums2[j];
                j++;
            } else {
                max_score += Math.max(sum1, sum2) + nums1[i];
                sum1 = 0;
                sum2 = 0;
                i++;
                j++;
            }
        }

        while (i < nums1.length) {
            sum1 += nums1[i];
            i++;
        }

        while (j < nums2.length) {
            sum2 += nums2[j];
            j++;
        }
            
        max_score += Math.max(sum1, sum2);

        return (int)(max_score % MOD);
    }
}

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Problem Info

Description

Problem 1537: Get the Maximum Score - Detailed explanation and solution approaches.

Topics

Array Two Pointers Dynamic Programming Greedy

Companies

IntuitMindtickle

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:1537

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

11 steps

1
Step 1: Initialize two pointers, `i` and `j`, to 0, representing the current index in `nums1` and `nums2`, respectively.
2
Step 2: Initialize `sum1` and `sum2` to 0 to store the sum of elements in `nums1` and `nums2` before a common element.
3
Step 3: Initialize `max_score` to 0 to store the maximum score obtained.
4
Step 4: While `i` is less than the length of `nums1` and `j` is less than the length of `nums2`:
5
Step 5: If `nums1[i]` is less than `nums2[j]`, add `nums1[i]` to `sum1` and increment `i`.
6
Step 6: Else if `nums1[i]` is greater than `nums2[j]`, add `nums2[j]` to `sum2` and increment `j`.
7
Step 7: Else (if `nums1[i]` equals `nums2[j]`): Add `max(sum1, sum2) + nums1[i]` to `max_score`, reset `sum1` and `sum2` to 0, and increment both `i` and `j`.
8
Step 8: While `i` is less than the length of `nums1`, add `nums1[i]` to `sum1` and increment `i`.
9
Step 9: While `j` is less than the length of `nums2`, add `nums2[j]` to `sum2` and increment `j`.
10
Step 10: Add `max(sum1, sum2)` to `max_score`.
11
Step 11: Return `max_score` modulo 10^9 + 7.

💡

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)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the while loops, which traverse each array at most once. The space complexity is constant because we only use a few variables to store sums and indices, irrespective of the input array sizes.

📖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 DP approach could potentially be used, but it's less efficient. We'd need to maintain a DP table storing the maximum score at each index of the arrays. This would likely involve more complex logic and higher space complexity.

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

Trade-offs: The two-pointer approach is much more efficient in both time and space because it leverages the sorted property of the arrays. DP would not be the right approach for this problem

✨

Key Insights

1Insight 1: Since the arrays are sorted, we can use a two-pointer approach to efficiently traverse them.
2Insight 2: We need to keep track of the sums of the elements encountered in each array before a common element is found.
3Insight 3: At each common element, we choose the path with the larger sum and add the common element's value to the total score. Reset the path sums after each merge point.

📚

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.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: One of the arrays is empty. The solution handles this correctly because if either array is empty, the corresponding while loop will not execute, and the score will correctly accumulate from the non-empty array.
Edge case 2: There are no common elements between the two arrays. The two pointers will proceed until one of the arrays is exhausted, and the sums will be accumulated correctly. The final `max(sum1, sum2)` will choose the array with the greater sum.
Edge case 3: Very large numbers in the input arrays. Using `long` (Java, C++) or equivalent in Python prevents integer overflow when calculating sums before taking the modulo.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. The sums of the elements can be very large, so using `int` instead of `long` (or equivalent) can lead to incorrect results. Always using `long` or equivalent is recommended during the accumulation of sums.

✗Common mistake 2: Not taking the modulo at the end. The problem asks for the result modulo 10^9 + 7. Failing to apply this modulo operation will lead to incorrect results for larger inputs.

✗Common mistake 3: Incorrectly handling edge cases where one of the arrays is completely traversed before the other. The while loops after the main loop are crucial for summing up the remaining elements.

✅

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: Data synchronization. Imagine merging sorted logs from two different servers. The common elements represent synchronization points, and maximizing the score could represent prioritizing log entries based on their importance.

🏦Real-world scenario 2: Route Optimization. Consider two sorted lists of cities on two possible routes. Common cities allow switching routes. Maximizing the 'score' which could be a function of travel time or cost, can lead to optimal route selection.

🎮Real-world scenario 3: Bioinformatics. Aligning two DNA sequences represented as sorted arrays of nucleotides. Common subsequences can indicate homology, and maximizing the similarity score (analogous to the path sum) aids in identifying evolutionary relationships.

💡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: September 6, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 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 "Get the Maximum Score"?

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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