What is the time complexity of Minimum Time to Make Array Sum At Most x?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Minimum Time to Make Array Sum At Most x?

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

What are the key insights for solving Minimum Time to Make Array Sum At Most x?

Insight 1: Sorting the indices based on the corresponding `nums2` values is crucial. Processing indices with smaller `nums2` values later allows them to contribute less to the overall sum over time. Insight 2: Dynamic Programming can be used to find the maximum reduction in sum achievable by zeroing out a certain number of elements. This helps determine if the target sum `x` can be reached within a given time.

Which companies ask LeetCode 2809. Minimum Time to Make Array Sum At Most x in interviews?

This problem is frequently asked at Jane Street.

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Minimum Time to Make Array Sum At Most x - LeetCode 2809 Solution

Q: How to solve Minimum Time to Make Array Sum At Most x?

The solution uses a dynamic programming approach combined with sorting to find the optimal strategy. First, the indices are sorted based on the values in `nums2`. Then, a DP table is built to find the maximum possible sum reduction for each number of elements that can be zeroed out. Finally, it iterates through possible times and checks if the sum of `nums1` after that time, minus the maximum possible reduction achievable in that time, is less than or equal to `x`. If it is, the time is returned. If no such time is found within the allowed limit, -1 is returned.

Minimum Time to Make Array Sum At Most x - Complete Solution Guide

Minimum Time to Make Array Sum At Most x is LeetCode problem 2809, 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 0-indexed integer arrays nums1 and nums2 of equal length. Every second, for all indices 0 <= i < nums1.length , value of nums1[i] is incremented by nums2[i] . After this is done, you can do the following operation: Choose an index 0 <= i < nums1.length and make nums1[i] = 0 . You are also given an integer x . Return the minimum time in which you can make the sum of all elements of nums1 to be less than or equal to x , or -1 if this is not possible. Example 1: Input: nums1 = [1,

Detailed Explanation

The problem asks us to find the minimum number of seconds (time) required to make the sum of elements in `nums1` less than or equal to `x`. Every second, each element `nums1[i]` increases by `nums2[i]`. We also have the option to set any `nums1[i]` to 0 after the increment. The goal is to find the optimal strategy of choosing indices to set to zero at each second to achieve the target sum in the least amount of time. If it's impossible to achieve the target sum, return -1.

Solution Approach

The solution uses a dynamic programming approach combined with sorting to find the optimal strategy. First, the indices are sorted based on the values in `nums2`. Then, a DP table is built to find the maximum possible sum reduction for each number of elements that can be zeroed out. Finally, it iterates through possible times and checks if the sum of `nums1` after that time, minus the maximum possible reduction achievable in that time, is less than or equal to `x`. If it is, the time is returned. If no such time is found within the allowed limit, -1 is returned.

Step-by-Step Algorithm

Step 1: Create pairs of (nums2[i], nums1[i]) and sort them in ascending order based on the nums2[i] values.
Step 2: Initialize a DP table `dp` of size `n + 1` with all values set to 0. `dp[i]` will store the maximum reduction in sum achievable by zeroing out `i` elements.
Step 3: Iterate through the sorted pairs. For each pair (n2, n1) at index `i`, iterate through the DP table from `i` down to 1. Update `dp[j]` as `max(dp[j], dp[j-1] + n1 + j * n2)`. This means either we don't zero out the current element, so `dp[j]` remains the same, or we zero out the current element, in which case we add the initial value n1 and the time elapsed so far j multiplied by the rate of increment n2.
Step 4: Calculate the initial sum of `nums1` (sum1) and the sum of `nums2` (sum2).
Step 5: If `sum1 <= x`, return 0 because no time is needed.
Step 6: Iterate through possible times `t` from 1 to `n`. Calculate the total sum at time `t` as `sum1 + t * sum2`.
Step 7: Subtract the maximum reduction achievable in time `t` (which is `dp[t]`) from the total sum at time `t`.
Step 8: If the resulting sum is less than or equal to `x`, return `t` as the minimum time.
Step 9: If no time `t` is found such that the condition in Step 8 is met, return -1.

Key Insights

Insight 1: Sorting the indices based on the corresponding `nums2` values is crucial. Processing indices with smaller `nums2` values later allows them to contribute less to the overall sum over time.
Insight 2: Dynamic Programming can be used to find the maximum reduction in sum achievable by zeroing out a certain number of elements. This helps determine if the target sum `x` can be reached within a given time.
Insight 3: The problem can be transformed into a knapsack-like problem where we want to maximize the value (reduction in sum) by choosing a subset of items (indices) to zero out, given a limited 'capacity' (number of seconds).
Insight 4: If the initial sum of nums1 is already <= x, the answer is 0. Also if after a certain time t, the sum is still greater than x, further increment will only make it larger so we don't need to go beyond n.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Sorting.

Companies

Asked at: Jane Street.

Minimum Time to Make Array Sum At Most x | LeetCode 2809

Hard

Array Dynamic Programming Sorting

Time: O(n^2)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;

class Solution {
    public int minimumTime(List<Integer> nums1, List<Integer> nums2, int x) {
        int n = nums1.size();
        
        List<Pair> pairs = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            pairs.add(new Pair(nums2.get(i), nums1.get(i)));
        }
        
        pairs.sort(Comparator.comparingInt(p -> p.n2));
        
        int[] dp = new int[n + 1];
        
        for (int i = 1; i <= n; i++) {
            Pair pair = pairs.get(i - 1);
            int n2 = pair.n2;
            int n1 = pair.n1;
            for (int j = i; j > 0; j--) {
                dp[j] = Math.max(dp[j], dp[j - 1] + n1 + j * n2);
            }
        }
        
        int sum1 = 0;
        for (int num : nums1) {
            sum1 += num;
        }
        int sum2 = 0;
        for (int num : nums2) {
            sum2 += num;
        }
        
        if (sum1 <= x) {
            return 0;
        }
        
        for (int t = 1; t <= n; t++) {
            int total_sum_at_t = sum1 + t * sum2;
            int max_reduction = dp[t];
            if (total_sum_at_t - max_reduction <= x) {
                return t;
            }
        }
        
        return -1;
    }
    
    private static class Pair {
        int n2;
        int n1;
        
        public Pair(int n2, int n1) {
            this.n2 = n2;
            this.n1 = n1;
        }
    }
}

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Problem Info

Description

Problem 2809: Minimum Time to Make Array Sum At Most x - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Sorting

Companies

Jane Street

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2809

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

9 steps

1
Step 1: Create pairs of (nums2[i], nums1[i]) and sort them in ascending order based on the nums2[i] values.
2
Step 2: Initialize a DP table `dp` of size `n + 1` with all values set to 0. `dp[i]` will store the maximum reduction in sum achievable by zeroing out `i` elements.
3
Step 3: Iterate through the sorted pairs. For each pair (n2, n1) at index `i`, iterate through the DP table from `i` down to 1. Update `dp[j]` as `max(dp[j], dp[j-1] + n1 + j * n2)`. This means either we don't zero out the current element, so `dp[j]` remains the same, or we zero out the current element, in which case we add the initial value n1 and the time elapsed so far j multiplied by the rate of increment n2.
4
Step 4: Calculate the initial sum of `nums1` (sum1) and the sum of `nums2` (sum2).
5
Step 5: If `sum1 <= x`, return 0 because no time is needed.
6
Step 6: Iterate through possible times `t` from 1 to `n`. Calculate the total sum at time `t` as `sum1 + t * sum2`.
7
Step 7: Subtract the maximum reduction achievable in time `t` (which is `dp[t]`) from the total sum at time `t`.
8
Step 8: If the resulting sum is less than or equal to `x`, return `t` as the minimum time.
9
Step 9: If no time `t` is found such that the condition in Step 8 is met, return -1.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The outer loop in the DP construction iterates `n` times, and the inner loop iterates up to `n` times as well, resulting in O(n^2) time complexity for the DP part. The other parts of the code (sorting and the final check) contribute either O(n log n) or O(n) time complexity, which are dominated by the O(n^2) DP complexity. The space complexity comes from the DP table of size O(n) and the `pairs` array, also O(n).

📖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

Branch and Bound

A branch and bound approach might be considered, where we explore different combinations of zeroing out elements at each time step. However, the search space grows exponentially, making it inefficient for larger inputs.

Time: ExponentialSpace: Exponential in worst case

Trade-offs: Generally much slower than the dynamic programming approach due to the exponential search space. Potentially could perform better than DP for small values of n if good pruning strategies are employed, but is difficult to implement correctly.

✨

Key Insights

1Insight 1: Sorting the indices based on the corresponding `nums2` values is crucial. Processing indices with smaller `nums2` values later allows them to contribute less to the overall sum over time.
2Insight 2: Dynamic Programming can be used to find the maximum reduction in sum achievable by zeroing out a certain number of elements. This helps determine if the target sum `x` can be reached within a given time.
3Insight 3: The problem can be transformed into a knapsack-like problem where we want to maximize the value (reduction in sum) by choosing a subset of items (indices) to zero out, given a limited 'capacity' (number of seconds).
4Insight 4: If the initial sum of nums1 is already <= x, the answer is 0. Also if after a certain time t, the sum is still greater than x, further increment will only make it larger so we don't need to go beyond n.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `nums1` is already small enough (sum(nums1) <= x). The solution correctly returns 0 in this case.
Edge case 2: It is impossible to make the sum small enough. The solution correctly returns -1 in this case after checking all possible times up to `n`.
Edge case 3: Empty input arrays. Although constraints specify the array length is at least 1, a robust solution should handle empty input arrays gracefully (which this implementation does not do explicitly, but the behavior would depend on the behavior of sum() on an empty list and would likely throw an exception).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the indices based on `nums2`. This leads to a suboptimal strategy and potentially incorrect results.

✗Common mistake 2: Not considering the time elapsed (`t`) when calculating the sum reduction. The value of each element increases over time, which must be factored into the reduction.

✗Common mistake 3: Incorrectly implementing the DP update rule. It's crucial to consider both the initial value `nums1[i]` and the increase `nums2[i]` over time.

✗Common mistake 4: Early return before checking all valid times. The optimal solution might involve zeroing out elements later rather than sooner.

✅

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 where you have a limited budget (x) and resources (nums1) that grow over time (nums2). You can choose to 'reset' certain resources (zeroing them out) to stay within your budget. The goal is to find the minimum number of time periods to achieve your budget target.

🏦Real-world scenario 2: Network traffic management where packets (nums1) arrive with increasing rates (nums2). You can drop certain packets (zeroing them out) to prevent congestion (stay below x). The goal is to find the minimum time to ensure no congestion.

💡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 18, 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

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

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

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

@AlgoMasterJun 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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 Time to Make Array Sum At Most x"?

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

🤖 Need help with this problem?