What is the time complexity of Minimum Value to Get Positive Step by Step Sum?

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

What data structures are used to solve Minimum Value to Get Positive Step by Step Sum?

This problem uses Array, Prefix Sum. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Minimum Value to Get Positive Step by Step Sum?

Insight 1: The minimum `startValue` is directly related to the minimum cumulative sum encountered while iterating through the `nums` array. If the minimum cumulative sum is negative, we need to adjust the `startValue` to compensate. Insight 2: A single pass through the array is sufficient to solve the problem using prefix sum technique. We keep track of the running sum and the minimum sum encountered so far.

Which companies ask LeetCode 1413. Minimum Value to Get Positive Step by Step Sum in interviews?

This problem is frequently asked at Swiggy.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Minimum Value to Get Positive Step by Step Sum - LeetCode 1413 Solution

Minimum Value to Get Positive Step by Step Sum - Complete Solution Guide

Minimum Value to Get Positive Step by Step Sum is LeetCode problem 1413, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an array of integers nums , you start with an initial positive value startValue . In each iteration, you calculate the step by step sum of startValue plus elements in nums (from left to right). Return the minimum positive value of startValue such that the step by step sum is never less than 1. Example 1: Input: nums = [-3,2,-3,4,2] Output: 5 Explanation: If you choose startValue = 4, in the third iteration your step by step sum is less than 1. step by step sum startValue = 4 | startValue =

Detailed Explanation

The problem asks you to find the minimum positive starting value (`startValue`) such that the cumulative sum of `startValue` and the elements in the input array `nums` remains positive at every step. The cumulative sum is calculated iteratively, adding each element of `nums` to the running total. The goal is to ensure this running total never drops below 1. The input is an array of integers (`nums`), and the output is a single integer representing the minimum positive `startValue`.

Solution Approach

The solution employs a prefix sum approach. It iterates through the `nums` array, maintaining a running sum (`currentSum`) and tracking the minimum sum encountered (`minSum`). After iterating, if `minSum` is non-negative, a `startValue` of 1 is sufficient. Otherwise, the minimum positive `startValue` is calculated as `1 - minSum` (or equivalently, `abs(minSum) + 1`). This ensures that even with the minimum cumulative sum, the running total will always be at least 1.

Step-by-Step Algorithm

Step 1: Initialize `minSum` and `currentSum` to 0.
Step 2: Iterate through the `nums` array. In each iteration, add the current element (`num`) to `currentSum`.
Step 3: Update `minSum` to be the minimum of `minSum` and `currentSum`.
Step 4: After the loop, if `minSum` is greater than or equal to 0, return 1 (because the cumulative sum is always positive).
Step 5: Otherwise, return `1 - minSum` (or `abs(minSum) + 1`), which is the minimum positive `startValue` needed to ensure the cumulative sum remains positive.

Key Insights

Insight 1: The minimum `startValue` is directly related to the minimum cumulative sum encountered while iterating through the `nums` array. If the minimum cumulative sum is negative, we need to adjust the `startValue` to compensate.
Insight 2: A single pass through the array is sufficient to solve the problem using prefix sum technique. We keep track of the running sum and the minimum sum encountered so far.
Insight 3: The minimum positive `startValue` is one more than the absolute value of the minimum cumulative sum. This ensures that the cumulative sum remains positive throughout the iteration.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Prefix Sum.

Companies

Asked at: Swiggy.

Minimum Value to Get Positive Step by Step Sum | LeetCode 1413

Easy

Array Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int minStartValue(int[] nums) {
        int minSum = 0;
        int currentSum = 0;
        for (int num : nums) {
            currentSum += num;
            minSum = Math.min(minSum, currentSum);
        }
        return Math.max(1, 1 - minSum);
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1413: Minimum Value to Get Positive Step by Step Sum - Detailed explanation and solution approaches.

Topics

Array Prefix Sum

Companies

Swiggy

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1413

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

5 steps

1
Step 1: Initialize `minSum` and `currentSum` to 0.
2
Step 2: Iterate through the `nums` array. In each iteration, add the current element (`num`) to `currentSum`.
3
Step 3: Update `minSum` to be the minimum of `minSum` and `currentSum`.
4
Step 4: After the loop, if `minSum` is greater than or equal to 0, return 1 (because the cumulative sum is always positive).
5
Step 5: Otherwise, return `1 - minSum` (or `abs(minSum) + 1`), which is the minimum positive `startValue` needed to ensure the cumulative sum remains positive.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the `nums` array once, taking linear time. It uses only a few constant extra variables, resulting in constant space complexity.

📖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

Mathematical Approach

Calculate the cumulative sum of the array, find its minimum value and add 1-minimum to get the starting value. This approach avoids explicit iteration and min tracking

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

Trade-offs: This is functionally equivalent but might be slightly less readable for beginners. The iterative approach in the solution is generally preferred for clarity.

✨

Key Insights

1Insight 1: The minimum `startValue` is directly related to the minimum cumulative sum encountered while iterating through the `nums` array. If the minimum cumulative sum is negative, we need to adjust the `startValue` to compensate.
2Insight 2: A single pass through the array is sufficient to solve the problem using prefix sum technique. We keep track of the running sum and the minimum sum encountered so far.
3Insight 3: The minimum positive `startValue` is one more than the absolute value of the minimum cumulative sum. This ensures that the cumulative sum remains positive throughout the iteration.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: The input array is empty. The solution correctly handles this by returning 1 because an empty array implies that the starting value must be at least 1 to remain positive.
Edge case 2: All elements in the array are positive. The solution correctly returns 1, as a starting value of 1 is sufficient.
Edge case 3: All elements are negative. The solution will correctly calculate the necessary starting value to offset the negative numbers

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the minimum cumulative sum is non-negative. In this case, the `startValue` should be 1, not 0.

✗Common mistake 2: Incorrectly calculating the required `startValue`. Some might mistakenly subtract `minSum` directly from 1 without considering that `startValue` must be positive.

✅

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: Maintaining a minimum balance in a bank account. The `nums` array could represent daily transactions (positive for deposits, negative for withdrawals). The `startValue` would be the initial balance required to ensure the account never falls below zero.

🏦Real-world scenario 2: Resource management in a system. The `nums` array might represent resource consumption or generation over time. The `startValue` would represent the initial resource level needed to guarantee sufficient resources at all times.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: November 1, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

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

@DevExpert92Jun 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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 Value to Get Positive Step by Step Sum"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether prefix sum techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?