What is the time complexity of Minimum Size Subarray Sum?

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

What data structures are used to solve Minimum Size Subarray Sum?

This problem uses Array, Binary Search, Sliding Window, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Size Subarray Sum?

Insight 1: The problem requires finding a *subarray*, meaning a contiguous sequence of elements within the original array. Insight 2: Since the numbers are positive, the sum of a subarray increases monotonically as we add more elements to the right. This allows us to use a sliding window approach.

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Minimum Size Subarray Sum - LeetCode 209 Solution

Q: How to solve Minimum Size Subarray Sum?

The solution uses a sliding window approach. We maintain a window defined by two pointers, `left` and `right`. We expand the window by incrementing the `right` pointer and adding the corresponding element to the `current_sum`. If the `current_sum` becomes greater than or equal to the `target`, we update the `min_len` with the current window size. Then, we shrink the window from the left by incrementing the `left` pointer and subtracting the corresponding element from the `current_sum`, continuing the process until `current_sum` is less than `target`. This process is repeated until the `right` pointer reaches the end of the array.

Q: Which companies ask LeetCode 209. Minimum Size Subarray Sum in interviews?

This problem is frequently asked at Autodesk, DE Shaw, Darwinbox, DoorDash, Goldman Sachs and 3 more companies.

Minimum Size Subarray Sum - Complete Solution Guide

Minimum Size Subarray Sum is LeetCode problem 209, a Medium 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 positive integers nums and a positive integer target , return the minimal length of a subarray whose sum is greater than or equal to target . If there is no such subarray, return 0 instead. Example 1: Input: target = 7, nums = [2,3,1,2,4,3] Output: 2 Explanation: The subarray [4,3] has the minimal length under the problem constraint. Example 2: Input: target = 4, nums = [1,4,4] Output: 1 Example 3: Input: target = 11, nums = [1,1,1,1,1,1,1,1] Output: 0 Constraints: 1 <= target

Detailed Explanation

The problem asks us to find the minimum length of a contiguous subarray within a given array of positive integers (`nums`) whose sum is greater than or equal to a given positive integer (`target`). If no such subarray exists, we should return 0.

Solution Approach

The solution uses a sliding window approach. We maintain a window defined by two pointers, `left` and `right`. We expand the window by incrementing the `right` pointer and adding the corresponding element to the `current_sum`. If the `current_sum` becomes greater than or equal to the `target`, we update the `min_len` with the current window size. Then, we shrink the window from the left by incrementing the `left` pointer and subtracting the corresponding element from the `current_sum`, continuing the process until `current_sum` is less than `target`. This process is repeated until the `right` pointer reaches the end of the array.

Step-by-Step Algorithm

Step 1: Initialize `min_len` to `n + 1`, `left` to 0, and `current_sum` to 0. `min_len` is initialized to a value greater than the possible length of any subarray in `nums`, acting as a sentinel value.
Step 2: Iterate through the array `nums` using the `right` pointer.
Step 3: Add the element at the `right` index to the `current_sum`.
Step 4: While the `current_sum` is greater than or equal to the `target`, update `min_len` to the minimum of its current value and the length of the current window (`right - left + 1`).
Step 5: Subtract the element at the `left` index from the `current_sum` and increment the `left` pointer to shrink the window from the left.
Step 6: Repeat steps 4 and 5 until `current_sum` is less than `target`.
Step 7: After the loop finishes, if `min_len` is still equal to `n + 1`, it means no subarray was found that meets the criteria, so return 0. Otherwise, return `min_len`.

Key Insights

Insight 1: The problem requires finding a *subarray*, meaning a contiguous sequence of elements within the original array.
Insight 2: Since the numbers are positive, the sum of a subarray increases monotonically as we add more elements to the right. This allows us to use a sliding window approach.
Insight 3: We need to efficiently update the sum of the current subarray and compare it with the target value to determine if the window needs to be shrunk from the left.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Sliding Window, Prefix Sum.

Companies

Asked at: Autodesk, DE Shaw, Darwinbox, DoorDash, Goldman Sachs, Nvidia, Oracle, Yandex.

Minimum Size Subarray Sum | LeetCode 209

Medium

Array Binary Search Sliding Window Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int n = nums.length;
        int min_len = n + 1;
        int left = 0;
        int current_sum = 0;
        
        for (int right = 0; right < n; right++) {
            current_sum += nums[right];
            
            while (current_sum >= target) {
                min_len = Math.min(min_len, right - left + 1);
                current_sum -= nums[left];
                left++;
            }
        }
        
        if (min_len == n + 1) {
            return 0;
        } else {
            return min_len;
        }
    }
}

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Problem Info

Description

Problem 209: Minimum Size Subarray Sum - Detailed explanation and solution approaches.

Topics

Array Binary Search Sliding Window Prefix Sum

Companies

AutodeskDE ShawDarwinboxDoorDashGoldman SachsNvidiaOracleYandex

Stats

Difficulty:Medium

Languages:4

Companies:8

Problem #:209

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: Initialize `min_len` to `n + 1`, `left` to 0, and `current_sum` to 0. `min_len` is initialized to a value greater than the possible length of any subarray in `nums`, acting as a sentinel value.
2
Step 2: Iterate through the array `nums` using the `right` pointer.
3
Step 3: Add the element at the `right` index to the `current_sum`.
4
Step 4: While the `current_sum` is greater than or equal to the `target`, update `min_len` to the minimum of its current value and the length of the current window (`right - left + 1`).
5
Step 5: Subtract the element at the `left` index from the `current_sum` and increment the `left` pointer to shrink the window from the left.
6
Step 6: Repeat steps 4 and 5 until `current_sum` is less than `target`.
7
Step 7: After the loop finishes, if `min_len` is still equal to `n + 1`, it means no subarray was found that meets the criteria, so return 0. Otherwise, return `min_len`.

💡

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 time complexity is O(n) because each element in the array is visited at most twice (once by the right pointer and once by the left pointer). The space complexity is O(1) because the algorithm uses a fixed amount of extra space regardless of the input size.

📖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

Binary Search with Prefix Sum

Calculate the prefix sum of the array. Then, for each index 'i', use binary search to find the smallest index 'j' such that prefixSum[j] - prefixSum[i] >= target. This gives an O(n log n) solution.

Time: O(n log n)Space: O(n)

Trade-offs: This approach trades off space complexity for a potentially faster time complexity in some cases, but in general sliding window is simpler and faster. Prefix Sum requires O(n) additional space for the prefix sum array.

✨

Key Insights

1Insight 1: The problem requires finding a *subarray*, meaning a contiguous sequence of elements within the original array.
2Insight 2: Since the numbers are positive, the sum of a subarray increases monotonically as we add more elements to the right. This allows us to use a sliding window approach.
3Insight 3: We need to efficiently update the sum of the current subarray and compare it with the target value to determine if the window needs to be shrunk from the left.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: The array is empty. The algorithm correctly handles this case since the loop won't execute and it will return 0 if no valid subarray is found.
Edge case 2: No subarray sum is greater than or equal to the target. In this case, `min_len` will remain `n + 1` and the function will return 0.
Edge case 3: The array contains a single element that is greater than or equal to the target. The algorithm will correctly find the subarray of length 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not initializing `min_len` to a value greater than the maximum possible length of the subarray (e.g., setting it to `Integer.MAX_VALUE` in Java or `n + 1` in Python). This can lead to incorrect results if no subarray is found.

✗Common mistake 2: Incorrectly shrinking the window from the left before updating `min_len`. The window must be shrunk only after finding a valid subarray whose sum meets or exceeds the target.

✅

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: Finding the minimum amount of contiguous data that needs to be transmitted to meet a certain bandwidth requirement. The `nums` array could represent the size of data packets, and the `target` could represent the required bandwidth.

🏦Real-world scenario 2: In financial analysis, determining the shortest period of consecutive positive gains that exceeds a target profit. The `nums` array could represent daily gains, and the `target` could represent the desired profit.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: June 30, 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 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 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.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 Size Subarray Sum"?

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

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