What is the time complexity of Find Minimum in Rotated Sorted Array?

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

What data structures are used to solve Find Minimum in Rotated Sorted Array?

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

Which companies ask LeetCode 153. Find Minimum in Rotated Sorted Array in interviews?

This problem is frequently asked at Cisco, FreshWorks, Goldman Sachs, IBM, Nutanix and 9 more companies.

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Find Minimum in Rotated Sorted Array - LeetCode 153 Solution

Q: What are the key insights for solving Find Minimum in Rotated Sorted Array?

Insight 1: The key observation is that the rotated array is still *partially* sorted. This allows us to use binary search to narrow down the search space. Insight 2: The comparison between the middle element and the rightmost element helps determine which half of the array contains the minimum value. If nums[mid] > nums[right], the minimum element is in the right half; otherwise, it's in the left half (including mid).

Find Minimum in Rotated Sorted Array - Complete Solution Guide

Find Minimum in Rotated Sorted Array is LeetCode problem 153, 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

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become: [4,5,6,7,0,1,2] if it was rotated 4 times. [0,1,2,4,5,6,7] if it was rotated 7 times. Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]] . Given the sorted rotated array nums of unique elements, return the minimum element of this array . You must write an algorithm that r

Detailed Explanation

The problem asks us to find the minimum element in a sorted array that has been rotated between 1 and n times. A rotated array is one where some portion of the array has been moved from the beginning to the end, or vice versa. For example, [0, 1, 2, 4, 5, 6, 7] rotated 4 times becomes [4, 5, 6, 7, 0, 1, 2]. The input is a list of unique integers that is sorted and rotated. The output should be the smallest element in the rotated array. The core challenge is to find this minimum element efficiently, specifically in O(log n) time complexity.

Solution Approach

The provided solution utilizes a modified binary search algorithm. The algorithm maintains two pointers, 'left' and 'right', that initially point to the start and end of the array, respectively. The core idea is to repeatedly divide the search space in half based on the relationship between the middle element and the rightmost element. This process continues until 'left' and 'right' converge at the index of the minimum element.

Step-by-Step Algorithm

Step 1: Initialize 'left' to 0 and 'right' to the length of the array minus 1.
Step 2: While 'left' is less than 'right', calculate the middle index 'mid' as left + (right - left) // 2 (or (left + right) / 2 to prevent overflow).
Step 3: Compare nums[mid] with nums[right].
Step 4: If nums[mid] > nums[right], the minimum element lies in the right half of the array, so update 'left' to mid + 1.
Step 5: Otherwise (if nums[mid] <= nums[right]), the minimum element lies in the left half (including mid), so update 'right' to mid.
Step 6: Repeat steps 2-5 until 'left' and 'right' are equal.
Step 7: Return nums[left] (or nums[right], as they are equal at this point), which is the minimum element.

Key Insights

Insight 1: The key observation is that the rotated array is still *partially* sorted. This allows us to use binary search to narrow down the search space.
Insight 2: The comparison between the middle element and the rightmost element helps determine which half of the array contains the minimum value. If nums[mid] > nums[right], the minimum element is in the right half; otherwise, it's in the left half (including mid).
Insight 3: The algorithm leverages the 'sorted' aspect of the array by checking if array is rotated. If not rotated, first element is the minimum.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: Cisco, FreshWorks, Goldman Sachs, IBM, Nutanix, PayPal, Paytm, ServiceNow, Synopsys, Tesla, TikTok, Yahoo, Yandex, eBay.

Find Minimum in Rotated Sorted Array | LeetCode 153

Medium

Array Binary Search

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

Solution Code

Available in:

class Solution {
    public int findMin(int[] nums) {
        int left = 0;
        int right = nums.length - 1;

        while (left < right) {
            int mid = left + (right - left) / 2;
            
            // If the middle element is greater than the rightmost element,
            // the minimum element must be in the right half.
            if (nums[mid] > nums[right]) {
                left = mid + 1;
            }
            // Otherwise, the minimum element is in the left half,
            // including the middle element.
            else {
                right = mid;
            }
        }
        
        // When the loop terminates, left and right will point to the minimum element.
        return nums[left];
    }
}

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Problem Info

Description

Problem 153: Find Minimum in Rotated Sorted Array - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

CiscoFreshWorksGoldman SachsIBMNutanixPayPalPaytmServiceNow

Stats

Difficulty:Medium

Languages:4

Companies:14

Problem #:153

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 'left' to 0 and 'right' to the length of the array minus 1.
2
Step 2: While 'left' is less than 'right', calculate the middle index 'mid' as left + (right - left) // 2 (or (left + right) / 2 to prevent overflow).
3
Step 3: Compare nums[mid] with nums[right].
4
Step 4: If nums[mid] > nums[right], the minimum element lies in the right half of the array, so update 'left' to mid + 1.
5
Step 5: Otherwise (if nums[mid] <= nums[right]), the minimum element lies in the left half (including mid), so update 'right' to mid.
6
Step 6: Repeat steps 2-5 until 'left' and 'right' are equal.
7
Step 7: Return nums[left] (or nums[right], as they are equal at this point), which is the minimum element.

💡

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(log n)

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(log n) because the algorithm repeatedly divides the search space in half with each iteration of the while loop. Binary search is inherently a logarithmic time algorithm. The space complexity is O(1) because the algorithm uses only a constant amount of extra space, regardless of the size of the input array. Only a few variables ('left', 'right', 'mid') are used.

📖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

Linear Search

Iterate through the array and keep track of the minimum element seen so far.

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

Trade-offs: This approach is simpler to implement but has a linear time complexity, making it less efficient than the binary search approach for larger arrays. Useful when the array size is very small or when the problem does not mandate O(log n) time complexity.

✨

Key Insights

1Insight 1: The key observation is that the rotated array is still *partially* sorted. This allows us to use binary search to narrow down the search space.
2Insight 2: The comparison between the middle element and the rightmost element helps determine which half of the array contains the minimum value. If nums[mid] > nums[right], the minimum element is in the right half; otherwise, it's in the left half (including mid).
3Insight 3: The algorithm leverages the 'sorted' aspect of the array by checking if array is rotated. If not rotated, first element is the minimum.

📚

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!

Edge Cases

Edge case 1: Array is not rotated at all (already sorted). In this case, the condition `nums[mid] > nums[right]` will never be true, and 'right' will continuously be updated to 'mid', eventually converging on the first element, which is indeed the minimum.
Edge case 2: Array has only one element. The while loop will not execute, and the algorithm will directly return the only element in the array, which is the minimum.
Edge case 3: Array with only two elements. The algorithm correctly handles the various relative orderings of the two elements after a rotation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly updating 'left' and 'right' pointers. For example, setting 'right' to 'mid - 1' when nums[mid] <= nums[right] would skip over the potential minimum element.

✗Common mistake 2: Failing to consider the case where the array is not rotated and is already sorted. The binary search approach correctly handles this, while a naive approach might fail.

✗Common mistake 3: Integer overflow when calculating the middle index. Using `mid = left + (right - left) // 2` prevents potential overflow when `left` and `right` are very large.

✗Common mistake 4: Confusing the comparison condition in the while loop. Using `left <= right` would lead to an infinite loop.

✅

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: Searching for a specific data point in a large, sorted database that has been partitioned and rearranged for efficiency. Database systems can use variations of this binary search technique.

🏦Real-world scenario 2: Finding the start of a repeating sequence in a circular buffer. Circular buffers are often used in data streaming and signal processing applications.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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.

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 "Find Minimum in Rotated Sorted Array"?

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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