Find Minimum in Rotated Sorted Array in DSA Rust - Build from Scratch

Find Minimum in Rotated Sorted Array

📖 Scenario: Imagine you have a sorted list of numbers that has been rotated at some unknown point. For example, the list [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]. Your task is to find the smallest number in this rotated list.

🎯 Goal: You will write a Rust program that finds the minimum number in a rotated sorted array using a simple loop.

📋 What You'll Learn

Create a vector called nums with the exact values [4,5,6,7,0,1,2]

Create a variable called min_value and set it to the first element of nums

Use a for loop with variable num to iterate over nums

Inside the loop, update min_value if num is smaller

Print the min_value after the loop

💡 Why This Matters

🌍 Real World

Rotated sorted arrays appear in systems where data is shifted or rotated, such as circular buffers or rotated logs.

💼 Career

Understanding how to find minimums efficiently in rotated arrays is useful in technical interviews and real-world software that deals with rotated or shifted data.

1

Create the rotated sorted array

Create a vector called nums with the exact values [4,5,6,7,0,1,2].

DSA Rust

# Create a vector called nums with values [4,5,6,7,0,1,2]
// Your code here

Hint

Use let nums = vec![4,5,6,7,0,1,2]; to create the vector.

2

Initialize the minimum value variable

Create a variable called min_value and set it to the first element of nums using nums[0].

DSA Rust

fn main() {
    let nums = vec![4,5,6,7,0,1,2];
    // Create min_value and set it to nums[0]
    // Your code here
}

Hint

Use let mut min_value = nums[0]; to create a mutable variable for the minimum value.

3

Find the minimum value using a loop

Use a for loop with variable num to iterate over nums. Inside the loop, update min_value if num is smaller than min_value.

DSA Rust

fn main() {
    let nums = vec![4,5,6,7,0,1,2];
    let mut min_value = nums[0];
    // Use a for loop to check each num in nums
    // Update min_value if num is smaller
    // Your code here
}

Hint

Use for num in nums.iter() to loop. Compare *num with min_value and update if smaller.

4

Print the minimum value

Print the min_value using println!("{}", min_value).

DSA Rust

fn main() {
    let nums = vec![4,5,6,7,0,1,2];
    let mut min_value = nums[0];
    for num in nums.iter() {
        if *num < min_value {
            min_value = *num;
        }
    }
    // Print min_value
    // Your code here
}

Hint

Use println!("{}", min_value); to print the smallest number.

Practice

1. What is the main goal when finding the minimum in a rotated sorted array?

easy

A. To sort the array back to original order

B. To find the smallest element after the rotation point

C. To count how many times the array was rotated

D. To find the largest element in the array

Solution

1. Step 1: Understand the rotated sorted array

   A rotated sorted array is a sorted array shifted some positions, so the smallest element is the rotation point.

2. Step 2: Identify the minimum element

   The minimum element is the smallest value after the rotation point, which is what we want to find.

3. Final Answer:

   To find the smallest element after the rotation point -> Option B

4. Quick Check:

   Minimum element = smallest after rotation [OK]

Hint: Look for the smallest number where order breaks [OK]

Common Mistakes:

• Confusing minimum with maximum element
• Trying to count rotations instead of finding minimum
• Assuming array is fully sorted without rotation

2. Which Rust code snippet correctly declares a function to find the minimum in a rotated sorted array?

easy

A. fn find_min(nums: &[i32]) -> i32 { /* implementation */ }

B. function find_min(nums: Vec<i32>) -> i32 { }

C. def find_min(nums: &[i32]) -> i32 { }

D. fn find_min(nums: Vec<i32>) { return i32; }

Solution

1. Step 1: Check Rust function syntax

   Rust functions start with fn, parameters have types, and return type is after ->.

2. Step 2: Validate each option

   fn find_min(nums: &[i32]) -> i32 { /* implementation */ } uses correct Rust syntax with slice &[i32] and return type i32. Others use wrong keywords or syntax.

3. Final Answer:

   fn find_min(nums: &[i32]) -> i32 { /* implementation */ } -> Option A

4. Quick Check:

   Rust fn syntax with &[i32] and -> i32 [OK]

Hint: Rust functions use fn, param: type, -> return_type [OK]

Common Mistakes:

• Using 'function' keyword instead of 'fn'
• Mixing Python or JavaScript syntax
• Incorrect return type placement

3. What is the output of the following Rust code?

fn main() {
    let nums = vec![4,5,6,7,0,1,2];
    let min = find_min(&nums);
    println!("{}", min);
}

fn find_min(nums: &[i32]) -> i32 {
    let mut left = 0;
    let mut right = nums.len() - 1;
    while left < right {
        let mid = left + (right - left) / 2;
        if nums[mid] > nums[right] {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    nums[left]
}

medium

A. 0

B. 4

C. 7

D. 2

Solution

1. Step 1: Understand the binary search logic

   The code uses binary search to find the smallest element by comparing mid and right elements.

2. Step 2: Trace the loop with input [4,5,6,7,0,1,2]

   Mid moves to find the point where array rotates; finally left points to 0, the smallest element.

3. Final Answer:

   0 -> Option A

4. Quick Check:

   Minimum in rotated array = 0 [OK]

Hint: Binary search narrows to smallest element [OK]

Common Mistakes:

• Returning nums[mid] instead of nums[left]
• Confusing left and right pointers
• Not handling rotation correctly

4. Identify the bug in this Rust function to find the minimum in a rotated sorted array:

fn find_min(nums: &[i32]) -> i32 {
    let mut left = 0;
    let mut right = nums.len();
    while left < right {
        let mid = left + (right - left) / 2;
        if nums[mid] > nums[right] {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    nums[left]
}

medium

A. nums[left] should be nums[right] at the end

B. left should start at 1 instead of 0

C. mid calculation causes infinite loop

D. right should be nums.len() - 1 to avoid out-of-bounds

Solution

1. Step 1: Check initialization of right pointer

   right is set to nums.len(), which is out of array bounds (index max is len-1).

2. Step 2: Correct right initialization

   right should be nums.len() - 1 to avoid accessing nums[right] out of bounds.

3. Final Answer:

   right should be nums.len() - 1 to avoid out-of-bounds -> Option D

4. Quick Check:

   Array index max = len-1 [OK]

Hint: Array indices go from 0 to len-1 [OK]

Common Mistakes:

• Using nums.len() as index causes panic
• Starting left at 1 misses first element
• Changing mid calculation unnecessarily

5. Given a rotated sorted array that may contain duplicates, which modification is needed in the binary search to find the minimum correctly?

fn find_min(nums: &[i32]) -> i32 {
    let mut left = 0;
    let mut right = nums.len() - 1;
    while left < right {
        let mid = left + (right - left) / 2;
        if nums[mid] > nums[right] {
            left = mid + 1;
        } else if nums[mid] < nums[right] {
            right = mid;
        } else {
            right -= 1; // What is the reason for this?
        }
    }
    nums[left]
}

hard

A. To avoid infinite loop by always moving left

B. To speed up search by skipping elements

C. To handle duplicates by shrinking right when equal to mid

D. To sort the array during search

Solution

1. Step 1: Understand duplicates effect

   When nums[mid] equals nums[right], we cannot decide which side minimum lies, so we reduce right by one.

2. Step 2: Reason for right -= 1

   This safely shrinks the search space without missing the minimum, handling duplicates correctly.

3. Final Answer:

   To handle duplicates by shrinking right when equal to mid -> Option C

4. Quick Check:

   Duplicates need careful right pointer move [OK]

Hint: When equal, reduce right to handle duplicates [OK]

Common Mistakes:

• Ignoring duplicates causes wrong minimum
• Moving left instead of right when equal
• Assuming no duplicates in rotated array