Find Minimum in Rotated Sorted Array in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find the minimum element output

What is the output of this Rust code that finds the minimum element in a rotated sorted array?

DSA Rust

fn find_min(nums: Vec<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]
}

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

A 0

B 1

C 4

D 7

💡 Hint

Think about how the binary search narrows down the smallest element in a rotated array.

The code uses binary search to find the smallest element. Since the array is rotated, the smallest element is the pivot point where the order breaks. The code compares middle and right elements to decide which half to search next. The output is 0, the smallest element.

❓ Predict Output

intermediate

Output for single element array

What will be printed when running this Rust code with a single-element rotated array?

DSA Rust

fn find_min(nums: Vec<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]
}

fn main() {
    let nums = vec![10];
    println!("{}", find_min(nums));
}

A panic at index out of bounds

B 10

C 0

D 1

💡 Hint

Consider what happens when the array has only one element.

With one element, the function returns that element immediately without entering the loop. So the output is 10.

❓ Predict Output

advanced

Output with no rotation (sorted array)

What is the output of this Rust code when the input array is sorted but not rotated?

DSA Rust

fn find_min(nums: Vec<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]
}

fn main() {
    let nums = vec![1,2,3,4,5,6,7];
    println!("{}", find_min(nums));
}

A 1

B 7

C 4

D 0

💡 Hint

If the array is not rotated, the smallest element is the first one.

Since the array is sorted and not rotated, the smallest element is the first element, 1.

🔧 Debug

advanced

Identify the error in this Rust code

What error will this Rust code produce when trying to find the minimum in a rotated sorted array?

DSA Rust

fn find_min(nums: Vec<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]
}

fn main() {
    let nums = vec![3,4,5,1,2];
    println!("{}", find_min(nums));
}

A compiles and prints 3

B infinite loop

C panic: index out of bounds

D compiles and prints 1

💡 Hint

Check the value of right initialization and how it is used as an index.

The code sets right to nums.len(), which is out of bounds for indexing. Accessing nums[right] causes a panic.

🧠 Conceptual

expert

Why binary search works for finding minimum in rotated sorted array

Why does binary search correctly find the minimum element in a rotated sorted array?

A Because the array is reversed, binary search finds the maximum instead of minimum.

B Because the array is fully sorted, binary search always finds the minimum at the middle index.

C Because the array is unsorted, binary search randomly guesses the minimum element.

D Because the array is partially sorted, comparing middle and right elements helps discard half the search space each time.

💡 Hint

Think about how rotation affects sorted order and how comparisons guide the search.

The rotated array is sorted in two parts. Comparing middle and right elements tells which part contains the minimum, allowing binary search to discard half the array each step.

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