Find Minimum in Rotated Sorted Array in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the first element of the array.

DSA Rust

fn find_min(nums: &[i32]) -> i32 {
    nums[[1]]
}

Drag options to blanks, or click blank then click option'

A0

B1

Cnums.len() - 1

Dnums.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 1 instead of 0

Using nums.len() which is out of bounds

The minimum element in a non-rotated sorted array is the first element at index 0.

2fill in blank

medium

Complete the code to find the middle index between left and right.

DSA Rust

let mid = (left [1] right) / 2;

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A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Subtracting instead of adding

Multiplying instead of adding

To find the middle index, add left and right, then divide by 2.

3fill in blank

hard

Fix the error in the comparison to decide which half to search next.

DSA Rust

if nums[mid] [1] nums[right] {
    left = mid + 1;
} else {
    right = mid;
}

Drag options to blanks, or click blank then click option'

A<

B>

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>'

Using '==' which is incorrect for this logic

If nums[mid] is greater than nums[right], the minimum is in the right half, so move left to mid + 1.

4fill in blank

hard

Fill both blanks to complete the while loop condition and return statement.

DSA Rust

while [1] < [2] {
    let mid = (left + right) / 2;
    if nums[mid] > nums[right] {
        left = mid + 1;
    } else {
        right = mid;
    }
}
return nums[left];

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cmid

Dnums.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid in loop condition

Returning nums[right] instead of nums[left]

The loop continues while left is less than right. After the loop, nums[left] is the minimum.

5fill in blank

hard

Fill all three blanks to complete the function that finds the minimum in a rotated sorted array.

DSA Rust

fn find_min(nums: &[i32]) -> i32 {
    let (mut left, mut right) = (0, nums.len() [1] 1);
    while left [2] right {
        let mid = (left + right) / 2;
        if nums[mid] [3] nums[right] {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    nums[left]
}

Drag options to blanks, or click blank then click option'

A-

B<

C>

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using length instead of length - 1

Using <= instead of < in loop

Using < instead of > in comparison

Initialize right as length minus one. Loop while left is less than right. If nums[mid] is greater than nums[right], move left to mid + 1.

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