Why Find Minimum in Rotated Sorted Array in DSA Rust?

The Big Idea

What if you could find the smallest number in a twisted list without checking every single one?

The Scenario

Imagine you have a sorted list of numbers, but someone rotated it at some point, like a clock hand moving. You want to find the smallest number quickly.

Without a smart way, you might check every number one by one.

The Problem

Checking each number takes a long time if the list is big.

It's easy to make mistakes counting or missing the smallest number.

This slow method wastes time and effort.

The Solution

Using a smart search method, you can find the smallest number fast without checking all numbers.

This method looks at the middle and decides which half to search next, cutting the search in half each time.

Before vs After

✗ Before

for i in 0..nums.len() {
    if nums[i] < min {
        min = nums[i];
    }
}

✓ After

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

What It Enables

You can quickly find the smallest number in a rotated sorted list, even if it's very large.

Real Life Example

Think of a circularly shifted schedule where you want to find the earliest time slot quickly without checking every slot.

Key Takeaways

Manual search is slow and error-prone for rotated lists.

Binary search cuts the search space in half each step.

Efficiently finds the minimum in O(log n) time.

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