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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the first element as minimum if array is not rotated.

DSA Java

if (nums[0] < nums[nums.length - 1]) {
    return nums[1];
}

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

A[nums.length - 1]

B[1]

C[0]

D[2]

Attempts:

3 left

💡 Hint

Common Mistakes

Returning nums[1] instead of nums[0]

Returning the last element nums[nums.length - 1]

If the first element is less than the last, the array is not rotated, so the minimum is the first element nums[0].

2fill in blank

medium

Complete the code to calculate the middle index correctly.

DSA Java

int mid = [1] + (right - left) / 2;

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

Anums.length

Bleft

C0

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using right instead of left as base

Using 0 or nums.length which are incorrect indices

The middle index is calculated by adding left to half the distance between right and left to avoid overflow.

3fill in blank

hard

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

DSA Java

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

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

A<

B==

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of <

Using == or != which are incorrect for this logic

If nums[mid] is less than nums[right], the minimum is in the left half including mid, so we move right to mid.

4fill in blank

hard

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

DSA Java

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

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

Aleft

Bright

Cmid

Dnums.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid in the loop condition

Using nums.length which is out of index

The loop continues while left is less than right, narrowing down the search. Finally, nums[left] is the minimum.

5fill in blank

hard

Fill all three blanks to complete the binary search method to find minimum in rotated sorted array.

DSA Java

public int findMin(int[] nums) {
    int [1] = 0, [2] = nums.length - 1;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] < nums[right]) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return nums[[3]];
}

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

Aleft

Bright

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid or right in return instead of left

Incorrect initialization of left or right

Initialize left and right pointers, use binary search to narrow down, and return nums[left] as minimum.

Practice

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

easy

A. To find the largest element in the array

B. To find the smallest element after the rotation point

C. To count the number of rotations

D. To sort the array again

Solution

1. Step 1: Understand the rotated sorted array

   A rotated sorted array is a sorted array shifted at some pivot. The smallest element is the rotation point.

2. Step 2: Identify the minimum element

   The minimum element is the smallest value after the rotation, which is the point where the order breaks.

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 point where order breaks to find minimum [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 Java syntax correctly declares a method to find the minimum in a rotated sorted array?

easy

A. public int findMin(int[] nums) {}

B. int findMin(int nums) {}

C. void findMin(int[] nums) {}

D. public static findMin(int[] nums) {}

Solution

1. Step 1: Check method return type and parameters

   The method should return an int (the minimum value) and accept an int array parameter.

2. Step 2: Validate Java method syntax

   public int findMin(int[] nums) {} uses 'public int findMin(int[] nums) {}' which is correct Java syntax for this method.

3. Final Answer:

   public int findMin(int[] nums) {} -> Option A

4. Quick Check:

   Correct method signature = public int findMin(int[] nums) {} [OK]

Hint: Method must return int and accept int array parameter [OK]

Common Mistakes:

• Missing return type or using void incorrectly
• Incorrect parameter syntax
• Missing access modifier or static keyword incorrectly

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

int[] nums = {4,5,6,7,0,1,2};
System.out.println(findMin(nums));

public static int findMin(int[] nums) {
    int left = 0, right = nums.length - 1;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] > nums[right]) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return nums[left];
}

medium

A. 2

B. 4

C. 7

D. 0

Solution

1. Step 1: Understand the binary search logic

   The code uses binary search to find the minimum by comparing middle and right elements, narrowing the search space.

2. Step 2: Trace the loop with given array

   Array: [4,5,6,7,0,1,2] - mid=3 (value 7), 7 > 2, so left = mid+1 = 4 - left=4, right=6 - mid=5 (value 1), 1 <= 2, so right = mid = 5 - left=4, right=5 - mid=4 (value 0), 0 <= 1, so right = mid = 4 Loop ends when left == right == 4, nums[4] = 0.

3. Final Answer:

   0 -> Option D

4. Quick Check:

   Minimum element found = 0 [OK]

Hint: Compare mid and right to narrow search for minimum [OK]

Common Mistakes:

• Returning nums[mid] instead of nums[left]
• Confusing comparison direction in binary search
• Not updating left or right correctly

4. Identify the error in this code snippet for finding minimum in rotated sorted array:

public static int findMin(int[] nums) {
    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] > nums[right]) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return nums[left];
}

medium

A. The return statement should be 'nums[right]'

B. The mid calculation is incorrect

C. The loop condition should be 'left < right' instead of 'left <= right'

D. The array should be sorted before searching

Solution

1. Step 1: Analyze the loop condition

   The loop uses 'left <= right', which can cause infinite loop or index errors because left can go beyond right.

2. Step 2: Correct the loop condition

   Changing the condition to 'left < right' ensures the loop stops when left meets right, correctly identifying the minimum index.

3. Final Answer:

   The loop condition should be 'left < right' instead of 'left <= right' -> Option C

4. Quick Check:

   Loop condition must be 'left < right' [OK]

Hint: Use 'left < right' to avoid infinite loops in binary search [OK]

Common Mistakes:

• Using 'left <= right' causing infinite loops
• Incorrectly adjusting right pointer
• Returning wrong array element after loop

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

int[] nums = {2,2,2,0,1,2};

hard

A. Add a condition to reduce right by one when nums[mid] equals nums[right]

B. Ignore duplicates and proceed as usual

C. Sort the array before searching

D. Use linear search instead of binary search

Solution

1. Step 1: Understand the effect of duplicates

   Duplicates can cause ambiguity when nums[mid] equals nums[right], making it unclear which side to search.

2. Step 2: Modify binary search to handle duplicates

   When nums[mid] == nums[right], reduce right by one to shrink search space safely without missing minimum.

3. Final Answer:

   Add a condition to reduce right by one when nums[mid] equals nums[right] -> Option A

4. Quick Check:

   Handle duplicates by shrinking right pointer [OK]

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

Common Mistakes:

• Ignoring duplicates causing wrong minimum
• Sorting array which is inefficient
• Switching to linear search unnecessarily