Find Minimum in Rotated Sorted Array in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C++

int findMin(vector<int>& nums) {
    return nums[[1]];
}

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Anums.size()

B1

Cnums.size() - 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 1 instead of 0

Using nums.size() 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 in binary search.

DSA C++

int left = 0, right = nums.size() - 1;
while (left < right) {
    int mid = (left [1] right) / 2;
    // rest of code
}

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A+

B+ +

Attempts:

3 left

💡 Hint

Common Mistakes

Using only left or right

Using bit shifts incorrectly

The middle index is calculated as (left + right) / 2.

3fill in blank

hard

Fix the error in the condition to check if mid element is greater than the right element.

DSA C++

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

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A<

B==

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of >

Using == which is incorrect here

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 binary search loop and return the minimum element.

DSA C++

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

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Aleft

Bright

Cmid

Dnums.size()

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid in the condition

Using nums.size() which is out of bounds

The loop continues while left is less than right to narrow down the search space.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each number to its square if the number is less than 5.

DSA C++

auto squares = std::unordered_map<int, int>{{
    { [1], [2] } for (int num : nums) if (num [3] 5)
}};

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Anum

Bnum * num

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<'

Using num instead of num * num for value

This comprehension maps each number to its square only if the number is less than 5.