Search in Rotated Sorted Array in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the binary search loop.

DSA C++

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

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Amid

Bright

Cleft

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid instead of left in the loop condition

Using right < left instead of <=.

The binary search continues while the left index is less than or equal to the right index.

2fill in blank

medium

Complete the code to check if the middle element is the target.

DSA C++

if (nums[[1]] == target) {
    return mid;
}

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Amid

Bright

Cleft

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using left or right instead of mid to check the element.

We compare the middle element nums[mid] with the target to check for a match.

3fill in blank

hard

Fix the error in the condition to check if the left half is sorted.

DSA C++

if (nums[left] <= nums[[1]]) {
    // Left half is sorted
}

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Aright

Bmid

Cleft

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing nums[left] with nums[right] instead of nums[mid].

We compare nums[left] with nums[mid] to check if the left half is sorted.

4fill in blank

hard

Fill both blanks to update the search boundaries correctly when the left half is sorted.

DSA C++

if (target >= nums[left] && target < nums[mid]) {
    right = [1];
} else {
    left = [2];
}

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

Amid - 1

Bmid + 1

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right updates.

Using mid instead of mid ± 1.

If the target is in the left sorted half, move right to mid - 1; otherwise, move left to mid + 1.

5fill in blank

hard

Fill all three blanks to update the search boundaries when the right half is sorted.

DSA C++

else {
    if (target > nums[mid] && target <= nums[[1]]) {
        left = [2];
    } else {
        right = [3];
    }
}

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

Aright

Bmid + 1

Cmid - 1

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using left instead of right in comparison.

Incorrect boundary updates.

If the target is in the right sorted half, move left to mid + 1; otherwise, move right to mid - 1.