Find Peak Element Using Binary Search in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the middle index for binary search.

DSA C++

int mid = low [1] (high - low) / 2;

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A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid = (low + high) / 2 which can cause overflow.

Using subtraction instead of addition.

We calculate mid as low plus half the distance between high and low: mid = low + (high - low) / 2.

2fill in blank

medium

Complete the condition to check if mid is a peak element.

DSA C++

if ((mid == 0 || nums[mid] > nums[mid - 1]) && (mid == n - 1 || nums[mid] > nums[[1]]))

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Amid + 1

Bn - 1

Cmid

Dmid - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid - 1 instead of mid + 1.

Comparing with mid instead of mid + 1.

We check if the element at mid is greater than the next element by comparing nums[mid] > nums[mid + 1].

3fill in blank

hard

Fix the error in updating the high pointer when the left neighbor is greater.

DSA C++

if (nums[mid] < nums[[1]]) {
    high = mid [2] 1;
}

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Amid - 1

Bmid + 1

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid + 1 instead of mid - 1.

Using addition instead of subtraction.

If the left neighbor is greater, we move high to mid - 1 to search the left half.

4fill in blank

hard

Fill both blanks to update low pointer when right neighbor is greater and to return the peak index.

DSA C++

else {
    low = mid [1] 1;
}

return [2];

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A+

B-

Cmid

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition for low.

Returning low instead of mid.

When the right neighbor is greater, we move low to mid + 1. The peak index is mid when found.

5fill in blank

hard

Fill all three blanks to complete the binary search loop and return the peak element index.

DSA C++

while (low [1] high) {
    int mid = low + (high - low) / 2;
    if ((mid == 0 || nums[mid] > nums[mid - 1]) && (mid == n - 1 || nums[mid] > nums[mid + 1])) {
        return [2];
    } else if (mid > 0 && nums[mid] < nums[mid - 1]) {
        high = mid [3] 1;
    } else {
        low = mid + 1;
    }
}
return low;

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A<=

Bmid

C-

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of <= in the loop condition.

Returning low instead of mid inside the loop.

Using addition instead of subtraction when updating high.

The loop continues while low is less than or equal to high. Return mid when peak found. Decrease high by 1 from mid when left neighbor is greater.