Binary Search on Answer Technique in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the binary search boundaries for the answer.

DSA Go

low, high := 0, [1]

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Alen(arr)

Bmath.MaxInt64

C1000000

Darr[len(arr)-1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using array length as upper bound which may be too small

Using last element of array which may not represent max answer

We set high to a large number as the upper bound for the answer search space.

2fill in blank

medium

Complete the code to calculate the middle value in binary search safely.

DSA Go

mid := low + (high - [1]) / 2

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Amid

Blow

Chigh

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using (low + high) / 2 directly which can overflow

Using wrong variable in subtraction

We subtract low from high to avoid overflow, then add low back.

3fill in blank

hard

Fix the error in the condition to check if mid is a valid answer.

DSA Go

if canAchieve(arr, [1]) {

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Amid

Bhigh

Clow

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Passing high or low instead of mid

Passing mid + 1 which is not the current candidate

We check if the current middle value mid can achieve the goal.

4fill in blank

hard

Fill both blanks to update the binary search boundaries correctly.

DSA Go

if canAchieve(arr, mid) {
    high = [1]
} else {
    low = [2]
}

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Amid

Bmid + 1

Cmid - 1

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to mid instead of mid + 1 causing infinite loop

Setting high to mid - 1 which may skip valid answers

If mid works, we try smaller or equal values by setting high = mid. Otherwise, we try larger values by setting low = mid + 1.

5fill in blank

hard

Fill all three blanks to return the final answer after binary search.

DSA Go

for low < high {
    mid := low + (high - low) / 2
    if canAchieve(arr, mid) {
        high = [1]
    } else {
        low = [2]
    }
}
return [3]

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Amid

Blow

Cmid + 1

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Returning mid instead of low after loop

Incorrectly updating boundaries causing infinite loop

Inside the loop, update high to mid if condition is true, else update low to mid + 1. After the loop, low holds the smallest valid answer.