Binary Search Iterative Approach in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the middle index in the binary search loop.

DSA Javascript

let mid = Math.floor(([1] + high) / 2);

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

A0

Bmid

Chigh

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid instead of low causes incorrect calculation.

Using high alone does not consider the lower bound.

The middle index is calculated by averaging the low and high indices. Here, low is the correct variable to use.

2fill in blank

medium

Complete the code to update the low index when the target is greater than the middle element.

DSA Javascript

if (arr[mid] < target) {
  [1] = mid + 1;
}

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

Amid

Blow

Chigh

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Updating high instead of low.

Setting low to mid instead of mid + 1.

If the target is greater than the middle element, we move the low index to one position after mid.

3fill in blank

hard

Fix the error in the loop condition to correctly run the binary search.

DSA Javascript

while ([1] <= high) {
  let mid = Math.floor((low + high) / 2);
  // rest of code
}

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

Atarget

Bmid

Clow

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid in the condition causes infinite loops.

Using target in the condition is invalid.

The loop should continue while low is less than or equal to high. Using mid or others is incorrect.

4fill in blank

hard

Fill both blanks to correctly update the high index when the target is less than the middle element.

DSA Javascript

if (arr[mid] > target) {
  [1] = mid [2] 1;
}

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

Ahigh

Blow

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Updating low instead of high.

Adding 1 instead of subtracting 1.

When the target is less than the middle element, we move the high index to one less than mid.

5fill in blank

hard

Fill both blanks to return the index if the target is found, or -1 if not found.

DSA Javascript

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

// After loop
return [2];

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

Amid

B-1

Clow

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Returning low or high instead of mid.

Returning 0 instead of -1 when not found.

Return mid when target is found. Return -1 if not found after loop ends.

Practice

1. What is the main purpose of the binary search iterative approach in a sorted array?

easy

A. To reverse the array elements

B. To find the position of a target value efficiently by repeatedly dividing the search interval in half

C. To sort the array elements in ascending order

D. To find the maximum value in the array

Solution

1. Step 1: Understand binary search goal

   Binary search aims to find a target value's position quickly in a sorted array by dividing the search space repeatedly.

2. Step 2: Identify the iterative approach

   The iterative method uses a loop to narrow down the search range until the target is found or the range is empty.

3. Final Answer:

   To find the position of a target value efficiently by repeatedly dividing the search interval in half -> Option B

4. Quick Check:

   Binary search = efficient position search [OK]

Hint: Binary search halves the search space each step [OK]

Common Mistakes:

• Confusing binary search with sorting algorithms
• Thinking it works on unsorted arrays
• Assuming it finds maximum or minimum values

2. Which of the following is the correct way to initialize the mid index in an iterative binary search in JavaScript?

easy

A. const mid = Math.ceil((start + end) / 2);

B. const mid = (start + end) / 2;

C. const mid = Math.floor((start + end) / 2);

D. const mid = start + end;

Solution

1. Step 1: Understand index calculation

   Mid index must be an integer index, so we use Math.floor to round down the average of start and end.

2. Step 2: Check options for correct syntax

   const mid = Math.floor((start + end) / 2); uses Math.floor((start + end) / 2), which correctly calculates the middle index as an integer.

3. Final Answer:

   const mid = Math.floor((start + end) / 2); -> Option C

4. Quick Check:

   Mid index = Math.floor average [OK]

Hint: Use Math.floor to get integer mid index [OK]

Common Mistakes:

• Using division without floor, resulting in float index
• Using Math.ceil which can cause out-of-range index
• Adding start and end without division

3. What will be the output of the following JavaScript code?

function binarySearch(arr, target) {
  let start = 0;
  let end = arr.length - 1;
  while (start <= end) {
    const mid = Math.floor((start + end) / 2);
    if (arr[mid] === target) return mid;
    else if (arr[mid] < target) start = mid + 1;
    else end = mid - 1;
  }
  return -1;
}

console.log(binarySearch([2, 4, 6, 8, 10], 6));

medium

A. 2

B. 3

C. -1

D. 4

Solution

1. Step 1: Trace the binary search steps

   Array is [2,4,6,8,10], target is 6. Initial start=0, end=4, mid=2. arr[2] = 6 matches target.

2. Step 2: Return index when found

   Since arr[mid] equals target, function returns mid = 2 immediately.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Target 6 found at index 2 [OK]

Hint: Check mid index value first for quick match [OK]

Common Mistakes:

• Confusing index with value
• Off-by-one errors in start/end updates
• Returning value instead of index

4. Identify the bug in the following iterative binary search code snippet:

function binarySearch(arr, target) {
  let start = 0;
  let end = arr.length - 1;
  while (start < end) {
    const mid = Math.floor((start + end) / 2);
    if (arr[mid] === target) return mid;
    else if (arr[mid] < target) start = mid + 1;
    else end = mid - 1;
  }
  return -1;
}

medium

A. The loop condition should be start <= end instead of start < end

B. The mid calculation should use Math.ceil instead of Math.floor

C. The function should return arr[mid] instead of mid

D. The start should be initialized to 1 instead of 0

Solution

1. Step 1: Analyze loop condition impact

   The loop uses start < end, which misses the case when start equals end, possibly skipping the last element check.

2. Step 2: Correct loop condition for full search

   Changing to start <= end ensures the loop checks all elements including when start equals end.

3. Final Answer:

   The loop condition should be start <= end instead of start < end -> Option A

4. Quick Check:

   Loop condition must include equality [OK]

Hint: Use <= to include last element in search [OK]

Common Mistakes:

• Using < excludes last element check
• Changing mid calculation unnecessarily
• Returning value instead of index

5. Given a sorted array with possible duplicate values, how can you modify the iterative binary search to find the first occurrence of a target value?

function binarySearchFirst(arr, target) {
  let start = 0;
  let end = arr.length - 1;
  let result = -1;
  while (start <= end) {
    const mid = Math.floor((start + end) / 2);
    if (arr[mid] === target) {
      result = mid;
      end = mid - 1; // Continue searching left
    } else if (arr[mid] < target) {
      start = mid + 1;
    } else {
      end = mid - 1;
    }
  }
  return result;
}

hard

A. Search right side by moving start to mid + 1 when target is found

B. Return immediately when target is found at mid

C. Change mid calculation to use Math.ceil instead of Math.floor

D. Keep track of found index and continue searching left by moving end to mid - 1

Solution

1. Step 1: Understand duplicate handling

   To find first occurrence, we must not stop at first match but continue searching left side.

2. Step 2: Use result variable and adjust end pointer

   Store current match index in result, then move end to mid - 1 to search left side for earlier occurrence.

3. Final Answer:

   Keep track of found index and continue searching left by moving end to mid - 1 -> Option D

4. Quick Check:

   Track index + search left for first occurrence [OK]

Hint: On match, move left to find first occurrence [OK]

Common Mistakes:

• Stopping search immediately on first match
• Searching right side instead of left
• Ignoring duplicates in binary search