Binary Search Iterative Approach in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

int start = [1];

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

A0

B-1

C1

Darray.length

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from 1 instead of 0 causes skipping the first element.

Using array.length causes index out of bounds.

The start index in binary search always begins at 0, the first element of the array.

2fill in blank

medium

Complete the code to calculate the middle index safely to avoid overflow.

DSA Java

int mid = start + (end [1] start) / 2;

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A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' causes overflow for large indexes.

Using '*' or '/' here is incorrect for midpoint calculation.

To avoid overflow, subtract start from end before dividing by 2, then add start back.

3fill in blank

hard

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

DSA Java

while (start [1] end) {

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

A<=

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' skips checking when start equals end.

Using '>' or '>=' causes the loop to never run or run incorrectly.

The loop should run while start is less than or equal to end to include all elements.

4fill in blank

hard

Fill both blanks to update the search range after comparing mid value.

DSA Java

if (arr[mid] < target) {
    start = mid [1] 1;
} else {
    end = mid [2] 1;
}

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

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' when increasing start causes infinite loop.

Using '+' when decreasing end causes wrong search range.

If mid value is less than target, move start to mid + 1; else move end to mid - 1.

5fill in blank

hard

Fill both blanks to return the correct index or -1 if not found.

DSA Java

if (arr[mid] == target) {
    return [1];
}
// After loop
return [2];

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

Amid

B-1

Cstart

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Returning start or end instead of mid causes wrong index.

Not returning -1 causes incorrect result when target is missing.

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

Practice

1. What is the main purpose of the mid variable in the iterative binary search algorithm?

easy

A. To find the middle index of the current search range

B. To store the target value to search

C. To keep track of the number of iterations

D. To hold the final result index

Solution

1. Step 1: Understand the role of mid in binary search

   The mid variable calculates the middle index between low and high pointers to split the search range.

2. Step 2: Purpose of mid

   It helps decide whether to search the left or right half by comparing the middle element with the target.

3. Final Answer:

   To find the middle index of the current search range -> Option A

4. Quick Check:

   mid = middle index [OK]

Hint: Mid splits search range to half each step [OK]

Common Mistakes:

• Confusing mid with target value
• Thinking mid counts iterations
• Assuming mid stores final answer

2. Which of the following is the correct way to calculate the middle index mid in an iterative binary search to avoid integer overflow?

easy

A. mid = low + (high - low) / 2

B. mid = (low + high) / 2

C. mid = (low - high) / 2

D. mid = (low * high) / 2

Solution

1. Step 1: Understand integer overflow risk

   Adding low and high directly can exceed integer limits if both are large.

2. Step 2: Safe calculation method

   Using low + (high - low) / 2 prevents overflow by subtracting before adding.

3. Final Answer:

   mid = low + (high - low) / 2 -> Option A

4. Quick Check:

   Safe mid calculation avoids overflow [OK]

Hint: Use low + (high - low)/2 to prevent overflow [OK]

Common Mistakes:

• Using (low + high)/2 causing overflow
• Using subtraction incorrectly
• Multiplying low and high

3. What will be the output of the following Java code snippet?

int[] arr = {2, 4, 6, 8, 10};
int target = 6;
int low = 0, high = arr.length - 1;
int result = -1;
while (low <= high) {
  int mid = low + (high - low) / 2;
  if (arr[mid] == target) {
    result = mid;
    break;
  } else if (arr[mid] < target) {
    low = mid + 1;
  } else {
    high = mid - 1;
  }
}
System.out.println(result);

medium

A. 4

B. 3

C. 2

D. -1

Solution

1. Step 1: Trace the binary search steps

   Initial low=0, high=4, mid=2. arr[2]=6 equals target 6, so result=2 and loop breaks.

2. Step 2: Output the result

   Result is 2, which is the index of target 6 in the array.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Index of 6 in arr = 2 [OK]

Hint: Check mid index value matches target first [OK]

Common Mistakes:

• Confusing index with value
• Not breaking loop on match
• Incorrect mid calculation

4. Identify the error in this iterative binary search code snippet and choose the correct fix:

int binarySearch(int[] arr, int target) {
  int low = 0, high = arr.length - 1;
  while (low < high) {
    int mid = low + (high - low) / 2;
    if (arr[mid] == target) return mid;
    else if (arr[mid] < target) low = mid + 1;
    else high = mid - 1;
  }
  return -1;
}

medium

A. Remove else if and use only if statements

B. Change mid calculation to (low + high) / 2

C. Return low instead of -1 at the end

D. Change while condition to low <= high

Solution

1. Step 1: Analyze loop condition

   The condition low < high misses the case when low equals high, possibly skipping the last element.

2. Step 2: Correct loop condition

   Changing to low <= high ensures the last element is checked.

3. Final Answer:

   Change while condition to low <= high -> Option D

4. Quick Check:

   Loop must include low == high [OK]

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

Common Mistakes:

• Using < instead of <= in loop
• Incorrect mid calculation
• Returning wrong index on failure

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?

hard

A. When arr[mid] == target, return mid immediately

B. When arr[mid] == target, update result and move high to mid - 1

C. When arr[mid] == target, move low to mid + 1

D. Ignore duplicates and return any matching index

Solution

1. Step 1: Understand duplicates in binary search

   Standard binary search returns any matching index, not necessarily the first occurrence.

2. Step 2: Modify search to find first occurrence

   When arr[mid] == target, store mid as potential answer and continue searching left by setting high = mid - 1.

3. Final Answer:

   When arr[mid] == target, update result and move high to mid - 1 -> Option B

4. Quick Check:

   Move left on match to find first occurrence [OK]

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

Common Mistakes:

• Returning immediately on first match
• Moving right on match
• Ignoring duplicates