Binary Search Iterative Approach in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Binary Search for Existing Element

What is the output of the following TypeScript code that uses iterative binary search to find the index of the number 7 in the sorted array?

DSA Typescript

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

const arr = [1, 3, 5, 7, 9, 11];
console.log(binarySearch(arr, 7));

A 4

B 3

C -1

D 2

💡 Hint

Remember that array indices start at 0 and the array is sorted.

The number 7 is at index 3 in the array [1, 3, 5, 7, 9, 11]. The binary search finds it and returns its index.

❓ Predict Output

intermediate

Output of Binary Search for Non-Existing Element

What will be the output of this TypeScript code when searching for the number 8 in the sorted array?

DSA Typescript

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

const arr = [2, 4, 6, 10, 12];
console.log(binarySearch(arr, 8));

A 0

B 2

C 3

D -1

💡 Hint

If the target is not found, the function returns -1.

The number 8 is not in the array, so the binary search returns -1 to indicate it was not found.

🧠 Conceptual

advanced

Why Use Iterative Binary Search Instead of Recursive?

Which of the following is the main advantage of using an iterative binary search over a recursive binary search?

A Iterative binary search uses less memory because it avoids the overhead of recursive calls.

B Iterative binary search is easier to write and understand than recursive binary search.

C Iterative binary search always finds the target faster than recursive binary search.

D Iterative binary search can handle unsorted arrays better than recursive binary search.

💡 Hint

Think about how function calls use memory.

Recursive calls add overhead to the call stack, which uses more memory. Iterative approach uses a loop and constant memory.

🔧 Debug

advanced

Identify the Bug in Binary Search Implementation

What is the bug in this iterative binary search code that causes it to fail for some inputs?

DSA Typescript

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

A The right boundary should be initialized to arr.length - 1, not arr.length.

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

C The loop condition should be left < right instead of left <= right.

D The function should return left instead of -1 when target is not found.

💡 Hint

Check the initial value of the right pointer and array indexing.

Array indices go from 0 to length - 1. Using arr.length as right causes out-of-bound access.

🚀 Application

expert

Binary Search to Find First Occurrence in Sorted Array with Duplicates

Given a sorted array with duplicate elements, which iterative binary search modification correctly finds the index of the first occurrence of the target number?

DSA Typescript

function firstOccurrence(arr: number[], target: number): number {
  let left = 0;
  let right = arr.length - 1;
  let result = -1;
  while (left <= right) {
    const mid = Math.floor((left + right) / 2);
    if (arr[mid] === target) {
      result = mid;
      right = mid - 1;
    } else if (arr[mid] < target) {
      left = mid + 1;
    } else {
      right = mid - 1;
    }
  }
  return result;
}

const arr = [1, 2, 2, 2, 3, 4];
console.log(firstOccurrence(arr, 2));

A 2

B 3

C 1

D -1

💡 Hint

When you find the target, keep searching to the left to find the first occurrence.

The first occurrence of 2 is at index 1. The code updates result and moves right to mid - 1 to find earlier occurrences.

Practice

1. What is the main purpose of the Binary Search Iterative Approach in a sorted array?

easy

A. To find the maximum value in the array

B. To sort the array elements in ascending order

C. To reverse the array elements

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

Solution

1. Step 1: Understand the goal of binary search

   Binary search is used to find a target value in a sorted array by dividing the search space repeatedly.

2. Step 2: Identify the iterative approach

   The iterative approach 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 D

4. Quick Check:

   Binary search = efficient target search [OK]

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

Common Mistakes:

• Confusing binary search with sorting algorithms
• Thinking binary search works on unsorted arrays
• Assuming it finds max/min values

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

easy

A. const mid = (low + high) / 2;

B. const mid = Math.floor((low + high) / 2);

C. const mid = Math.ceil((low + high) / 2);

D. const mid = low + high;

Solution

1. Step 1: Understand index calculation in binary search

   The middle index must be an integer index, so we use Math.floor to round down the division result.

2. Step 2: Check each option

   const mid = (low + high) / 2; returns a float, const mid = Math.floor((low + high) / 2); correctly floors the value, const mid = Math.ceil((low + high) / 2); rounds up which can cause errors, const mid = low + high; sums indices without division.

3. Final Answer:

   const mid = Math.floor((low + high) / 2); -> Option B

4. Quick Check:

   Use Math.floor for middle index [OK]

Hint: Always floor the average of low and high [OK]

Common Mistakes:

• Using division without flooring causing float index
• Using ceil which may skip elements
• Adding indices without dividing

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

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

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

medium

A. 2

B. 3

C. -1

D. 4

Solution

1. Step 1: Trace the binary search for target 6

   Initial low=0, high=4, mid=2 (element 6). arr[mid] equals target, so return 2.

2. Step 2: Confirm output

   Function returns index 2, which is printed.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Target 6 found at index 2 [OK]

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

Common Mistakes:

• Off-by-one errors in index calculation
• Returning wrong index when target found
• Confusing return value with element value

4. Identify the bug in this iterative binary search code snippet and select the correct fix:

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

medium

A. Change let high = arr.length; to let high = arr.length - 1;

B. Change low <= high to low < high

C. Change mid = Math.floor((low + high) / 2); to mid = Math.ceil((low + high) / 2);

D. Remove the else before if (arr[mid] < target)

Solution

1. Step 1: Identify the initialization error

   High should be the last valid index, which is arr.length - 1, not arr.length.

2. Step 2: Understand impact of wrong high value

   Setting high to arr.length causes out-of-bound access at arr[mid] when mid equals arr.length.

3. Final Answer:

   Change let high = arr.length; to let high = arr.length - 1; -> Option A

4. Quick Check:

   High index must be last valid array index [OK]

Hint: High index = array length minus one [OK]

Common Mistakes:

• Setting high to array length causing out-of-bounds
• Changing loop condition incorrectly
• Using ceil instead of floor for mid

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] equals target, update low to mid + 1 and continue searching right

B. Return mid immediately when arr[mid] equals target

C. When arr[mid] equals target, update high to mid - 1 and continue searching left

D. Use recursion instead of iteration

Solution

1. Step 1: Understand the problem with duplicates

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

2. Step 2: Modify search to find first occurrence

   When target found at mid, move high to mid - 1 to continue searching left side for earlier occurrence.

3. Step 3: Keep track of potential answer

   Store mid as potential answer and continue until low > high.

4. Final Answer:

   When arr[mid] equals target, update high to mid - 1 and continue searching left -> Option C

5. Quick Check:

   Move left on match to find first occurrence [OK]

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

Common Mistakes:

• Returning immediately on first match
• Searching right side on match
• Ignoring duplicates in binary search