Binary Search Iterative Approach in DSA Go - Execution Trace

Concept Flow - Binary Search Iterative Approach

Start with sorted array and target

↓

Set low = 0, high = len-1

↓

While low <= high

↓

Calculate mid = (low + high) / 2

↓

Compare array[mid

↓

high=mid-1

↓

Repeat loop

↓

If not found, return -1

This flow shows how binary search narrows down the search range by adjusting low and high pointers until the target is found or the range is empty.

Execution Sample

DSA Go

func binarySearch(arr []int, target int) int {
  low, high := 0, len(arr)-1
  for low <= high {
    mid := (low + high) / 2
    if arr[mid] == target {
      return mid
    } else if arr[mid] < target {
      low = mid + 1
    } else {
      high = mid - 1
    }
  }
  return -1
}

This code searches for target in a sorted array using an iterative binary search and returns the index or -1 if not found.

Execution Table

Step	Operation	low	high	mid	arr[mid]	Comparison	Action	Search Range Visual
1	Initialize low and high	0	6	-	-	-	Start search	[1, 3, 5, 7, 9, 11, 13]
2	Calculate mid	0	6	3	7	7 < 9?	arr[mid] < target, low = mid + 1	[1, 3, 5, (7), 9, 11, 13]
3	Update low	4	6	-	-	-	New range low=4, high=6	[9, 11, 13]
4	Calculate mid	4	6	5	11	11 > 9?	arr[mid] > target, high = mid - 1	[9, (11), 13]
5	Update high	4	4	-	-	-	New range low=4, high=4	[9]
6	Calculate mid	4	4	4	9	9 == 9?	Found target, return 4	[9]
7	End	4	4	4	9	Target found	Return index 4	[9]

💡 Target found at index 4, search ends.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
low	0	0	4	4	4	4	4
high	6	6	6	6	4	4	4
mid	-	3	-	5	-	4	4
arr[mid]	-	7	-	11	-	9	9

Key Moments - 3 Insights

Why do we update 'low' to mid + 1 instead of mid when arr[mid] < target?

Because arr[mid] is less than target and already checked, so we exclude mid by setting low to mid + 1 (see Step 2 and 3 in execution_table).

Why does the loop stop when low > high?

Because the search range becomes empty, meaning target is not in the array. The condition low <= high ensures we only search valid ranges (see Step 7 exit).

Why do we calculate mid as (low + high) / 2 each iteration?

To check the middle element of the current search range and decide which half to search next (see Steps 2, 4, 6).

Visual Quiz - 3 Questions

Look at the execution_table at Step 4, what is the value of 'mid' and 'arr[mid]'?

A mid=4, arr[mid]=9

B mid=5, arr[mid]=11

C mid=3, arr[mid]=7

D mid=6, arr[mid]=13

💡 Hint

Check the 'mid' and 'arr[mid]' columns in row Step 4 of execution_table.

At which step does the condition low <= high become false, ending the loop if target was not found?

A After Step 7

B Step 7

C Step 6

D Step 3

💡 Hint

Look at the exit_note and understand the loop condition from concept_flow.

If the target was 14 (not in array), how would the 'low' and 'high' pointers change after Step 6?

A low would be 4, high would be 4

B low would be 0, high would be 6

C low would be 7, high would be 6

D low would be 5, high would be 4

💡 Hint

Think about how low moves when arr[mid] < target and when the search ends (see variable_tracker).

Concept Snapshot

Binary Search Iterative Approach:
- Works on sorted arrays
- Use two pointers: low and high
- Calculate mid = (low + high) / 2
- Compare arr[mid] with target
- If equal, return mid
- If arr[mid] < target, move low to mid + 1
- Else, move high to mid - 1
- Repeat until low > high
- Return -1 if not found

Full Transcript

Binary Search Iterative Approach searches a sorted array by repeatedly dividing the search range in half. It starts with low at 0 and high at the last index. Each step calculates mid as the middle index. If the middle element equals the target, it returns the index. If the middle element is less than the target, it moves low to mid + 1 to search the right half. If greater, it moves high to mid - 1 to search the left half. This repeats until low exceeds high, meaning the target is not found. The execution table shows each step's low, high, mid, and comparison results, helping visualize how the search narrows down.

Practice

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

easy

A. To keep track of the number of iterations

B. To store the target value to search

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

D. To hold the entire array being searched

Solution

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

   The mid variable calculates the middle index between the current low and high indices to split the search range. By checking the element at mid, the algorithm decides whether to search the left or right half next, making the search efficient.

2. Final Answer:

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

3. Quick Check:

   mid = middle index [OK]

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

Common Mistakes:

• Confusing mid with target value
• Using mid as a counter instead of index
• Not updating mid inside the loop

2. Which of the following is the correct way to calculate the middle index mid in Go to avoid integer overflow?

easy

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

B. mid := (high - low) / 2

C. mid := (low - high) / 2

D. mid := low + (high - low) / 2

Solution

1. Step 1: Understand integer overflow risk and safe calculation

   Adding low + high directly can overflow if both are large integers. Calculating mid as low + (high - low) / 2 avoids overflow by subtracting first.

2. Final Answer:

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

3. Quick Check:

   Safe mid calculation avoids overflow [OK]

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

Common Mistakes:

• Using (low + high)/2 directly causing overflow
• Subtracting incorrectly causing negative mid
• Confusing order of operations

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

package main
import "fmt"

func binarySearch(arr []int, target int) int {
    low, high := 0, len(arr)-1
    for low <= high {
        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
}

func main() {
    arr := []int{2, 4, 6, 8, 10}
    result := binarySearch(arr, 6)
    fmt.Println(result)
}

medium

A. 3

B. 2

C. -1

D. 0

Solution

1. Step 1: Trace the binary search for target 6 and verify return

   Array is [2,4,6,8,10]. Initial low=0, high=4. mid=2. arr[2]=6 matches target. Function returns mid = 2 immediately.

2. Final Answer:

   2 -> Option B

3. Quick Check:

   Index of 6 in array = 2 [OK]

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

Common Mistakes:

• Returning value instead of index
• Off-by-one errors in mid calculation
• Not updating low/high correctly

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

func binarySearch(arr []int, target int) int {
    low, high := 0, len(arr)
    for low <= high {
        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. The initial value of high should be len(arr) - 1

B. The loop condition should be low < high

C. The mid calculation causes overflow

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

Solution

1. Step 1: Check high initialization and correct it

   High is set to len(arr), which is out of array index range (max index is len(arr)-1). Set high to len(arr) - 1 to avoid index out of range error when accessing arr[mid].

2. Final Answer:

   The initial value of high should be len(arr) - 1 -> Option A

3. Quick Check:

   High = len(arr)-1 fixes index error [OK]

Hint: High index must be last valid array index [OK]

Common Mistakes:

• Setting high to len(arr) causing runtime error
• Using wrong loop condition
• Returning value instead of index

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

hard

A. When arr[mid] == target, update high = mid - 1 and continue searching

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

C. When arr[mid] == target, update low = mid + 1 and continue searching

D. Use linear search after binary search to find first occurrence

Solution

1. Step 1: Understand duplicates issue and modify search logic

   Standard binary search returns any occurrence, not necessarily the first. When arr[mid] == target, move high to mid - 1 to continue searching left side for earlier occurrence, but store mid as potential answer.

2. Final Answer:

   When arr[mid] == target, update high = mid - 1 and continue searching -> Option A

3. Quick Check:

   Adjust high to find first occurrence [OK]

Hint: Move high left on match to find first occurrence [OK]

Common Mistakes:

• Returning immediately on first match
• Moving low right on match losing first occurrence
• Using linear search unnecessarily