Find Minimum in Rotated Sorted Array in DSA Go - Execution Trace

Concept Flow - Find Minimum in Rotated Sorted Array

Start

↓

Set low=0, high=len-1

↓

While low < high

↓

Calculate mid = (low+high)/2

↓

Compare nums[mid

↓

mid

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high = mid

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Repeat loop

↓

Return nums[low

We use two pointers low and high to narrow down the search space by comparing middle and high elements until low meets high, which points to the minimum.

Execution Sample

DSA Go

nums := []int{4,5,6,7,0,1,2}
low, high := 0, len(nums)-1
for low < high {
  mid := (low + high) / 2
  if nums[mid] > nums[high] {
    low = mid + 1
  } else {
    high = mid
  }
}
min := nums[low]

This code finds the minimum element in a rotated sorted array by binary searching with low and high pointers.

Execution Table

Step	Operation	low	high	mid	Compare nums[mid] and nums[high]	Pointer Changes	Array State
1	Initialize pointers	0	6	-	-	None	[4,5,6,7,0,1,2]
2	Calculate mid	0	6	3	nums[3]=7 > nums[6]=2	low = mid + 1 = 4	[4,5,6,7,0,1,2]
3	Calculate mid	4	6	5	nums[5]=1 <= nums[6]=2	high = mid = 5	[4,5,6,7,0,1,2]
4	Calculate mid	4	5	4	nums[4]=0 <= nums[5]=1	high = mid = 4	[4,5,6,7,0,1,2]
5	Check loop condition	4	4	-	-	Loop ends	[4,5,6,7,0,1,2]
6	Return minimum	4	4	-	-	Minimum found at index 4	[4,5,6,7,0,1,2]

💡 Loop ends when low equals high, pointing to the minimum element.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	Final
low	0	4	4	4	4	4
high	6	6	5	4	4	4
mid	-	3	5	4	-	-

Key Moments - 3 Insights

Why do we compare nums[mid] with nums[high] instead of nums[low]?

Comparing nums[mid] with nums[high] helps decide which side contains the minimum. If nums[mid] > nums[high], minimum is right side, so low moves up (see Step 2). Otherwise, minimum is left side or at mid, so high moves down (see Steps 3 and 4).

Why does the loop stop when low equals high?

When low equals high, the search space is one element, which must be the minimum (see Step 5). Continuing would not narrow the search further.

What happens if the array is not rotated?

If not rotated, nums[mid] will always be less or equal to nums[high], so high moves down until low reaches 0, returning the first element as minimum.

Visual Quiz - 3 Questions

Look at the execution_table at Step 3, what is the value of 'high' after the pointer change?

A 6

B 4

C 5

D 3

💡 Hint

Check the 'Pointer Changes' and 'high' column in Step 3 of execution_table.

At which step does the loop condition become false?

A Step 4

B Step 5

C Step 6

D Step 3

💡 Hint

Look at the 'Operation' and 'Pointer Changes' columns in Step 5 where loop ends.

If nums[mid] was always less than or equal to nums[high], how would 'low' change?

A low would stay the same

B low would increase each step

C low would decrease

D low would jump to high

💡 Hint

Refer to the 'Pointer Changes' column where low changes only if nums[mid] > nums[high].

Concept Snapshot

Find Minimum in Rotated Sorted Array:
- Use two pointers: low=0, high=len-1
- While low < high:
  - mid = (low+high)//2
  - If nums[mid] > nums[high], low = mid+1
  - Else, high = mid
- Return nums[low] as minimum
- Works by narrowing search space based on mid-high comparison

Full Transcript

This concept finds the minimum element in a rotated sorted array using a binary search approach. We start with two pointers, low at 0 and high at the last index. We calculate mid as the middle index between low and high. We compare the element at mid with the element at high. If nums[mid] is greater than nums[high], the minimum must be to the right of mid, so we move low to mid + 1. Otherwise, the minimum is at mid or to the left, so we move high to mid. We repeat this until low equals high, which points to the minimum element. This method efficiently finds the minimum in O(log n) time by narrowing the search space each step.

Practice

1. What is the main idea behind finding the minimum element in a rotated sorted array?

easy

A. Use a hash map to store elements and find minimum

B. Sort the array again to find the minimum

C. Traverse the array linearly to find the smallest element

D. Use binary search to find the point where order breaks

Solution

1. Step 1: Understand the rotated sorted array property

   A rotated sorted array is sorted but rotated at some pivot point, causing a break in order.

2. Step 2: Use binary search to find the break point

   Binary search helps find the point where the order breaks, which is where the minimum element lies.

3. Final Answer:

   Use binary search to find the point where order breaks -> Option D

4. Quick Check:

   Binary search finds minimum efficiently [OK]

Hint: Look for the pivot where order breaks using binary search [OK]

Common Mistakes:

• Trying to sort the array again wastes time
• Using linear search ignores the sorted property
• Using extra space unnecessarily

2. Which Go code snippet correctly initializes variables for binary search to find minimum in a rotated sorted array nums?

easy

A. start, end := -1, len(nums)-1

B. start, end := 0, len(nums)-1

C. start, end := 1, len(nums)

D. start, end := 0, len(nums)

Solution

1. Step 1: Understand array indexing in Go

   Arrays are zero-indexed, so start should be 0 and end should be last index (length - 1).

2. Step 2: Check each option for correct initialization

   start, end := 0, len(nums)-1 correctly sets start=0 and end=len(nums)-1, which is valid for binary search.

3. Final Answer:

   start, end := 0, len(nums)-1 -> Option B

4. Quick Check:

   Correct index range initialization [OK]

Hint: Start at 0 and end at length minus one for array indices [OK]

Common Mistakes:

• Using length as end index causes out-of-range errors
• Starting at 1 misses the first element
• Negative start index is invalid

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

nums := []int{4,5,6,7,0,1,2}
start, end := 0, len(nums)-1
for start < end {
  mid := (start + end) / 2
  if nums[mid] > nums[end] {
    start = mid + 1
  } else {
    end = mid
  }
}
fmt.Println(nums[start])

medium

A. 0

B. 4

C. 7

D. 2

Solution

1. Step 1: Trace the binary search loop

   Initial start=0, end=6. Mid=(0+6)/2=3, nums[3]=7 > nums[6]=2, so start=4.

2. Step 2: Continue loop until start meets end

   Now start=4, end=6. Mid=(4+6)/2=5, nums[5]=1 <= nums[6]=2, so end=5. Next mid=(4+5)/2=4, nums[4]=0 <= nums[5]=1, so end=4. Loop ends when start=4, end=4.

3. Final Answer:

   0 -> Option A

4. Quick Check:

   Minimum element found at index 4 is 0 [OK]

Hint: Compare mid with end to decide search side [OK]

Common Mistakes:

• Confusing mid calculation causing infinite loop
• Incorrectly updating start or end pointers
• Printing nums[end] instead of nums[start]

4. Identify the bug in this Go code to find minimum in rotated sorted array nums:

start, end := 0, len(nums)-1
for start <= end {
  mid := (start + end) / 2
  if nums[mid] > nums[end] {
    start = mid
  } else {
    end = mid - 1
  }
}
fmt.Println(nums[start])

medium

A. All of the above

B. Loop condition should be start < end, not <=

C. end should be set to mid, not mid - 1

D. start should be set to mid + 1, not mid

Solution

1. Step 1: Analyze loop condition and pointer updates

   The loop uses start <= end which can cause infinite loop when start == end. Also, start = mid does not move start forward properly, and end = mid - 1 skips the mid element incorrectly.

2. Step 2: Correct the loop and pointer updates

   Loop should be start < end. When nums[mid] > nums[end], start = mid + 1 to move forward. Else, end = mid to include mid in next search.

3. Final Answer:

   All of the above -> Option A

4. Quick Check:

   Proper loop and pointer updates avoid infinite loops [OK]

Hint: Use start < end and adjust pointers by +1 or no change [OK]

Common Mistakes:

• Using <= causes infinite loops
• Not moving start pointer forward correctly
• Skipping mid element by subtracting 1 from end

5. Given a rotated sorted array nums that may contain duplicates, which modification is needed in the binary search approach to find the minimum element?

hard

A. Ignore duplicates and apply normal binary search

B. If nums[mid] == nums[end], increase start by 1 to skip duplicate

C. If nums[mid] == nums[end], reduce end by 1 to skip duplicate

D. Sort the array first to handle duplicates

Solution

1. Step 1: Understand duplicates affect binary search

   Duplicates can cause nums[mid] == nums[end], making it unclear which side to search.

2. Step 2: Modify binary search to handle duplicates

   When nums[mid] == nums[end], safely reduce end by 1 to skip duplicate without missing minimum.

3. Final Answer:

   If nums[mid] == nums[end], reduce end by 1 to skip duplicate -> Option C

4. Quick Check:

   Handle duplicates by shrinking search space carefully [OK]

Hint: When equal, reduce end to skip duplicates safely [OK]

Common Mistakes:

• Ignoring duplicates causes wrong results
• Increasing start instead of reducing end misses minimum
• Sorting wastes time and is unnecessary