Search in Rotated Sorted Array in DSA C++ - Build from Scratch

Search in Rotated Sorted Array

📖 Scenario: Imagine you have a list of numbers that was originally sorted from smallest to largest. But then, someone rotated it at some point, so the order is mixed. You want to find if a certain number is in this rotated list.

🎯 Goal: You will write a program to find the position of a given number in a rotated sorted array using an efficient search method.

📋 What You'll Learn

Create a vector called nums with the exact values: {4, 5, 6, 7, 0, 1, 2}

Create an integer variable called target and set it to 0

Write a function called search that takes nums and target and returns the index of target in nums or -1 if not found

Print the result of calling search(nums, target)

💡 Why This Matters

🌍 Real World

Rotated sorted arrays appear in systems where data is shifted or rotated, such as circular buffers or rotated logs. Efficient search helps quickly find needed information.

💼 Career

Understanding how to search in rotated sorted arrays is useful for software engineers working on performance-critical applications, embedded systems, or interview coding challenges.

1

Create the rotated sorted array

Create a vector called nums with the exact values {4, 5, 6, 7, 0, 1, 2}.

DSA C++

#include 
#include 

int main() {
    // Create vector nums with values {4, 5, 6, 7, 0, 1, 2}
    // Your code here
    return 0;
}

Hint

Use std::vector<int> nums = {4, 5, 6, 7, 0, 1, 2}; to create the vector.

2

Set the target number to search

Create an integer variable called target and set it to 0.

DSA C++

#include 
#include 

int main() {
    std::vector nums = {4, 5, 6, 7, 0, 1, 2};
    // Create int variable target and set it to 0
    // Your code here
    return 0;
}

Hint

Use int target = 0; to create the target variable.

3

Write the search function for rotated sorted array

Write a function called search that takes const std::vector<int>& nums and int target and returns the index of target in nums or -1 if not found. Use a binary search approach adapted for rotated arrays.

DSA C++

#include 
#include 

// Write search function here
// Your code here

int main() {
    std::vector nums = {4, 5, 6, 7, 0, 1, 2};
    int target = 0;

    return 0;
}

Hint

Use binary search with checks on which half is sorted to decide where to search next.

4

Print the search result

Print the result of calling search(nums, target).

DSA C++

#include 
#include 

int search(const std::vector& nums, int target) {
    int left = 0;
    int right = (int)nums.size() - 1;
    while (left  nums = {4, 5, 6, 7, 0, 1, 2};
    int target = 0;

    // Print the index of target in nums
    // Your code here
    return 0;
}

Hint

Use std::cout << search(nums, target) << std::endl; to print the result.

Practice

1. What is the main challenge when searching for an element in a rotated sorted array compared to a normal sorted array?

easy

A. The array is not fully sorted, so normal binary search cannot be directly applied.

B. The array contains duplicate elements which confuse the search.

C. The array size is unknown, so we cannot set search boundaries.

D. The array elements are all negative numbers.

Solution

1. Step 1: Understand normal binary search assumption

   Binary search works efficiently on fully sorted arrays by comparing mid element to target.

2. Step 2: Identify the problem with rotated arrays

   Rotated sorted arrays are sorted but shifted, so the order is broken at the rotation point, making direct binary search invalid.

3. Final Answer:

   The array is not fully sorted, so normal binary search cannot be directly applied. -> Option A

4. Quick Check:

   Rotated array breaks sorted order = need modified search [OK]

Hint: Look for the rotation point to adjust binary search [OK]

Common Mistakes:

• Assuming array is fully sorted and applying normal binary search
• Ignoring rotation and searching linearly
• Confusing duplicates with rotation

2. Which of the following is the correct syntax to start a binary search loop for a rotated sorted array in C++?

easy

A. if (arr[mid] == target)

B. for (int i = 0; i < n; i++)

C. while (start <= end)

D. do { ... } while (start < end)

Solution

1. Step 1: Identify loop type for binary search

   Binary search uses a while loop with condition start <= end to narrow search range.

2. Step 2: Check options for correct loop syntax

   while (start <= end) uses while (start <= end), which is standard for binary search loops.

3. Final Answer:

   while (start <= end) -> Option C

4. Quick Check:

   Binary search loop uses while (start <= end) [OK]

Hint: Binary search loops use while (start <= end) [OK]

Common Mistakes:

• Using for loop instead of while for binary search
• Using incorrect loop condition like start < end
• Confusing if statement with loop

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

int arr[] = {4,5,6,7,0,1,2};
int target = 0;
int start = 0, end = 6;
int result = -1;
while (start <= end) {
  int mid = start + (end - start) / 2;
  if (arr[mid] == target) {
    result = mid;
    break;
  }
  if (arr[start] <= arr[mid]) {
    if (target >= arr[start] && target < arr[mid])
      end = mid - 1;
    else
      start = mid + 1;
  } else {
    if (target > arr[mid] && target <= arr[end])
      start = mid + 1;
    else
      end = mid - 1;
  }
}
std::cout << result << std::endl;

medium

A. 3

B. 0

C. -1

D. 4

Solution

1. Step 1: Trace initial variables and mid calculation

   start=0, end=6, mid=3 (element 7). arr[mid] != target (0).

2. Step 2: Check which half is sorted and adjust search

   arr[start]=4 <= arr[mid]=7, left half sorted. Target 0 not in [4,7), so search right half: start=mid+1=4.

3. Step 3: Next iteration mid=5 (element 1)

   arr[mid]=1 != target. arr[start]=0 <= arr[mid]=1, left half sorted. Target 0 in [0,1), so end=mid-1=4.

4. Step 4: Next iteration mid=4 (element 0)

   arr[mid]=0 == target, result=4, break loop.

5. Final Answer:

   4 -> Option D

6. Quick Check:

   Target 0 found at index 4 [OK]

Hint: Check sorted half and adjust start/end accordingly [OK]

Common Mistakes:

• Not updating start/end correctly based on sorted half
• Confusing indices and returning wrong index
• Missing break after finding target

4. Identify the bug in the following code snippet for searching in a rotated sorted array:

int search(int arr[], int n, int target) {
  int start = 0, end = n - 1;
  while (start < end) {
    int mid = start + (end - start) / 2;
    if (arr[mid] == target) return mid;
    if (arr[start] <= arr[mid]) {
      if (target >= arr[start] && target < arr[mid])
        end = mid - 1;
      else
        start = mid + 1;
    } else {
      if (target > arr[mid] && target <= arr[end])
        start = mid + 1;
      else
        end = mid - 1;
    }
  }
  return -1;
}

medium

A. The mid calculation is incorrect and can cause overflow.

B. The loop condition should be while (start <= end) instead of while (start < end).

C. The target comparison should use > instead of >=.

D. The function does not handle empty arrays.

Solution

1. Step 1: Analyze loop condition impact

   Using while (start < end) misses checking when start == end, possibly skipping last element.

2. Step 2: Correct loop condition for complete search

   Changing to while (start <= end) ensures all elements are checked, including when start equals end.

3. Final Answer:

   The loop condition should be while (start <= end) instead of while (start < end). -> Option B

4. Quick Check:

   Loop must cover start == end to check last element [OK]

Hint: Use while (start <= end) for full binary search coverage [OK]

Common Mistakes:

• Using while (start < end) and missing last element
• Ignoring overflow in mid calculation (not in this code though)
• Incorrect comparison operators

5. Given a rotated sorted array arr = {15, 18, 2, 3, 6, 12}, write a C++ function to find the index of the minimum element. What is the output of your function?

hard

A. 2

B. 0

C. 5

D. 3

Solution

1. Step 1: Understand the minimum element in rotated array

   The minimum element is the rotation point where order breaks. Here, 2 at index 2 is smallest.

2. Step 2: Use binary search to find minimum

   Check mid and compare with end to decide which half contains minimum. Eventually, index 2 is found.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Minimum element 2 is at index 2 [OK]

Hint: Compare mid element with end to find rotation point [OK]

Common Mistakes:

• Returning first element index instead of minimum
• Confusing maximum with minimum element
• Not handling rotation correctly