Search in Rotated Sorted Array 2 | Leetcode #4 | Modified Binary Search
Introduction:
Welcome back to another video! In this session, we'll tackle the "Search in Sorted Array II" problem. This problem revolves around searching for a target value in a sorted array that may contain duplicates. We'll employ the binary search algorithm with some modifications to efficiently handle the presence of duplicates. Let's dive into the problem statement, explain the approach, and provide a step-by-step solution.
Problem Statement:
The Search in Sorted Array II problem requires us to determine if a target value exists in a sorted array that may contain duplicates. We need to return a boolean value indicating whether the target is present or not.
Approach:
To solve this problem, we'll adapt the binary search algorithm to handle duplicates in the sorted array. We'll make slight modifications to account for the possibility of multiple occurrences of the target value. Let's explore the step-by-step solution using binary search with these adaptations.
Solution Steps:
Initialize two pointers, left and right, to represent the boundaries of the search space. Set left to 0 (the start of the array) and right to the length of the array minus one.
Enter a loop while left is less than or equal to right.
Calculate the middle index as mid by taking the average of left and right, rounded down to the nearest integer.
Compare the value at index mid with the target value.
If the value at mid is equal to the target, return true as the target exists in the array.
If the value at mid is equal to the value at left, it means we have encountered a duplicate.
Increment left by 1 to skip the duplicate element.
Continue to the next iteration without modifying right.
If the value at mid is greater than the value at left, it means the left half of the array is sorted without duplicates.
If the target value is within the range of the left half (i.e., target greater than or equal to nums[left] and target less than nums[mid]), set right to mid - 1 to search in the left half.
Otherwise, set left to mid + 1 to search in the right half.
If the value at mid is less than the value at left, it means the right half of the array is sorted without duplicates.
If the target value is within the range of the right half (i.e., target greater=nums[mid] and target less= nums[right]), set left to mid + 1 to search in the right half.
Otherwise, set right to mid - 1 to search in the left half.
If we exit the loop without finding the target, return false.
Explanation:
By modifying the binary search algorithm to handle duplicates, we can efficiently navigate the search space. When encountering duplicates, we skip them by incrementing left and continuing the search. This approach allows us to accurately determine the presence or absence of the target value in the sorted array. The time complexity remains O(log n), where n is the length of the array.
Conclusion:
The Search in Sorted Array II problem challenges us to find whether a target value exists in a sorted array with possible duplicates. By adapting the binary search algorithm to handle duplicates, we can efficiently navigate the search space and determine the presence or absence of the target value. I hope this explanation has provided you with a clear understanding of the problem and its solution using binary search with modifications
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