Search in Rotated Sorted Array II

class Solution {
public:
    bool searchElement(vector<int>& arr, int target) {
        int size = arr.size();
        if (size == 0) return false;
        int left = 0;
        int right = size - 1;
    
        while (left <= right) {
            int mid = left + (right - left) / 2;
    
            if (arr[mid] == target) {
                return true;
            }
    
            if (!canOptimizeSearch(arr, left, arr[mid])) {
                left++;
                continue;
            }
    
            // Determine the arrays the pivot and target belong to
            bool pivotInFirst = isPivotInFirst(arr, left, arr[mid]);
            bool targetInFirst = isPivotInFirst(arr, left, target);
            if (pivotInFirst ^ targetInFirst) {
                if (pivotInFirst) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            } else {
                if (arr[mid] < target) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
        }
        return false;
    }
    
    bool canOptimizeSearch(vector<int>& arr, int start, int val) {
        return arr[start] != val;
    }
    
    bool isPivotInFirst(vector<int>& arr, int start, int val) {
        return arr[start] <= val;
    }
};

This solution is designed to handle the problem of searching for a target element in a rotated sorted array, which might contain duplicates. Here, the implementation is done using C++ and revolves around a binary search technique with added conditions tailored to determine the array's division post-rotation.

Understand the key parts of the solution:

• Initialization and Edge Case Handling: Before proceeding with the main logic, the program ensures that the array is not empty. It sets left and right pointers to mark the boundary of the array segment being considered.

• Binary Search with Modified Conditions: The usual binary search mechanism is modified to account for the rotation and possible duplicates. The loop continues as long as left is less than or equal to right. A mid index is calculated each time to check if the target is found.

• Handling Duplicates: The function canOptimizeSearch is utilized to skip over duplicates by incrementally adjusting the left pointer when encountering elements that do not contribute to narrowing the search range.

• Segment Determination: The logic determines which segmented part of the array (before or after the pivot point) the mid value and the target value lie in, using isPivotInFirst. Based on the comparison of these segments (whether they match or not), the search bounds are adjusted.

• Adjustment Based on Comparisons: Using the information about segments and comparing the mid value directly with the target, the boundaries (left or right) are adjusted to narrow down the search effectively.

Overall, this implementation effectively navigates the complexities introduced by the rotated nature of the array and the presence of duplicates, providing a method to achieve logarithmic search time in many cases, barring the instances dominated by duplicates where it may deteriorate to a linear search.

class Solution {
    public boolean findElement(int[] array, int target) {
        int length = array.length;
        if (length == 0) return false;
        int end = length - 1;
        int start = 0;
    
        while (start <= end) {
            int mid = start + (end - start) / 2;
    
            if (array[mid] == target) {
                return true;
            }
    
            if (!isHelpful(array, start, array[mid])) {
                start++;
                continue;
            }
            boolean isPivotInFirst = isElementInFirstSegment(array, start, array[mid]);
            boolean isTargetInFirst = isElementInFirstSegment(array, start, target);
            if (isPivotInFirst ^ isTargetInFirst) {
                if (isPivotInFirst) {
                    start = mid + 1;
                } else {
                    end = mid - 1;
                }
            } else {
                if (array[mid] < target) {
                    start = mid + 1;
                } else {
                    end = mid - 1;
                }
            }
        }
        return false;
    }
    
    private boolean isHelpful(int[] arr, int start, int element) {
        return arr[start] != element;
    }
    
    private boolean isElementInFirstSegment(int[] arr, int start, int element) {
        return arr[start] <= element;
    }
}

The provided Java solution implements a method to determine if a target element exists in a rotated sorted array. Here's an overview of how the solution works:

• Define the method findElement that takes an integer array and a target integer to find in the array.
• Initialize pointers for the start and end of the array.
• Employ a while loop to execute as long as the start pointer is less than or equal to the end pointer:
  • Calculate the mid-point of the current segment of the array.
  • Check if the element at the mid-point is the target. If true, return true.
  • Determine if the current segment is informative (i.e., the element at the start does not equal the element at mid-point) by calling the isHelpful method.
  • If uninformative, increment the start pointer and skip further checks for the current iteration.
  • Determine the configuration of the array segment and the target's position relative to the mid-point:
    • Use the method isElementInFirstSegment to check if the mid-point and target elements are less than the start element.
    • Apply XOR to decide whether the search should continue in the first or second half of the current segment based on comparisons of the target and mid-point positions.
• If the target is not found by the end of the loop, return false.

This method effectively handles cases with duplicates through its informational check and adapts the binary search algorithm to accommodate the potential distortions in order caused by the rotated nature of the array. The operations inside the loop ensure that at each stage, the search space is halved, following the principles of binary search, but with adjustments to manage the rotation and duplicates in the array.

bool searchKeyExists(int* array, int arrayLength, int searchTarget) {
    if (arrayLength == 0) return false;
    int lastIndex = arrayLength - 1;
    int firstIndex = 0;
    while (firstIndex <= lastIndex) {
        int middleIndex = firstIndex + (lastIndex - firstIndex) / 2;
        if (array[middleIndex] == searchTarget) return true;
        if (!isBinaryHelpful(array, firstIndex, array[middleIndex])) {
            firstIndex++;
            continue;
        }
        bool pivotPosition = isInFirstArray(array, firstIndex, array[middleIndex]);
        bool targetPosition = isInFirstArray(array, firstIndex, searchTarget);
        if (pivotPosition ^ targetPosition) {
            if (pivotPosition) {
                firstIndex = middleIndex + 1;
            } else {
                lastIndex = middleIndex - 1;
            }
        } else {
            if (array[middleIndex] < searchTarget) {
                firstIndex = middleIndex + 1;
            } else {
                lastIndex = middleIndex - 1;
            }
        }
    }
    return false;
}
bool isBinaryHelpful(int* array, int startIdx, int element) {
    return array[startIdx] != element;
}
bool isInFirstArray(int* array, int startIdx, int element) {
    return array[startIdx] <= element;
}

The implementation provided in C handles the search for a specified target in a rotated sorted array where the array might contain duplicates. Let's break down the solution approach and key components of the code:

1. Start by checking if the array length is zero, returning false since there are no elements to search within.
2. Initialize indices to represent the beginning (firstIndex) and the end (lastIndex) of array sections being checked during each step.
3. The main loop continues until firstIndex exceeds lastIndex, adjusting these indices based on comparisons with the middle element and the search target.
4. Calculate middleIndex using the average of firstIndex and lastIndex, effectively splitting the array section into halves.
5. If the element at middleIndex equals the search target, return true, indicating that the target is found.
6. Utilize helper functions isBinaryHelpful and isInFirstArray to decide how to adjust the indices:
   • isBinaryHelpful checks if the array can still be considered binary-search friendly by ensuring that the start element of the current segment and the middle element are different.
   • isInFirstArray determines whether an element is part of the first subarray formed after the array's rotation point based on relative values at startIdx.
7. Use logic based on bit-exclusive OR (^) to handle situations where the search target and middle element lie in different subarrays formed due to the rotation of a sorted array.
8. Adjust firstIndex and lastIndex based on comparisons of values relative to middleIndex, handling both scenarios where values are greater or smaller than the search target.
9. If after the termination of the while loop the target hasn't been found, return false.

This implementation is efficient in conducting searches even in complex scenarios introduced by array rotations and duplications, adapting binary search techniques accordingly.

var findTarget = function (elements, value) {
    let length = elements.length;
    if (length == 0) return false;
    let high = length - 1;
    let low = 0;
    while (low <= high) {
        let middle = low + Math.floor((high - low) / 2);
        if (elements[middle] == value) {
            return true;
        }
        if (!canReduceSearch(elements, low, elements[middle])) {
            low++;
            continue;
        }
        let isPivotInFirst = isElementInFirst(elements, low, elements[middle]);
        let isTargetInFirst = isElementInFirst(elements, low, value);
        if (isPivotInFirst ^ isTargetInFirst) {
            if (isPivotInFirst) {
                low = middle + 1;
            } else {
                high = middle - 1;
            }
        } else {
            if (elements[middle] < value) {
                low = middle + 1;
            } else {
                high = middle - 1;
            }
        }
    }
    return false;
};
    
function canReduceSearch(elements, start, element) {
    return elements[start] != element;
}
    
function isElementInFirst(elements, start, element) {
    return elements[start] <= element;
}

Searching for a target value in a rotated sorted array can be efficiently performed using binary search, even when duplicates exist, as demonstrated in the provided JavaScript implementation. The function findTarget leverages binary search by continuously halving the search space until the target is found or the search bounds are exhausted.

• First, the function checks if the array is empty, immediately returning false if so.
• It then initializes pointers (low, high) to the start and end of the array, respectively.

The core of the algorithm lies in adjusting these pointers based on the comparison of the middle element of the current search range against the target. If the middle element matches the target, the function returns true.

To manage duplicates and rotations:

• The canReduceSearch function checks if the current search can be simplified by verifying that the start element is different from the mid element.
• The function isElementInFirst helps determine if the element (mid or target) is in the first half of the rotated array based on its value relative to the start element.

Depending on the relationship between the middle element, target, and array's start:

• If they belong to different halves, you adjust low or high to focus on the potential half containing the target.
• Otherwise, you adjust based on whether the target is greater or less than the middle element.

This method continuously narrows down the search space until low exceeds high, indicating the target is not in the array, at which point it returns false.

This approach efficiently handles the complexities introduced by the rotation and duplicates within the sorted array, ensuring a logarithmic time complexity under most circumstances, similar to standard binary search but with added checks to handle the array's rotated nature and duplicate values.

class Finder:
    def locate(self, elements, key):
        count = len(elements)
        if count == 0:
            return False
        low = 0
        high = count - 1
        while low <= high:
            mid = low + (high - low) // 2
            if elements[mid] == key:
                return True
            if not self.isHelpful(elements, low, elements[mid]):
                low += 1
                continue
            inFirstArray = self.isFirstArray(elements, low, elements[mid])
            targetInFirstArray = self.isFirstArray(elements, low, key)
            if inFirstArray ^ targetInFirstArray:
                if inFirstArray:
                    low = mid + 1
                else:
                    high = mid - 1
            else:
                if elements[mid] < key:
                    low = mid + 1
                else:
                    high = mid - 1
        return False
    
    def isHelpful(self, elements, start, elem):
        return elements[start] != elem
    
    def isFirstArray(self, elements, start, elem):
        return elements[start] <= elem

The provided code defines a class Finder which includes methods to help determine if a specific key can be found in a rotated sorted array. This problem is significant as it deals with a sorted array that may have been rotated some number of times, and might contain duplicates as well.

Here's a breakdown of how the solution works:

• The main method locate begins by checking whether the input list elements is empty. If so, it returns False immediately, indicating that the key is not found.

• If not empty, the method uses a while loop to perform a modified binary search:

  • It calculates the middle index mid. If the element at this index matches the key, the method returns True, indicating the key has been found.
  • The method then uses the helper method isHelpful to check if the current mid element provides any useful information. If not, it increments the low pointer to skip this unhelpful element.

• The helper method isFirstArray checks if the mid element and either bounds (low or key) belong to the first sorted sub-array.

• Based on whether the mid element and the key belong to the same part of the rotated array, the search space is adjusted accordingly:

  • If they belong to different parts, adjust the boundary setters (low or high) to narrow down the search region.
  • Adjust the search bounds within the part where the key might exist, based on comparison of mid element with the key.

• The function concludes by returning False if the loop completes without finding the key.

This solution efficiently handles searches in a rotated sorted array, even with the added complexity of possible duplicate values, relying primarily on strategic use of binary search principles and additional logical checks to ensure correct partitioning and searching of the array segments.