878168. Search in an almost Sorted Array
Here is a detailed Markdown blog post for the problem "Search in an almost Sorted Array":
Search in an Almost Sorted Array
Summary
This problem involves searching for an element in an array that is almost sorted. The array may have some elements out of order, but overall it remains mostly sorted. We need to find a way to efficiently search the array and return the index of the target element.
The key concepts involved are binary search and sorting.
Detailed Explanation
To solve this problem, we can start by recognizing that the array is almost sorted, meaning most elements are in order. This allows us to use binary search as our foundation. However, when an element is out of order, we need to handle it differently.
Here's a step-by-step breakdown of the solution:
- Initialization: Start by initializing two pointers,
lowandhigh, to represent the range of elements in the array. - Binary Search: Perform a binary search on the array using the
lowandhighpointers as our bounds. For each iteration, calculate the mid-point and check if it's equal to the target element. - Handling Out-of-Order Elements: If the mid-point is not equal to the target element, we need to handle out-of-order elements. Check if the element before or after the mid-point is in order. If it's in order, update our
lowandhighpointers accordingly. - Repeat Binary Search: Continue performing binary searches with updated
lowandhighpointers until we find the target element. - Return Index: Once we've found the target element, return its index.
Time complexity: O(log n) (where n is the size of the array) Space complexity: O(1)
Optimized Solutions
Java
public int searchAlmostSortedArray(int[] arr, int target) {
int low = 0;
int high = arr.length - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (arr[mid] == target) {
return mid;
}
if (mid > 0 && arr[mid - 1] < target && target < arr[mid]) {
// element before mid is out of order, move high pointer
high = mid - 1;
} else if (mid < arr.length - 1 && arr[mid + 1] > target && target > arr[mid]) {
// element after mid is out of order, move low pointer
low = mid + 1;
} else {
// binary search on the remaining range
low = mid + 1;
high = high - (high - low) / 2;
}
}
return -1; // not found
}