在排序和旋转的数组中进行搜索

My first attempt would be to find using binary search the number of rotations applied - this can be done by finding the index n where a[n] > a[n + 1] using the usual binary search mechanism. Then do a regular binary search while rotating all indexes per shift found.

You can do 2 binary searches: first to find the index i such that arr[i] > arr[i+1].

Apparently, (arr\[1], arr[2], ..., arr[i]) and (arr[i+1], arr[i+2], ..., arr[n]) are both sorted arrays.

Then if arr[1] <= x <= arr[i], you do binary search at the first array, else at the second.

The complexity O(logN)

EDIT: the code.

If you know that the array has been rotated s to the right, you can simply do a binary search shifted s to the right. This is O(lg N)

By this, I mean, initialize the left limit to s and the right to (s-1) mod N, and do a binary search between these, taking a bit of care to work in the correct area.

If you don't know how much the array has been rotated by, you can determine how big the rotation is using a binary search, which is O(lg N), then do a shifted binary search, O(lg N), a grand total of O(lg N) still.

If you know how (far) it was rotated you can still do a binary search.

The trick is that you get two levels of indices: you do the b.s. in a virtual 0..n-1 range and then un-rotate them when actually looking up a value.

You don't need to rotate the array first. You can use binary search on the rotated array (with some modifications).

Let N be the number you are searching for:

Read the first number (arr[start]) and the number in the middle of the array (arr[end]):

• if arr[start] > arr[end] --> the first half is not sorted but the second half is sorted:

  • if arr[end] > N --> the number is in index: (middle + N - arr[end])

  • if N repeat the search on the first part of the array (see end to be the middle of the first half of the array etc.)

(the same if the first part is sorted but the second one isn't)

This can be done in O(logN) using a slightly modified binary search.

The interesting property of a sorted + rotated array is that when you divide it into two halves, atleast one of the two halves will always be sorted.

Let input array arr = [4,5,6,7,8,9,1,2,3]
number of elements  = 9
mid index = (0+8)/2 = 4


[4,5,6,7,8,9,1,2,3]
^
left   mid  right

as seem right sub-array is not sorted while left sub-array is sorted.

If mid happens to be the point of rotation them both left and right sub-arrays will be sorted.

[6,7,8,9,1,2,3,4,5]
^

But in any case one half(sub-array) must be sorted.

We can easily know which half is sorted by comparing start and end element of each half.

Once we find which half is sorted we can see if the key is present in that half - simple comparison with the extremes.

If the key is present in that half we recursively call the function on that half
else we recursively call our search on the other half.

We are discarding one half of the array in each call which makes this algorithm O(logN).

Pseudo code:

function search( arr[], key, low, high)


mid = (low + high) / 2


// key not present
if(low > high)
return -1


// key found
if(arr[mid] == key)
return mid


// if left half is sorted.
if(arr[low] <= arr[mid])


// if key is present in left half.
if (arr[low] <= key && arr[mid] >= key)
return search(arr,key,low,mid-1)


// if key is not present in left half..search right half.
else
return search(arr,key,mid+1,high)
end-if


// if right half is sorted.
else
// if key is present in right half.
if(arr[mid] <= key && arr[high] >= key)
return search(arr,key,mid+1,high)


// if key is not present in right half..search in left half.
else
return search(arr,key,low,mid-1)
end-if
end-if


end-function

The key here is that one sub-array will always be sorted, using which we can discard one half of the array.

short mod_binary_search( int m, int *arr, short start, short end)
{


if(start <= end)
{
short mid = (start+end)/2;


if( m == arr[mid])
return mid;
else
{
//First half is sorted
if(arr[start] <= arr[mid])
{
if(m < arr[mid] && m >= arr[start])
return mod_binary_search( m, arr, start, mid-1);
return mod_binary_search( m, arr, mid+1, end);
}


//Second half is sorted
else
{
if(m > arr[mid] && m < arr[start])
return mod_binary_search( m, arr, mid+1, end);
return mod_binary_search( m, arr, start, mid-1);
}
}
}
return -1;
}

The accepted answer has a bug when there are duplicate elements in the array. For example, arr = {2,3,2,2,2} and 3 is what we are looking for. Then the program in the accepted answer will return -1 instead of 1.

This interview question is discussed in detail in the book 'Cracking the Coding Interview'. The condition of duplicate elements is specially discussed in that book. Since the op said in a comment that array elements can be anything, I am giving my solution as pseudo code in below:

function search( arr[], key, low, high)


if(low > high)
return -1
    

mid = (low + high) / 2
    

if(arr[mid] == key)
return mid


// if the left half is sorted.
if(arr[low] < arr[mid]) {


// if key is in the left half
if (arr[low] <= key && key <= arr[mid])
// search the left half
return search(arr,key,low,mid-1)
else
// search the right half
return search(arr,key,mid+1,high)
end-if


// if the right half is sorted.
else if(arr[mid] < arr[high])
// if the key is in the right half.
if(arr[mid] <= key && arr[high] >= key)
return search(arr,key,mid+1,high)
else
return search(arr,key,low,mid-1)
end-if
   

else if(arr[mid] == arr[low])
       

if(arr[mid] != arr[high])
// Then elements in left half must be identical.
// Because if not, then it's impossible to have either arr[mid] < arr[high] or arr[mid] > arr[high]
// Then we only need to search the right half.
return search(arr, mid+1, high, key)
else
// arr[low] = arr[mid] = arr[high], we have to search both halves.
result = search(arr, low, mid-1, key)
if(result == -1)
return search(arr, mid+1, high, key)
else
return result
end-if
end-function

int rotated_binary_search(int A[], int N, int key) {
int L = 0;
int R = N - 1;


while (L <= R) {
// Avoid overflow, same as M=(L+R)/2
int M = L + ((R - L) / 2);
if (A[M] == key) return M;


// the bottom half is sorted
if (A[L] <= A[M]) {
if (A[L] <= key && key < A[M])
R = M - 1;
else
L = M + 1;
}
// the upper half is sorted
else {
if (A[M] < key && key <= A[R])
L = M + 1;
else
R = M - 1;
}
}
return -1;
}

Reply for the above mentioned post "This interview question is discussed in detail in the book 'Cracking the Coding Interview'. The condition of duplicate elements is specially discussed in that book. Since the op said in comment that array elements can be anything, I am giving my solution as pseudo code in below:"

Your solution is O(n) !! (The last if condition where you check both halves of the array for a single condition makes it a sol of linear time complexity )

I am better off doing a linear search than getting stuck in a maze of bugs and segmentation faults during a coding round.

I dont think there is a better solution than O(n) for a search in a rotated sorted array (with duplicates)

First, you need to find the shift constant, k. This can be done in O(lgN) time. From the constant shift k, you can easily find the element you're looking for using a binary search with the constant k. The augmented binary search also takes O(lgN) time The total run time is O(lgN + lgN) = O(lgN)

To find the constant shift, k. You just have to look for the minimum value in the array. The index of the minimum value of the array tells you the constant shift. Consider the sorted array [1,2,3,4,5].

The possible shifts are:
[1,2,3,4,5] // k = 0
[5,1,2,3,4] // k = 1
[4,5,1,2,3] // k = 2
[3,4,5,1,2] // k = 3
[2,3,4,5,1] // k = 4
[1,2,3,4,5] // k = 5%5 = 0

To do any algorithm in O(lgN) time, the key is to always find ways to divide the problem by half. Once doing so, the rest of the implementation details is easy

Below is the code in C++ for the algorithm

// This implementation takes O(logN) time
// This function returns the amount of shift of the sorted array, which is
// equivalent to the index of the minimum element of the shifted sorted array.
#include <vector>
#include <iostream>
using namespace std;


int binarySearchFindK(vector<int>& nums, int begin, int end)
{
int mid = ((end + begin)/2);
// Base cases
if((mid > begin && nums[mid] < nums[mid-1]) || (mid == begin && nums[mid] <= nums[end]))
return mid;
// General case
if (nums[mid] > nums[end])
{
begin = mid+1;
return binarySearchFindK(nums, begin, end);
}
else
{
end = mid -1;
return binarySearchFindK(nums, begin, end);
}
}
int getPivot(vector<int>& nums)
{
if( nums.size() == 0) return -1;
int result = binarySearchFindK(nums, 0, nums.size()-1);
return result;
}


// Once you execute the above, you will know the shift k,
// you can easily search for the element you need implementing the bottom


int binarySearchSearch(vector<int>& nums, int begin, int end, int target, int pivot)
{
if (begin > end) return -1;
int mid = (begin+end)/2;
int n = nums.size();
if (n <= 0) return -1;


while(begin <= end)
{
mid = (begin+end)/2;
int midFix = (mid+pivot) % n;
if(nums[midFix] == target)
{
return midFix;
}
else if (nums[midFix] < target)
{
begin = mid+1;
}
else
{
end = mid - 1;
}
}
return -1;
}
int search(vector<int>& nums, int target) {
int pivot = getPivot(nums);
int begin = 0;
int end = nums.size() - 1;
int result = binarySearchSearch(nums, begin, end, target, pivot);
return result;
}

Hope this helps!=)
Soon Chee Loong,
University of Toronto

public class PivotedArray {


//56784321 first increasing than decreasing
public static void main(String[] args) {
// TODO Auto-generated method stub
int [] data ={5,6,7,8,4,3,2,1,0,-1,-2};


System.out.println(findNumber(data, 0, data.length-1,-2));


}


static int findNumber(int data[], int start, int end,int numberToFind){


if(data[start] == numberToFind){
return start;
}


if(data[end] == numberToFind){
return end;
}
int mid = (start+end)/2;
if(data[mid] == numberToFind){
return mid;
}
int idx = -1;
int midData = data[mid];
if(numberToFind < midData){
if(midData > data[mid+1]){
idx=findNumber(data, mid+1, end, numberToFind);
}else{
idx =  findNumber(data, start, mid-1, numberToFind);
}
}


if(numberToFind > midData){
if(midData > data[mid+1]){
idx =  findNumber(data, start, mid-1, numberToFind);


}else{
idx=findNumber(data, mid+1, end, numberToFind);
}
}
return idx;
}


}

Here is a simple (time,space)efficient non-recursive O(log n) python solution that doesn't modify the original array. Chops down the rotated array in half until I only have two indices to check and returns the correct answer if one index matches.

def findInRotatedArray(array, num):


lo,hi = 0, len(array)-1
ix = None




while True:




if hi - lo <= 1:#Im down to two indices to check by now
if (array[hi] == num):  ix = hi
elif (array[lo] == num): ix = lo
else: ix = None
break


mid = lo + (hi - lo)/2
print lo, mid, hi


#If top half is sorted and number is in between
if array[hi] >= array[mid] and num >= array[mid] and num <= array[hi]:
lo = mid


#If bottom half is sorted and number is in between
elif array[mid] >= array[lo] and num >= array[lo] and num <= array[mid]:
hi = mid




#If top half is rotated I know I need to keep cutting the array down
elif array[hi] <= array[mid]:
lo = mid


#If bottom half is rotated I know I need to keep cutting down
elif array[mid] <= array[lo]:
hi = mid


print "Index", ix

Another approach that would work with repeated values is to find the rotation and then do a regular binary search applying the rotation whenever we access the array.

test = [3, 4, 5, 1, 2]
test1 = [2, 3, 2, 2, 2]


def find_rotated(col, num):
pivot = find_pivot(col)
return bin_search(col, 0, len(col), pivot, num)


def find_pivot(col):
prev = col[-1]
for n, curr in enumerate(col):
if prev > curr:
return n
prev = curr
raise Exception("Col does not seem like rotated array")


def rotate_index(col, pivot, position):
return (pivot + position) % len(col)


def bin_search(col, low, high, pivot, num):
if low > high:
return None
mid = (low + high) / 2
rotated_mid = rotate_index(col, pivot, mid)
val = col[rotated_mid]
if (val == num):
return rotated_mid
elif (num > val):
return bin_search(col, mid + 1, high, pivot, num)
else:
return bin_search(col, low, mid - 1,  pivot, num)


print(find_rotated(test, 2))
print(find_rotated(test, 4))
print(find_rotated(test1, 3))

Try this solution

bool search(int *a, int length, int key)
{
int pivot( length / 2 ), lewy(0), prawy(length);
if (key > a[length - 1] || key < a[0]) return false;
while (lewy <= prawy){
if (key == a[pivot]) return true;
if (key > a[pivot]){
lewy = pivot;
pivot += (prawy - lewy) / 2 ? (prawy - lewy) / 2:1;}
else{
prawy = pivot;
pivot -= (prawy - lewy) / 2 ? (prawy - lewy) / 2:1;}}
return false;
}

For a rotated array with duplicates, if one needs to find the first occurrence of an element, one can use the procedure below (Java code):

public int mBinarySearch(int[] array, int low, int high, int key)
{
if (low > high)
return -1; //key not present


int mid = (low + high)/2;


if (array[mid] == key)
if (mid > 0 && array[mid-1] != key)
return mid;


if (array[low] <= array[mid]) //left half is sorted
{
if (array[low] <= key && array[mid] >= key)
return mBinarySearch(array, low, mid-1, key);
else //search right half
return mBinarySearch(array, mid+1, high, key);
}
else //right half is sorted
{
if (array[mid] <= key && array[high] >= key)
return mBinarySearch(array, mid+1, high, key);
else
return mBinarySearch(array, low, mid-1, key);
}


}

This is an improvement to codaddict's procedure above. Notice the additional if condition as below:

if (mid > 0 && array[mid-1] != key)

My simple code :-

public int search(int[] nums, int target) {
int l = 0;
int r = nums.length-1;
while(l<=r){
int mid = (l+r)>>1;
if(nums[mid]==target){
return mid;
}
if(nums[mid]> nums[r]){
if(target > nums[mid] || nums[r]>= target)l = mid+1;
else r = mid-1;
}
else{
if(target <= nums[r] && target > nums[mid]) l = mid+1;
else r = mid -1;
}
}
return -1;
}

Time Complexity O(log(N)).

This code in C++ should work for all cases, Although It works with duplicates, please let me know if there's bug in this code.

#include "bits/stdc++.h"
using namespace std;
int searchOnRotated(vector<int> &arr, int low, int high, int k) {


if(low > high)
return -1;


if(arr[low] <= arr[high]) {


int p = lower_bound(arr.begin()+low, arr.begin()+high, k) - arr.begin();
if(p == (low-high)+1)
return -1;
else
return p;
}


int mid = (low+high)/2;


if(arr[low] <= arr[mid]) {


if(k <= arr[mid] && k >= arr[low])
return searchOnRotated(arr, low, mid, k);
else
return searchOnRotated(arr, mid+1, high, k);
}
else {


if(k <= arr[high] && k >= arr[mid+1])
return searchOnRotated(arr, mid+1, high, k);
else
return searchOnRotated(arr, low, mid, k);
}
}
int main() {


int n, k; cin >> n >> k;
vector<int> arr(n);
for(int i=0; i<n; i++) cin >> arr[i];
int p = searchOnRotated(arr, 0, n-1, k);
cout<<p<<"\n";
return 0;
}

Question: Search in Rotated Sorted Array

public class SearchingInARotatedSortedARRAY {
public static void main(String[] args) {
int[] a = { 4, 5, 6, 0, 1, 2, 3 };


System.out.println(search1(a, 6));


}


private static int search1(int[] a, int target) {
int start = 0;
int last = a.length - 1;
while (start + 1 < last) {
int mid = start + (last - start) / 2;


if (a[mid] == target)
return mid;
// if(a[start] < a[mid]) => Then this part of the array is not rotated
if (a[start] < a[mid]) {
if (a[start] <= target && target <= a[mid]) {
last = mid;
} else {
start = mid;
}
}
// this part of the array is rotated
else {
if (a[mid] <= target && target <= a[last]) {
start = mid;
} else {
last = mid;
}
}
} // while
if (a[start] == target) {
return start;
}
if (a[last] == target) {
return last;
}
return -1;
}
}

In Javascript

var search = function(nums, target,low,high) {
low= (low || low === 0) ? low : 0;


high= (high || high == 0) ? high : nums.length -1;


if(low > high)
return -1;


let mid = Math.ceil((low + high) / 2);




if(nums[mid] == target)
return mid;


if(nums[low] < nums[mid]) {
// if key is in the left half
if (nums[low] <= target && target <= nums[mid])
// search the left half
return search(nums,target,low,mid-1);
else
// search the right half
return search(nums,target,mid+1,high);
} else {
// if the key is in the right half.
if(nums[mid] <= target && nums[high] >= target)
return search(nums,target,mid+1,high)
else
return search(nums,target,low,mid-1)
}
};

Input: nums = [4,5,6,7,0,1,2], target = 0 Output: 4

import java.util.*;


class Main{
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int arr[]=new int[n];
int max=Integer.MIN_VALUE;
int min=Integer.MAX_VALUE;
int min_index=0,max_index=n;


for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
if(arr[i]>max){
max=arr[i];
max_index=i;
}
if(arr[i]<min){
min=arr[i];
min_index=i;
}


}


int element=sc.nextInt();
int index;
if(element>arr[n-1]){
index=Arrays.binarySearch(arr,0,max_index+1,element);
}
else {
index=Arrays.binarySearch(arr,min_index,n,element);
}
if(index>=0){
System.out.println(index);
}
else{
System.out.println(-1);
}
}


}

Here are my two cents:

• If the array does not contain duplicates, one can find the solution in O(log(n)). As many people have shown it the case, a tweaked version of binary search can be used to find the target element.

• However, if the array contains duplicates, I think there is no way to find the target element in O(log(n)). Here is an example shows why I think O(log(n)) is not possible. Consider the two arrays below:

a = [2,.....................2...........3,6,2......2]
b = [2.........3,6,2........2......................2]

All the dots are filled with the number 2. You can see that both arrays are sorted and rotated. If one wants to consider binary search, then they have to cut the search domain by half every iteration -- this is how we get O(log(n)). Let us assume we are searching for the number 3. In the frist case, we can see it hiding in the right side of the array, and on the second case it is hiding in the second side of the array. Here is what we know about the array at this stage:

• left = 0
• right = length - 1;
• mid = left + (right - left) / 2;
• arr[mid] = 2;
• arr[left] = 2;
• arr[right] = 2;
• target = 3;

This is all the information we have. We can clearly see it is not enough to make a decision to exclude one half of the array. As a result of that, the only way is to do linear search. I am not saying we can't optimize that O(n) time, all I am saying is that we can't do O(log(n)).

There is something i don't like about binary search because of mid, mid-1 etc that's why i always use binary stride/jump search

How to use it on a rotated array? use twice(once find shift and then use a .at() to find the shifted index -> original index)

Or compare the first element, if it is less than first element, it has to be near the end

do a backwards jump search from end, stop if any pivot tyoe leement is found

if it is > start element just do a normal jump search :)

Implemented using C#

public class Solution {
public int Search(int[] nums, int target) {
if (nums.Length == 0) return -1;
int low = 0;
int high = nums.Length - 1;
while (low <= high)
{
int mid = (low + high) / 2;
if (nums[mid] == target) return mid;
if (nums[low] <= nums[mid]) // 3 4 5 6 0 1 2
{
if (target >= nums[low] && target <= nums[mid])
high = mid;
else
low = mid + 1;
}
else // 5 6 0 1 2 3 4
{
if (target >= nums[mid] && target <= nums[high])
low= mid;
else
high = mid - 1;
}
}
return -1;
}
}

Swift Solution 100% working tested

 func searchInArray(A:[Int],key:Int)->Int{
for i in 0..<A.count{
if key == A[i] {
print(i)
return i
}
}
print(-1)
return -1
}

There is a simple idea to solve this problem in O(logN) complexity with binary search. The idea is,

If the middle element is greater than the left element, then the left part is sorted. Otherwise, the right part is sorted.

Once the sorted part is determined, all you need is to check if the value falls under that sorted part or not. If not, you can divide the unsorted part and find the sorted part from that (the unsorted part) and continue binary search.

For example, consider the image below. An array can be left rotated or right rotated.
Below image shows the relation of the mid element compared with the left most one and how this relates to which part of the array is purely sorted.
If you see the image, you find that the mid element is >= the left element and in that case, the left part is purely sorted.
An array can be left rotated by number of times, like once, twice, thrice and so on. Below image shows that for each rotation, the property of if mid >= left, left part is sorted still prevails.

More explanation with images can be found in below link. (Disclaimer: I am associated with this blog) https://foolishhungry.com/search-in-rotated-sorted-array/.

Hope this will be helpful.

Happy coding! :)

Search An Element In A Sorted And Rotated Array In Java

package yourPackageNames;


public class YourClassName {


public static void main(String[] args) {
int[] arr = {3, 4, 5, 1, 2};
//  int arr[]={16,19,21,25,3,5,8,10};
int key = 1;
searchElementAnElementInRotatedAndSortedArray(arr, key);
}


public static void searchElementAnElementInRotatedAndSortedArray(int[] arr, int key) {
int mid = arr.length / 2;
int pivotIndex = 0;
int keyIndex = -1;
boolean keyIndexFound = false;
boolean pivotFound = false;


for (int rightSide = mid; rightSide < arr.length - 1; rightSide++) {
if (arr[rightSide] > arr[rightSide + 1]) {
pivotIndex = rightSide;
pivotFound = true;
System.out.println("1st For Loop - PivotFound: " + pivotFound + ". Pivot is: " + arr[pivotIndex] + ". Pivot Index is: " + pivotIndex);
break;
}
}
if (!pivotFound) {
for (int leftSide = 0; leftSide < arr.length - mid; leftSide++) {
if (arr[leftSide] > arr[leftSide + 1]) {
pivotIndex = leftSide;
pivotFound = true;
System.out.println("2nd For Loop - PivotFound: " + pivotFound + ". Pivot is: " + arr[pivotIndex] + ". Pivot Index is: " + pivotIndex);
break;
}
}
}
for (int i = 0; i <= pivotIndex; i++) {
if (arr[i] == key) {
keyIndex = i;
keyIndexFound = true;
break;
}
}
if (!keyIndexFound) {
for (int i = pivotIndex; i < arr.length; i++) {
if (arr[i] == key) {
keyIndex = i;
break;
}
}
}
System.out.println(keyIndex >= 0 ? key + " found at index: " + keyIndex : key + " was not found in the array.");
}

}