Problems with Internal path length function - c++

I've been asked to compute the average depth of a node in both a binary search tree, and an AVL tree. Through some research, I found that the average depth of a tree is the internal path length divided by the number of nodes in a tree, and that the internal path length (the sum of the path lengths of every node in the tree) is given by this recurrence:
D(1) = 0, D(N) = D(i) + D(N − i − 1) + N − 1
where D(N) is a tree with N nodes, D(i), is the IPL of the left subtree, and D(N-i-1) is the IPL of the right subtree.
Using that, I wrote this function:
int internalPathLength(Node *t, int& sum) const{
if(t == nullptr || (t->left == nullptr && t->right == nullptr)) {
return 0;
}
else {
int a = 0;
sum += internalPathLength(t->left, sum) + internalPathLength(t->right, sum) + (countNodes(t,a)-1);
cout << sum << endl;
return sum;
}
This function gives me, with a binary search tree of 565 nodes, an IPL of 1,264,875,230 and an average depth of 2,238,717, a preposterously high number. Using it on an AVL tree of similar size gives me an IPL of -1,054,188,525 and an average depth of -1,865,820, which is a negative number on top of being preposterously high. Is there something wrong my interpretation/implementation of the recurrence? what else can I try? Or are the values I'm getting in the normal range for this computation after all?

The problem is that you pass sum by reference, so it gets incremented way too many times. You don't really need this sum at all. This should work:
int internalPathLength(Node *t) const{
if(t == nullptr || (t->left == nullptr && t->right == nullptr)) {
return 0;
}
else {
return internalPathLength(t->left) + internalPathLength(t->right) + countNodes(t) - 1;
}
}
This is not optimal, because your count function is probably also recursive.
You can count the nodes in each subtree in the same recursion and then use it. Like this:
int internalPathLength(Node *t, int &count) const{
if(t == nullptr) {
count = 0;
return 0;
}
else if(t->left == nullptr && t->right == nullptr){
count = 1;
return 0;
}
else {
count = 1;
int leftCount;
int rightCount;
int sum = internalPathLength(t->left, leftCount) + internalPathLength(t->right, rightCount);
count += leftCount + rightCount;
return sum + count - 1;
}
}

Related

counting number of elements less than X in a BST

I had implemented a BST for a multiset using the C++ code below, whereas each node contains the number of occurrence num of each distinct number data, and I try to find the number of elements less than certain value x, using the order function below.
It works, however, inefficient in terms of execution time.
Is there any method with better time complexity?
struct Node {
int data;
int height;
int num;
Node *left;
Node *right;
};
int order(Node *n, int x) {
int sum = 0;
if (n != NULL) {
if (n->data < x) {
sum += n->num;
sum += order(n->right, x);
}
sum += order(n->left, x);
}
return sum;
}
You can bring the algorithm down to O(logN) time by storing in each node the number of elements in the subtree of which it is the root. Then you'd only have to recurse on one of the two children of each node (go left if x < node->data, right if x > node->data), which if the tree is balanced only takes logarithmic time.
struct Node {
int data;
int height;
int num;
int size; // numer of elements in the subtree starting at this node
Node *left;
Node *right;
};
int order(Node *n, int x) {
if(n == NULL) return 0;
// elements less than n->data make up the whole left subtree
if (x == n->data) {
return n->left ? n->left->size : 0;
}
// even smaller? just recurse left
else if (x < n->data) {
return order(n->left, x);
}
// bigger? take the whole left subtree and part of the right one
else {
return (n->left ? n->left->size : 0) + order(n->right, x);
}
}
Of course, now you have to keep track of the size, but this can be done very efficiently when updating the tree: simply recalculate the size (n->left->size + n->right->size + 1) of each modified node in an insertion or deletion.
If you can add the size to your structure, I highly recommend using Dario Petrillo’s answer.
If you have to stick to your structure, you can reduce the number of instructions and recursions.
int count_all(Node* n) {
int acc = n->num;
if (n->left != NULL) acc += count_all(n->left);
if (n->right != NULL) acc += count_all(n->right);
return acc;
}
int order(Node *n, int x) {
if (n == NULL) return 0;
// Find the first left node which is < x
while (n->data >= x) {
n = n->left;
if (n == NULL) return 0;
}
assert(n != NULL && n->data < x);
int sum = n->num;
// Grab everything left because all of them are < x
if (n->left != NULL) sum += count_all(n->left);
// Some of the right nodes may be < x, some may not
// Repeat the algorithm to find out
if (n->right != NULL) sum += order(n->right, x);
return sum;
}
This reduces the number of recursions when the root is bigger than x and you want to quickly find the next left node that satisfies n->data < x. It also removes a ton of unnecessary comparisons to x for the left side of a tree where you already know that everything is < x.

Improving the performance of this search?

Is there way to do the following search using a faster way? The items on A array are sorted in DESC order.
int find_pos(int A[], int value, int items_no, bool only_exact_match)
{
for(int i = 0; i < items_no; i++)
if(value == A[i] || (value > A[i] && !only_exact_match))
return i;
return -1;
}
You can use std::lower_bound algorithm in your case. It performs binary search with O(log N), as other people wrote. It will be something like this:
int find_pos(int A[], int value, int items_no, bool only_exact_match)
{
const int *pos_ptr = std::lower_bound(A, A + items_no, value, std::greater<int>());
const ptrdiff_t pos = pos_ptr - A;
if (pos >= items_no)
return -1;
if (*pos_ptr != value && only_exact_match)
return -1;
return pos;
}
A binary search
int left = 0;
int right = items_no; // Exclusive
while (left < right) {
int mid = (left + right) / 2;
if (value == A[mid])
return mid;
if (value < A[mid]) {
left = mid + 1;
} else {
right = mid;
}
}
return only_exact_match ? -1 : right - 1; // The greater
Because your array is sorted, you can search in steps, akin to a bisection. First, check the midpoint against your value. If it's equal, you have your answer. If it's greater, your value is in the lower half of the array. If not, your value is on the upper half. Repeat this process by bisecting the remaining elements of the array until you find your value, or run out of elements. As for your second if clause, if no matching value is found, the closest smaller element is element i+1, if that exists (i.e. you are not at the end of the array).

How do I do the following recursive function?

Ok, so I have a regular Node list, with members info and next.
I need to use a function, recursively, to calculate the average, and then compare if each node is bigger than the average or not.
int Acount(NodeType* Node, int sum, int& avg){
if (Node == NULL){//last call
avg = sum / avg;
return 0;
}
else {
return (Acount(Node->next, sum + Node->info, ++avg) + (Node->info > avg ? 1 : 0));
}
}
Which is quite simple. Problem is the value returned is always 0.
The problem appears to be with
(Node->info > avg ? 1 : 0));
I've done the tests and when I do the following:
return (Acount(Node->next, sum + Node->info, ++avg) + Node->info;
or
return (Acount(Node->next, sum + Node->info, ++avg) + avg;
Results meet expectations. As in, I'm getting the sum of the Node->info in the first case, and I'm getting average*number of nodes in the second case.
Point of this, I've proved that the function is working perfectly.
Yet when it comes to
(Node->info > avg ? 1 : 0));
Appears to be problematic, which is quite peculiar. if I place for example:
(Node->info == 5 ? 1 : 0));
And there is only one 5 in the nodes, then the function returns 1. So everything is working as intended, yet I keep getting a 0.
The following are the main functions and additional functions for the Node.
#include <iostream>
using std::cout;
using std::cin;
using std::endl;
struct NodeType{
int info;
NodeType *next;
};
//pre: first node passed is not NULL
int Acount(NodeType* Node, int sum, int& avg){
if (Node == NULL){//last call
avg = sum / avg;
return 0;
}
else {
return (Acount(Node->next, sum + Node->info, ++avg) + (Node->info > avg ? 1 : 0));
}
}
void fill(NodeType*& Node){
NodeType *temp;
Node = new NodeType;
Node->info = 0;
Node->next = NULL;
temp = Node;
for (int i = 1; i < 10; i++){
temp->next = new NodeType;
temp = temp->next;
temp->info = i;
temp->next = NULL;
}
}
void print(NodeType* Node){
NodeType *temp = Node;
while (temp != NULL){
cout << temp->info << " ";
temp = temp->next;
}
cout << endl;
}
void Delete(NodeType* Node){
NodeType *temp;
while (Node != NULL){
temp = Node;
Node = Node->next;
delete temp;
}
}
void main(){
int sum = 0, avg = 0;
NodeType *Node;
fill(Node);
print(Node);
cout << Acount(Node, sum, avg) << endl;
Delete(Node);
}
In C++ there is no concept of left-to-right (or right-to-left) evaluation order of expressions. Operator priorities will control associativity, but in the case of f1() + f2() there is no guarantee that f1() is invoked before f2() (and viceversa). It may depend on the compiler or other.
My suggestion is to split the expression into 2 distinct statements as follows:
int tmp = Acount(Node->next, sum + Node->info, ++avg);
return tmp + (Node->info > avg ? 1 : 0);
I am not sure if your code has defined behaviour. But, this line
return (Acount(Node->next, sum + Node->info, ++avg) + (Node->info > avg ? 1 : 0));
depends on if the left summand or the right summand is calculated first.
If it is the left one, then Acount goes down the recursion an incrementing avg until avg equals the number of elements in the list (here 10 when starting from zero called by the main routine). Note, that avg is passed by reference. Thus, when the recursion goes back up, this term in the right summand
Node->info > avg
will never be true because Node->info is set in the fill routine to values smaller then the number of elements.
I don't think your method will work.
In this statement:
return (Acount(Node->next, sum + Node->info, ++avg) + (Node->info > avg ? 1 : 0))
You don't know when the second term has be evaluated. It's not defined in C++.

How to heapify the minheap using an array in C++?

This program should work correctly but it doesn't! assume you are building a minheap by inserting nmubers into an array. Each time of insertion should be followed by Heapify function to make sure that the sort of numbers do not violate the minheap rule. This is what I wrote but there is something wrong with it and I couldn't make it!
int P(int i) //returning the index of parent
{
if (i % 2 == 0) { i = ((i - 2) / 2); }
else { i = ((i - 1) / 2); }
return i;
}
void Heapify(double A[], int i)//putting the smallest value in the root because we have a min heap
{
if (P(i) != NULL && A[i] < A[P(i)])
{
temp = A[P(i)];
A[P(i)] = A[i];
A[i] = temp;
Heapify(A, P(i));
}
}
Generally speaking, your heapify function doesn't seem to take a minimum of both left and right branches into consideration. Let me show you an ideal, working implementation (object-oriented, so you might want to pass the heap as a parameter). You can find the exact pseudocode all over the internet, so I'm not really presenting anything unique.
void Heap::Heapify (int i)
{
int l = left(i);
int r = right(i);
int lowest;
if (l < heap_size && heap[l] -> value < heap[i] -> value )
lowest = l;
else
lowest = i;
if (r < heap_size && heap[r] -> value < heap[lowest] -> value)
lowest = r;
if (lowest != i)
{
swap (heap[i], heap[lowest]);
Heapify(lowest);
}
}
where
int left ( int i ) { return 2 * i; }
int right ( int i ) { return 2 * i + 1; }
As you can see, an algorithm first checks which one of left and right children have lower value. That value is swapped with current value. That is everything there is to it.

Searching in a sorted and rotated array

While preparing for an interview I stumbled upon this interesting question:
You've been given an array that is sorted and then rotated.
For example:
Let arr = [1,2,3,4,5], which is sorted
Rotate it twice to the right to give [4,5,1,2,3].
Now how best can one search in this sorted + rotated array?
One can unrotate the array and then do a binary search. But that is no better than doing a linear search in the input array, as both are worst-case O(N).
Please provide some pointers. I've googled a lot on special algorithms for this but couldn't find any.
I understand C and C++.
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.
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
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.
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.
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;
}
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.
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)
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)
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;
}
}
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;
}
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
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)
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! :)
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
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;
}
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;
}
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;
}
}
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.");
}
}
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))
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)).
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;
}
}
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
}