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Finding multiple entries with binary search
(15 answers)
Closed 3 years ago.
I've been tasked with creating a method that will print all the indices where value x is found in a sorted array.
I understand that if we just scanned through the array from 0 to N (length of array) it would have a running time of O(n) worst case. Since the array that will be passed into the method will be sorted, I'm assuming that I can take advantage of using a Binary Search since this will be O(log n). However, this only works if the array has unique values. Since the Binary Search will finish after the first "find" of a particular value. I was thinking of doing a Binary Search for finding x in the sorted array, and then checking all values before and after this index, but then if the array contained all x values, it doesn't seem like it would be that much better.
I guess what I'm asking is, is there a better way to find all the indices for a particular value in a sorted array that is better than O(n)?
public void PrintIndicesForValue42(int[] sortedArrayOfInts)
{
// search through the sortedArrayOfInts
// print all indices where we find the number 42.
}
Ex: sortedArray = { 1, 13, 42, 42, 42, 77, 78 } would print: "42 was found at Indices: 2, 3, 4"
You will get the result in O(lg n)
public static void PrintIndicesForValue(int[] numbers, int target) {
if (numbers == null)
return;
int low = 0, high = numbers.length - 1;
// get the start index of target number
int startIndex = -1;
while (low <= high) {
int mid = (high - low) / 2 + low;
if (numbers[mid] > target) {
high = mid - 1;
} else if (numbers[mid] == target) {
startIndex = mid;
high = mid - 1;
} else
low = mid + 1;
}
// get the end index of target number
int endIndex = -1;
low = 0;
high = numbers.length - 1;
while (low <= high) {
int mid = (high - low) / 2 + low;
if (numbers[mid] > target) {
high = mid - 1;
} else if (numbers[mid] == target) {
endIndex = mid;
low = mid + 1;
} else
low = mid + 1;
}
if (startIndex != -1 && endIndex != -1){
for(int i=0; i+startIndex<=endIndex;i++){
if(i>0)
System.out.print(',');
System.out.print(i+startIndex);
}
}
}
Well, if you actually do have a sorted array, you can do a binary search until you find one of the indexes you're looking for, and from there, the rest should be easy to find since they're all next to each-other.
once you've found your first one, than you go find all the instances before it, and then all the instances after it.
Using that method you should get roughly O(lg(n)+k) where k is the number of occurrences of the value that you're searching for.
EDIT:
And, No, you will never be able to access all k values in anything less than O(k) time.
Second edit: so that I can feel as though I'm actually contributing something useful:
Instead of just searching for the first and last occurrences of X than you can do a binary search for the first occurence and a binary search for the last occurrence. which will result in O(lg(n)) total. once you've done that, you'll know that all the between indexes also contain X(assuming that it's sorted)
You can do this by searching checking if the value is equal to x , AND checking if the value to the left(or right depending on whether you're looking for the first occurrence or the last occurrence) is equal to x.
public void PrintIndicesForValue42(int[] sortedArrayOfInts) {
int index_occurrence_of_42 = left = right = binarySearch(sortedArrayOfInts, 42);
while (left - 1 >= 0) {
if (sortedArrayOfInts[left-1] == 42)
left--;
}
while (right + 1 < sortedArrayOfInts.length) {
if (sortedArrayOfInts[right+1] == 42)
right++;
}
System.out.println("Indices are from: " + left + " to " + right);
}
This would run in O(log(n) + #occurrences)
Read and understand the code. It's simple enough.
Below is the java code which returns the range for which the search-key is spread in the given sorted array:
public static int doBinarySearchRec(int[] array, int start, int end, int n) {
if (start > end) {
return -1;
}
int mid = start + (end - start) / 2;
if (n == array[mid]) {
return mid;
} else if (n < array[mid]) {
return doBinarySearchRec(array, start, mid - 1, n);
} else {
return doBinarySearchRec(array, mid + 1, end, n);
}
}
/**
* Given a sorted array with duplicates and a number, find the range in the
* form of (startIndex, endIndex) of that number. For example,
*
* find_range({0 2 3 3 3 10 10}, 3) should return (2,4). find_range({0 2 3 3
* 3 10 10}, 6) should return (-1,-1). The array and the number of
* duplicates can be large.
*
*/
public static int[] binarySearchArrayWithDup(int[] array, int n) {
if (null == array) {
return null;
}
int firstMatch = doBinarySearchRec(array, 0, array.length - 1, n);
int[] resultArray = { -1, -1 };
if (firstMatch == -1) {
return resultArray;
}
int leftMost = firstMatch;
int rightMost = firstMatch;
for (int result = doBinarySearchRec(array, 0, leftMost - 1, n); result != -1;) {
leftMost = result;
result = doBinarySearchRec(array, 0, leftMost - 1, n);
}
for (int result = doBinarySearchRec(array, rightMost + 1, array.length - 1, n); result != -1;) {
rightMost = result;
result = doBinarySearchRec(array, rightMost + 1, array.length - 1, n);
}
resultArray[0] = leftMost;
resultArray[1] = rightMost;
return resultArray;
}
Another result for log(n) binary search for leftmost target and rightmost target. This is in C++, but I think it is quite readable.
The idea is that we always end up when left = right + 1. So, to find leftmost target, if we can move right to rightmost number which is less than target, left will be at the leftmost target.
For leftmost target:
int binary_search(vector<int>& nums, int target){
int n = nums.size();
int left = 0, right = n - 1;
// carry right to the greatest number which is less than target.
while(left <= right){
int mid = (left + right) / 2;
if(nums[mid] < target)
left = mid + 1;
else
right = mid - 1;
}
// when we are here, right is at the index of greatest number
// which is less than target and since left is at the next,
// it is at the first target's index
return left;
}
For the rightmost target, the idea is very similar:
int binary_search(vector<int>& nums, int target){
while(left <= right){
int mid = (left + right) / 2;
// carry left to the smallest number which is greater than target.
if(nums[mid] <= target)
left = mid + 1;
else
right = mid - 1;
}
// when we are here, left is at the index of smallest number
// which is greater than target and since right is at the next,
// it is at the first target's index
return right;
}
I came up with the solution using binary search, only thing is to do the binary search on both the sides if the match is found.
public static void main(String[] args) {
int a[] ={1,2,2,5,5,6,8,9,10};
System.out.println(2+" IS AVAILABLE AT = "+findDuplicateOfN(a, 0, a.length-1, 2));
System.out.println(5+" IS AVAILABLE AT = "+findDuplicateOfN(a, 0, a.length-1, 5));
int a1[] ={2,2,2,2,2,2,2,2,2};
System.out.println(2+" IS AVAILABLE AT = "+findDuplicateOfN(a1, 0, a1.length-1, 2));
int a2[] ={1,2,3,4,5,6,7,8,9};
System.out.println(10+" IS AVAILABLE AT = "+findDuplicateOfN(a2, 0, a2.length-1, 10));
}
public static String findDuplicateOfN(int[] a, int l, int h, int x){
if(l>h){
return "";
}
int m = (h-l)/2+l;
if(a[m] == x){
String matchedIndexs = ""+m;
matchedIndexs = matchedIndexs+findDuplicateOfN(a, l, m-1, x);
matchedIndexs = matchedIndexs+findDuplicateOfN(a, m+1, h, x);
return matchedIndexs;
}else if(a[m]>x){
return findDuplicateOfN(a, l, m-1, x);
}else{
return findDuplicateOfN(a, m+1, h, x);
}
}
2 IS AVAILABLE AT = 12
5 IS AVAILABLE AT = 43
2 IS AVAILABLE AT = 410236578
10 IS AVAILABLE AT =
I think this is still providing the results in O(logn) complexity.
A Hashmap might work, if you're not required to use a binary search.
Create a HashMap where the Key is the value itself, and then value is an array of indices where that value is in the array. Loop through your array, updating each array in the HashMap for each value.
Lookup time for the indices for each value will be ~ O(1), and creating the map itself will be ~ O(n).
Find_Key(int arr[], int size, int key){
int begin = 0;
int end = size - 1;
int mid = end / 2;
int res = INT_MIN;
while (begin != mid)
{
if (arr[mid] < key)
begin = mid;
else
{
end = mid;
if(arr[mid] == key)
res = mid;
}
mid = (end + begin )/2;
}
return res;
}
Assuming the array of ints is in ascending sorted order; Returns the index of the first index of key occurrence or INT_MIN. Runs in O(lg n).
It is using Modified Binary Search. It will be O(LogN). Space complexity will be O(1).
We are calling BinarySearchModified two times. One for finding start index of element and another for finding end index of element.
private static int BinarySearchModified(int[] input, double toSearch)
{
int start = 0;
int end = input.Length - 1;
while (start <= end)
{
int mid = start + (end - start)/2;
if (toSearch < input[mid]) end = mid - 1;
else start = mid + 1;
}
return start;
}
public static Result GetRange(int[] input, int toSearch)
{
if (input == null) return new Result(-1, -1);
int low = BinarySearchModified(input, toSearch - 0.5);
if ((low >= input.Length) || (input[low] != toSearch)) return new Result(-1, -1);
int high = BinarySearchModified(input, toSearch + 0.5);
return new Result(low, high - 1);
}
public struct Result
{
public int LowIndex;
public int HighIndex;
public Result(int low, int high)
{
LowIndex = low;
HighIndex = high;
}
}
public void printCopies(int[] array)
{
HashMap<Integer, Integer> memberMap = new HashMap<Integer, Integer>();
for(int i = 0; i < array.size; i++)
if(!memberMap.contains(array[i]))
memberMap.put(array[i], 1);
else
{
int temp = memberMap.get(array[i]); //get the number of occurances
memberMap.put(array[i], ++temp); //increment his occurance
}
//check keys which occured more than once
//dump them in a ArrayList
//return this ArrayList
}
Alternatevely, instead of counting the number of occurances, you can put their indices in a arraylist and put that in the map instead of the count.
HashMap<Integer, ArrayList<Integer>>
//the integer is the value, the arraylist a list of their indices
public void printCopies(int[] array)
{
HashMap<Integer, ArrayList<Integer>> memberMap = new HashMap<Integer, ArrayList<Integer>>();
for(int i = 0; i < array.size; i++)
if(!memberMap.contains(array[i]))
{
ArrayList temp = new ArrayList();
temp.add(i);
memberMap.put(array[i], temp);
}
else
{
ArrayList temp = memberMap.get(array[i]); //get the lsit of indices
temp.add(i);
memberMap.put(array[i], temp); //update the index list
}
//check keys which return lists with length > 1
//handle the result any way you want
}
heh, i guess this will have to be posted.
int predefinedDuplicate = //value here;
int index = Arrays.binarySearch(array, predefinedDuplicate);
int leftIndex, rightIndex;
//search left
for(leftIndex = index; array[leftIndex] == array[index]; leftIndex--); //let it run thru it
//leftIndex is now the first different element to the left of this duplicate number string
for(rightIndex = index; array[rightIndex] == array[index]; rightIndex++); //let it run thru it
//right index contains the first different element to the right of the string
//you can arraycopy this [leftIndex+1, rightIndex-1] string or just print it
for(int i = leftIndex+1; i<rightIndex; i++)
System.out.println(array[i] + "\t");
Related
I'm trying to figure out the running time complexity of the findingDup algorithm because I'm unsure if it's O(n) or O(log n). My goal is to implement a sublinear algorithm that finds how many times an int value is duplicated. You can assume the given array int[] A is always sorted. If you have any additional questions please leave them below.
public class Controller {
public static void main(String[] args){
int[] A = {-1, 2, 3, 5, 6, 6, 6, 9, 10};
int value = 6;
System.out.println(findingDup(A, value));
}// end main
public static int findingDup(int[] a, int x){
int counter = 0;
int index = binarySearch(a, x); // index = 4
int leftIndex = index - 1; // leftIndex = 3
int rightIndex = index + 1; // rightIndex = 5
if(index == -1){
return 0;
}
else if(a[index] == x){
counter++;
}
// checking if all numbers are dups
if(a[0] == a[a.length - 1]){
return a.length;
}
if(leftIndex >= 0){
while(a[leftIndex] == x){
counter++;
leftIndex--;
if(leftIndex < 0){
break;
}
}
}
if(rightIndex <= a.length - 1){
while(a[rightIndex] == x){
counter++;
rightIndex++;
if(rightIndex > a.length - 1){
break;
}
}
}
return counter;
}// end method
public static int binarySearch(int[] a, int x){
int low = 0, high = a.length - 1;
while(low <= high){
int mid = (low + high) / 2;
if(a[mid] < x){
low = mid + 1;
}
else if(a[mid] > x){
high = mid - 1;
}
else{
return mid;
}
}
return -1;
}// End Method
}// end class
Your code is O(k + log n), where "k" is number of times the value is present on the list.
If the k = O(n) it degrades to O(n).
As an example, in the extreme case of the list being [6, 6, 6, 6, 6, ...] you will end-up processing all the elements.
You can still fix this problem by running more than one binary search.
First you run it to find first occurrence of "value", and then you run it again to find a first number larger than value (search for value+1).
Your binary search algorithm needs to be modified to return the first occurrence of the value, or larger value if the value cannot be found.
As of now it finds any occurrence, not guaranteed to be the first one nor the last one.
Your binary search has the following condition:
if (smaller) {...}
else if (larger) {...}
else {we have found it!}
So it can return any occurrence.
You should be looking for an index that:
a[mid - 1] < value && a[mid] >= value
mid-1 can be smaller than 0, so you need to check for that first.
If this is not the case, we haven't found the first occurrence, and need to move either left or right index.
We have an assignment to search for the minimum element of a sorted array that is shifted to the right afterwards. For example: [1, 5, 6, 19, 56, 101] becomes [19, 56, 101, 1, 5, 6]. The method should be implemented using a divide and conquer algorithm and it should have a better asymptotic time complexity than O(n).
EDIT: I forgot to add that the elements int the array are unique.
I already implemented a method and wanted to ask if this is better than O(n) and if there are ways to improve my method.
public class FindMinimum {
public void findMinimum(int[] arr) {
// the recursive method ends when the length of the array is smaller than 2
if (arr.length < 2) {
return;
}
int mid = arr.length / 2;
/*
* if the array length is greater or the same as two, check if the middle
* element is smaller as the element before that. And print the middle element
* if it's true.
*/
if (arr.length >= 2) {
if (arr[mid - 1] > arr[mid]) {
System.out.println("Minimum: " + arr[mid]);
return;
}
}
/*
* separate the array in two sub-arrays through the middle and start the method
* with those two arrays again.
*/
int[] leftArr = new int[mid];
int[] rightArr = new int[arr.length - mid];
for (int i = 0; i < mid; i++) {
leftArr[i] = arr[i];
}
for (int i = mid; i < arr.length; i++) {
rightArr[i - mid] = arr[i];
}
findMinimum(leftArr);
findMinimum(rightArr);
}
}
In Java you could use a List because than you can create a Sublist.
private Integer findMinimum(List<Integer> list) {
if (list.size() < 2)
return list.get(0);
int mid = list.size() / 2;
// create left and right list
List<Integer> leftList = list.subList(0, mid);
List<Integer> rightList = list.subList(mid, list.size());
if (leftList.get(leftList.size() - 1) <= rightList.get(rightList.size() - 1))
return findMin(leftList);
else
return findMin(rightList);
}
When you create a Sublist with Java there is no copy. So to create a new Sublist takes a complexity of O(1).
So the function has a complexity of O(logn).
This is my new solution, without copying the array anywhere.
public class FindMinimum {
public void findMinimum(int[] arr) {
findMinimumSub(arr, 0, arr.length - 1, 2);
}
private void findMinimumSub(int[] arr, int start, int end, int size) {
// the recursive method ends when the length of the array is smaller than 2
if ((end - start) < 2) {
if (arr[end] > arr[start])
System.out.println("Minimum: " + arr[start]);
else
System.out.println("Minimum: " + arr[end]);
return;
}
int mid = arr.length / size;
if (arr[start] > arr[end]) {
// right side
start += mid;
findMinimumSub(arr, start, end, size * 2);
}
else {
// left side
findMinimumSub(arr, start, mid, size * 2);
}
}
}
I was wondering if it is possible to find the closest lower element in a non-empty sorted array for an element that may be there or may not be there. Elements can be repeated also any number of times. All elements of the array +ve.
For example, if we had the values [2,5,6,7,7,8,9] and we are looking for the element closest lower to 6, it should return 5, because 5 is the biggest number in the array, that is smaller than 6.
Similarly, if we're looking for the element closest lower to 9, it should return 8, because 8 is the biggest number in the array, that is smaller than 9.
And if the closest lower element is not found, return -1 like if we're looking for the element closest lower to 1, it should return -1, because there is no such element which can be lower than 1. Here -1 represents that there's no such value is present in the array which is closest lower to the element
I have tried this below code. Is it all right? If I'm missing something, please help me. Java code will be more helpful.
static int find(int[] a, int target)
{
int n = a.length;
if(target <= a[0])
return -1;
if(target > a[n-1])
return a[n-1];
int i=0,j=n,mid=0;
while(i<j)
{
mid = (i+j)/2;
if(target <= a[mid])
{
if( mid >0 & target> a[mid-1] )
{
return a[mid-1];
}
j= mid;
}
else
{
if( mid<(n-1) & target > a[mid+1] )
{
return a[mid+1];
}
i= mid+1;
}
}
return mid;
}
Using streams:
import java.util.stream.IntStream;
public class FindNearestLowestValue {
public final static void main(String[] args) {
int[] array = {2,5,6,7,7,8,9};
int searchVal = 6;
// reverse the order so the first element of the filtered stream is the result
System.out.println(
IntStream.range(0, array.length)
.map(i -> array[array.length - 1 - i])
.filter(n -> n < searchVal)
.findFirst().orElse(-1)
);
}
}
There exists a standard binarySearch function.
static int find(int[] a, int target) {
int position = Arrays.binarySearch(a, target);
if (position >= 0) {
System.out.println("Found at index " + position);
} else {
int insertIndex = ~position;
System.out.println("Insert position at index " + insertIndex);
position = insertIndex;
}
return position;
}
When not found it delives the ones-complement of the insert position, as shown above. This means when the result is negative, the item is not found.
It does more or less what you did, but on not finding, it cleverly return a negative: ~ insert position (or -insert position - 1).
/**
* Search the next smaller array element.
* #param a the array sorted in ascending order.
* #param target the value to keep below.
* #return the greatest smaller element, or -1.
*/
static int findNextSmaller(int[] a, int target) {
int i= Arrays.binarySearch(a, target);
if (i >= 0) {
--i;
while (i>= 0 && a[i] == target) {
--i;
}
} else {
i = ~i;
--i;
}
return i == -1 ? -1 : a[i];
}
Or, as int is discrete:
static int findNextSmaller(int[] a, int target) {
int i= Arrays.binarySearch(a, target - 1);
if (i >= 0) {
return target - 1;
}
i = ~i;
--i;
return i == -1 ? -1 : a[i];
}
I am trying to write code to determine the n smallest item in an array. It's sad that I am struggling with this. Based on the algorithm from my college textbook from back in the day, this looks to be correct. However, obviously I am doing something wrong as it gives me a stack overflow exception.
My approach is:
Set the pivot to be at start + (end-start) / 2 (rather than start+end/2 to prevent overflow)
Use the integer at this location to be the pivot that I compare everything to
Iterate and swap everything around this pivot so things are sorted (sorted relative to the pivot)
If n == pivot, then I think I am done
Otherwise, if I want the 4 smallest element and pivot is 3, for example, then I need to look on the right side (or left side if I wanted the 2nd smallest element).
-
public static void main(String[] args) {
int[] elements = {30, 50, 20, 10};
quickSelect(elements, 3);
}
private static int quickSelect(int[] elements2, int k) {
return quickSelect(elements2, k, 0, elements2.length - 1);
}
private static int quickSelect(int[] elements, int k, int start, int end) {
int pivot = start + (end - start) / 2;
int midpoint = elements[pivot];
int i = start, j = end;
while (i < j) {
while (elements[i] < midpoint) {
i++;
}
while (elements[j] > midpoint) {
j--;
}
if (i <= j) {
int temp = elements[i];
elements[i] = elements[j];
elements[j] = temp;
i++;
j--;
}
}
// Guessing something's wrong here
if (k == pivot) {
System.out.println(elements[pivot]);
return pivot;
} else if (k < pivot) {
return quickSelect(elements, k, start, pivot - 1);
} else {
return quickSelect(elements, k, pivot + 1, end);
}
}
Edit: Please at least bother commenting why if you're going to downvote a valid question.
This won't fix the issue, but there are several problems with your code :
If you do not check for i < end and j > start in your whiles, you may run into out of bounds in some cases
You choose your pivot to be in the middle of the subarray, but nothing proves that it won't change position during partitioning. Then, you check for k == pivot with the old pivot position, which obviously won't work
Hope this helps a bit.
Alright so the first thing I did was rework how I get my pivot/partition point. The shortcoming, as T. Claverie pointed out, is that the pivot I am using isn't technically the pivot since the element's position changes during the partitioning phase.
I actually rewrote the partitioning code into its own method as below. This is slightly different.
I choose the first element (at start) as the pivot, and I create a "section" in front of this with items less than this pivot. Then, I swap the pivot's value with the last item in the section of values < the pivot. I return that final index as the point of the pivot.
This can be cleaned up more (create separate swap method).
private static int getPivot(int[] elements, int start, int end) {
int pivot = start;
int lessThan = start;
for (int i = start; i <= end; i++) {
int currentElement = elements[i];
if (currentElement < elements[pivot]) {
lessThan++;
int tmp = elements[lessThan];
elements[lessThan] = elements[i];
elements[i] = tmp;
}
}
int tmp = elements[lessThan];
elements[lessThan] = elements[pivot];
elements[pivot] = tmp;
return lessThan;
}
Here's the routine that's calls this:
private static int quickSelect(int[] elements, int k, int start, int end) {
int pivot = getPivot(elements, start, end);
if (k == (pivot - start + 1)) {
System.out.println(elements[pivot]);
return pivot;
} else if (k < (pivot - start + 1)) {
return quickSelect(elements, k, start, pivot - 1);
} else {
return quickSelect(elements, k - (pivot - start + 1), pivot + 1, end);
}
}
I am trying to search a sorted array for a number that i choose. if the array contains this number i want to return the indices of each occurrence. I need to write to methods to do this task. one Recursively and one iteratively. I have started the recursive method below.
public int findRecursive(T anEntry) {
return binarySearchRecursive(0, length - 1, anEntry);
}
private int binarySearchRecursive(int first, int last, T desiredItem)
{
int position = 0;
int mid = (first + last) / 2;
if (first > last || (desiredItem.compareTo(list[position]) != 0))
position = -(position + 1);
else if (desiredItem.equals(list[mid]))
position = mid;
else if (desiredItem.compareTo(list[mid]) < 0)
position = binarySearchRecursive(first, mid -1, desiredItem);
else
position = binarySearchRecursive(mid + 1, last, desiredItem);
return position;
}
Here is the array/output that should be expected
public static void main(String[] args) {
AList<Integer> testList = new AList<Integer>();
testList.add(1);
testList.add(3);
testList.add(2);
testList.add(5);
testList.add(7);
testList.add(2);
testList.add(4);
//* question 5
//* should output 1, -8, 4
// my out is returning -1, -1, -1
testList.sort();
System.out.println(testList.findRecursive(2));
System.out.println(testList.findRecursive(8));
System.out.println(testList.findRecursive(4));
System.out.println("");
There were a couple of small changes required to your recursive method in order to get the desired output -
private int binarySearchRecursive(int first, int last, T desiredItem)
{
int position = 0;
int mid = (first + last) / 2;
if (first > last)
position = -(last + 1);
else if (desiredItem.equals(list[mid]))
position = mid;
else if (desiredItem.compareTo(list[mid]) < 0)
position = binarySearchRecursive(first, mid -1, desiredItem);
else
position = binarySearchRecursive(mid + 1, last, desiredItem);
return position;
}
Both changes were in the termination area. When first > last you have exhausted the search, you don't need the comparison for desired item. Also in the termination you should return -(last+1), not -(position+1) as position will always be 0.