My code implements a basic binary search in a 1D array. Writing mid inside of while scope is giving the output. What I don't understand is when I declared start, end also outside of while loop which gives output without issue.
package com.company;
public class Main {
public static void main(String[] args) {
int arr[] = {1,23,34,45,56,67,78,88};
int target = 78;
int result = BinarySearch(arr, target);
System.out.println("Element found at index : " + result);
}
static int BinarySearch(int arr[], int target){
int start = 0;
int end = arr.length - 1;
// **writing mid here not giving any error, not even output**
int mid = start + (end - start) /2;
while(start <= end){
int mid = start + (end - start) /2; // **If mid written here, it gives the output**
if(target < arr[mid]){
end = mid - 1;
} else if(target > arr[mid]){
start = mid + 1;
}else{
we got the value (if its present.)
return mid;
}
}
// **if while didnt run (that means value not present in the array).**
return -1;
}
}
When it's outside the loop it is only evaluated for the first run of the loop, then never again. It is not updated, so target > arr[mid] is just always true and the loop always chooses that path, running in circles forever.
But really, you should get a debugger up and running and try stepping your code, before asking someone for help. It's the best teaching tool, since it shows, not tells.
This question already has answers here:
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");
I am supposed to create two methods: The first method, bSearch, is a binary search algorithm. The second method, insertInOrder, is supposed to take the index we got from the bSearch method and find that index in the array, open up a spot at that index by shifting all other elements of that array over, and insert the key/ int at that index.
The ints will be received from a text file containing 25 ints. I am not to make any copies or extra arrays to do this, and I am supposed to encode and then decode the index in the insertInOrder method. In these methods, key will be the current int read from the file of ints, and count will count the number of ints which have been received. I am basically building my own sort method, I can't call any outside methods, and the array is not to be out of order at any time.
I have filled in these methods, but I am a shaky on my understanding. I think my problem is that in the bSearch method, because I can't get it to return anything but 0. I can't get it to return the new value of mid, which is the index where the key/ int is supposed to be inserted. Than you for your help. The code is below:
static void insertInOrder( int[] arr, int count, int key )
{
int index=bSearch( arr, count, key );
int encodedIndex = -(index+1);
int decodedIndex = -(index+1);
for (int i = index; i > 0; i--)
{
arr[i] = arr[i-1];
}
arr[index]=key;
}
static int bSearch(int[] a, int count, int key)
{
int lo = 0;
int hi = a.length-1;
int index = 0;
boolean found = false;
while (!found && lo <= hi)
{
int mid = lo + ((hi - lo) / 2);
if (a[mid] == key)
{
found = true;
index = mid;
}
else if (a[mid] > key)
{
hi = mid -1;
}
else
{
lo = mid +1;
}
}
return index;
}
Your binary search method works correctly for a pre-filled array.
The problem is during insertion into the array, the binary search returns 0 if the value doesn't exist in the array instead of providing the correct insert position.
In this case, you probably need track how many values are current used in the array. Is that what the value "count" is for? In that case, you should start your "hi" value at count instead of the max length and you need to return the count as the index if you've reached the end of the array without finding a value.
Update: You can use this binary search to return insertion position.
int lo = 0;
int hi = count-1;
int index = 0;
boolean found = false;
while (!found && lo < hi)
{
int mid = lo + ((hi - lo) / 2);
if (a[mid] == key)
{
found = true;
index = mid;
}
else if (a[mid] > key)
{
hi = mid;
index = hi;
}
else
{
lo = mid +1;
index = lo;
}
}
return index;
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 create a binary search function and I keep getting stuck in a loop. I am confined to using such magnets. The program already gives me elements to search for. The code below creates an infinite loop.
public class Student < T extends Comparable <? super T >> {
public int binarySearchIter(T[] data, T key) {
int first = 0;
int last = data.length - 1;
int mid, result;
mid = (first + last) / 2;
result = key.compareTo(data[mid]);
while (first <= last) {
if (result == -1) {
return -1;
} else if (result > 0) {
first = mid + 1;
} else {
last = mid - 1;
}
}
return mid;
}
The contract of the compareTo method is to return a positive integer, 0 or a negative integer depending on whether the number is greater, equal or less than the given number. Those numbers might or might not be -1, so you must not rely on that.
Therefore, you should change your while loop like this:
If the result is negative, the key is before the middle element so we update last to before the middle index.
If the result is positive, the key is after the middle element so we update first to after the middle index.
If the result is zero, we just found the correct index.
Also, you are not updating the middle element once you have updated the first and last variable so you should move that code inside the while loop.
public static <T extends Comparable <T>> int binarySearchIter(T[] data, T key) {
int first = 0;
int last = data.length - 1;
int mid, result;
while (first <= last) {
mid = (first + last) / 2;
result = key.compareTo(data[mid]);
if (result < 0) {
last = mid - 1;
} else if (result > 0) {
first = mid + 1;
} else {
return mid;
}
}
return -1;
}
If this is supposed to be a binary search, while loop should start just above mid = (first + last) / 2;
Plus the note from Tunaki.