I have a working recursive solution to a DP problem. I wish to memoize it.
Currently it depends on two states: the index i and a boolean variable true or false.
Could someone please point out how I could memoize it? Specifically, how I could initialize the memoization table (dp)?
I am confused because if I initialize the second state with false, I wouldn't be able to differentiate between the false that is due to initialization, versus the one where it is actually the value of the state.
Could someone please provide some advise?
Thanks.
To clarify further: This is how I declare the dp table right now:
vector<vector < bool > > dp;
How do I initialize the inner vector<bool>? I don't think I can set it to either true or false since I wouldn't be able to distinguish later if that is the value generated while executing (solving the problem) or the initialization value.
Edit: Adding the code:
class Solution {
public:
unordered_map<int, int> m1, m2;
vector<int> n1, n2;
vector<vector<int>> v;
int helper(int i, bool parsingNums1) {
if((parsingNums1 && i>=n1.size()) || (!parsingNums1 && i>=n2.size())) return v[i][parsingNums1]=0;
if(v[i][parsingNums1]!=-1) return v[i][parsingNums1];
int ans=0;
if(parsingNums1) {
//we are traversing path 1
//see if we can switch to path 2
if(m2.find(n1[i])!=m2.end())
ans=n1[i] + helper(m2[n1[i]]+1, false);
ans=max(ans, n1[i] + helper(i+1, true));
}
if(!parsingNums1) {
//we are traversing path 2
//see if we can switch to path 1
if(m1.find(n2[i])!=m1.end())
ans=n2[i] + helper(m1[n2[i]]+1, true);
ans=max(ans, n2[i] + helper(i+1, false));
}
return v[i][parsingNums1]=ans;
}
int maxSum(vector<int>& nums1, vector<int>& nums2) {
for(int i=0; i<nums1.size(); i++)
m1[nums1[i]]=i;
for(int i=0; i<nums2.size(); i++)
m2[nums2[i]]=i;
n1=nums1;
n2=nums2;
v.resize((nums1.size()>nums2.size()?nums1.size()+1:nums2.size()+1), vector<int>(2,-1));
return max(helper(0, true), helper(0, false))%(int)(1e9+7);
}
};
I am solving this LeetCode question: https://leetcode.com/problems/get-the-maximum-score/
There are 2 easy methods for handling this.
Declare another vector<vector < bool > > is_stored which is initialised as 0 and when dp[i][j] is calculated, mark is_stored[i][j] as 1. So when you are checking wether the particular state is being memorized, you can look into the is_stored.
Use vector< vector < int > > instead of vector< vector < bool > > and initialise -1 to every state to mark as not memorised.
Another way to store values is using
Map<String, Boolean> map = new HashMap<String, Boolean>(); // just a java version
then you can create key by appending i & j and store respective boolean value to that key. for example
String key = i + ',' + j;
// To validate if we calculated data before
if(map.containsKeys(key)) return map.get(key);
// To store/memoize values
boolean result = someDPmethod(); map.put(key, result);
In C# you can use nullable value types.
A nullable value type T? represents all values of its underlying value type T and an additional null value. For example, you can assign any of the following three values to a bool? variable: true, false, or null. An underlying value type T cannot be a nullable value type itself.
You can use null for indication of unvisited or unprocessed dp states.
you can simulate this in C++ by Initializing your dp memo to
vector<vector<bool*>> dp( m, vector<bool*>(n, nullptr) );
now you can use nullptr as an indicator for unprocessed dp states.
What does ArrayIndexOutOfBoundsException mean and how do I get rid of it?
Here is a code sample that triggers the exception:
String[] names = { "tom", "bob", "harry" };
for (int i = 0; i <= names.length; i++) {
System.out.println(names[i]);
}
Your first port of call should be the documentation which explains it reasonably clearly:
Thrown to indicate that an array has been accessed with an illegal index. The index is either negative or greater than or equal to the size of the array.
So for example:
int[] array = new int[5];
int boom = array[10]; // Throws the exception
As for how to avoid it... um, don't do that. Be careful with your array indexes.
One problem people sometimes run into is thinking that arrays are 1-indexed, e.g.
int[] array = new int[5];
// ... populate the array here ...
for (int index = 1; index <= array.length; index++)
{
System.out.println(array[index]);
}
That will miss out the first element (index 0) and throw an exception when index is 5. The valid indexes here are 0-4 inclusive. The correct, idiomatic for statement here would be:
for (int index = 0; index < array.length; index++)
(That's assuming you need the index, of course. If you can use the enhanced for loop instead, do so.)
if (index < 0 || index >= array.length) {
// Don't use this index. This is out of bounds (borders, limits, whatever).
} else {
// Yes, you can safely use this index. The index is present in the array.
Object element = array[index];
}
See also:
The Java Tutorials - Language Basics - Arrays
Update: as per your code snippet,
for (int i = 0; i<=name.length; i++) {
The index is inclusive the array's length. This is out of bounds. You need to replace <= by <.
for (int i = 0; i < name.length; i++) {
From this excellent article: ArrayIndexOutOfBoundsException in for loop
To put it briefly:
In the last iteration of
for (int i = 0; i <= name.length; i++) {
i will equal name.length which is an illegal index, since array indices are zero-based.
Your code should read
for (int i = 0; i < name.length; i++)
^
It means that you are trying to access an index of an array which is not valid as it is not in between the bounds.
For example this would initialize a primitive integer array with the upper bound 4.
int intArray[] = new int[5];
Programmers count from zero. So this for example would throw an ArrayIndexOutOfBoundsException as the upper bound is 4 and not 5.
intArray[5];
What causes ArrayIndexOutOfBoundsException?
If you think of a variable as a "box" where you can place a value, then an array is a series of boxes placed next to each other, where the number of boxes is a finite and explicit integer.
Creating an array like this:
final int[] myArray = new int[5]
creates a row of 5 boxes, each holding an int. Each of the boxes has an index, a position in the series of boxes. This index starts at 0 and ends at N-1, where N is the size of the array (the number of boxes).
To retrieve one of the values from this series of boxes, you can refer to it through its index, like this:
myArray[3]
Which will give you the value of the 4th box in the series (since the first box has an index of 0).
An ArrayIndexOutOfBoundsException is caused by trying to retrieve a "box" that does not exist, by passing an index that is higher than the index of the last "box", or negative.
With my running example, these code snippets would produce such an exception:
myArray[5] //tries to retrieve the 6th "box" when there is only 5
myArray[-1] //just makes no sense
myArray[1337] //way to high
How to avoid ArrayIndexOutOfBoundsException
In order to prevent ArrayIndexOutOfBoundsException, there are some key points to consider:
Looping
When looping through an array, always make sure that the index you are retrieving is strictly smaller than the length of the array (the number of boxes). For instance:
for (int i = 0; i < myArray.length; i++) {
Notice the <, never mix a = in there..
You might want to be tempted to do something like this:
for (int i = 1; i <= myArray.length; i++) {
final int someint = myArray[i - 1]
Just don't. Stick to the one above (if you need to use the index) and it will save you a lot of pain.
Where possible, use foreach:
for (int value : myArray) {
This way you won't have to think about indexes at all.
When looping, whatever you do, NEVER change the value of the loop iterator (here: i). The only place this should change value is to keep the loop going. Changing it otherwise is just risking an exception, and is in most cases not necessary.
Retrieval/update
When retrieving an arbitrary element of the array, always check that it is a valid index against the length of the array:
public Integer getArrayElement(final int index) {
if (index < 0 || index >= myArray.length) {
return null; //although I would much prefer an actual exception being thrown when this happens.
}
return myArray[index];
}
To avoid an array index out-of-bounds exception, one should use the enhanced-for statement where and when they can.
The primary motivation (and use case) is when you are iterating and you do not require any complicated iteration steps. You would not be able to use an enhanced-for to move backwards in an array or only iterate on every other element.
You're guaranteed not to run out of elements to iterate over when doing this, and your [corrected] example is easily converted over.
The code below:
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i< name.length; i++) {
System.out.print(name[i] + "\n");
}
...is equivalent to this:
String[] name = {"tom", "dick", "harry"};
for(String firstName : name) {
System.out.println(firstName + "\n");
}
In your code you have accessed the elements from index 0 to the length of the string array. name.length gives the number of string objects in your array of string objects i.e. 3, but you can access only up to index 2 name[2],
because the array can be accessed from index 0 to name.length - 1 where you get name.length number of objects.
Even while using a for loop you have started with index zero and you should end with name.length - 1. In an array a[n] you can access form a[0] to a[n-1].
For example:
String[] a={"str1", "str2", "str3" ..., "strn"};
for(int i=0; i<a.length(); i++)
System.out.println(a[i]);
In your case:
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i<=name.length; i++) {
System.out.print(name[i] +'\n');
}
For your given array the length of the array is 3(i.e. name.length = 3). But as it stores element starting from index 0, it has max index 2.
So, instead of 'i**<=name.length' you should write 'i<**name.length' to avoid 'ArrayIndexOutOfBoundsException'.
So much for this simple question, but I just wanted to highlight a new feature in Java which will avoid all confusions around indexing in arrays even for beginners. Java-8 has abstracted the task of iterating for you.
int[] array = new int[5];
//If you need just the items
Arrays.stream(array).forEach(item -> { println(item); });
//If you need the index as well
IntStream.range(0, array.length).forEach(index -> { println(array[index]); })
What's the benefit? Well, one thing is the readability like English. Second, you need not worry about the ArrayIndexOutOfBoundsException
The most common case I've seen for seemingly mysterious ArrayIndexOutOfBoundsExceptions, i.e. apparently not caused by your own array handling code, is the concurrent use of SimpleDateFormat. Particularly in a servlet or controller:
public class MyController {
SimpleDateFormat dateFormat = new SimpleDateFormat("MM/dd/yyyy");
public void handleRequest(ServletRequest req, ServletResponse res) {
Date date = dateFormat.parse(req.getParameter("date"));
}
}
If two threads enter the SimplateDateFormat.parse() method together you will likely see an ArrayIndexOutOfBoundsException. Note the synchronization section of the class javadoc for SimpleDateFormat.
Make sure there is no place in your code that are accessing thread unsafe classes like SimpleDateFormat in a concurrent manner like in a servlet or controller. Check all instance variables of your servlets and controllers for likely suspects.
You are getting ArrayIndexOutOfBoundsException due to i<=name.length part. name.length return the length of the string name, which is 3. Hence when you try to access name[3], it's illegal and throws an exception.
Resolved code:
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i < name.length; i++) { //use < insteadof <=
System.out.print(name[i] +'\n');
}
It's defined in the Java language specification:
The public final field length, which contains the number of components
of the array. length may be positive or zero.
That's how this type of exception looks when thrown in Eclipse. The number in red signifies the index you tried to access. So the code would look like this:
myArray[5]
The error is thrown when you try to access an index which doesn't exist in that array. If an array has a length of 3,
int[] intArray = new int[3];
then the only valid indexes are:
intArray[0]
intArray[1]
intArray[2]
If an array has a length of 1,
int[] intArray = new int[1];
then the only valid index is:
intArray[0]
Any integer equal to the length of the array, or bigger than it: is out of bounds.
Any integer less than 0: is out of bounds;
P.S.: If you look to have a better understanding of arrays and do some practical exercises, there's a video here: tutorial on arrays in Java
For multidimensional arrays, it can be tricky to make sure you access the length property of the right dimension. Take the following code for example:
int [][][] a = new int [2][3][4];
for(int i = 0; i < a.length; i++){
for(int j = 0; j < a[i].length; j++){
for(int k = 0; k < a[j].length; k++){
System.out.print(a[i][j][k]);
}
System.out.println();
}
System.out.println();
}
Each dimension has a different length, so the subtle bug is that the middle and inner loops use the length property of the same dimension (because a[i].length is the same as a[j].length).
Instead, the inner loop should use a[i][j].length (or a[0][0].length, for simplicity).
For any array of length n, elements of the array will have an index from 0 to n-1.
If your program is trying to access any element (or memory) having array index greater than n-1, then Java will throw ArrayIndexOutOfBoundsException
So here are two solutions that we can use in a program
Maintaining count:
for(int count = 0; count < array.length; count++) {
System.out.println(array[count]);
}
Or some other looping statement like
int count = 0;
while(count < array.length) {
System.out.println(array[count]);
count++;
}
A better way go with a for each loop, in this method a programmer has no need to bother about the number of elements in the array.
for(String str : array) {
System.out.println(str);
}
ArrayIndexOutOfBoundsException whenever this exception is coming it mean you are trying to use an index of array which is out of its bounds or in lay man terms you are requesting more than than you have initialised.
To prevent this always make sure that you are not requesting a index which is not present in array i.e. if array length is 10 then your index must range between 0 to 9
ArrayIndexOutOfBounds means you are trying to index a position within an array that is not allocated.
In this case:
String[] name = { "tom", "dick", "harry" };
for (int i = 0; i <= name.length; i++) {
System.out.println(name[i]);
}
name.length is 3 since the array has been defined with 3 String objects.
When accessing the contents of an array, position starts from 0. Since there are 3 items, it would mean name[0]="tom", name[1]="dick" and name[2]="harry
When you loop, since i can be less than or equal to name.length, you are trying to access name[3] which is not available.
To get around this...
In your for loop, you can do i < name.length. This would prevent looping to name[3] and would instead stop at name[2]
for(int i = 0; i<name.length; i++)
Use a for each loop
String[] name = { "tom", "dick", "harry" };
for(String n : name) {
System.out.println(n);
}
Use list.forEach(Consumer action) (requires Java8)
String[] name = { "tom", "dick", "harry" };
Arrays.asList(name).forEach(System.out::println);
Convert array to stream - this is a good option if you want to perform additional 'operations' to your array e.g. filter, transform the text, convert to a map etc (requires Java8)
String[] name = { "tom", "dick", "harry" };
--- Arrays.asList(name).stream().forEach(System.out::println);
--- Stream.of(name).forEach(System.out::println);
ArrayIndexOutOfBoundsException means that you are trying to access an index of the array that does not exist or out of the bound of this array. Array indexes start from 0 and end at length - 1.
In your case
for(int i = 0; i<=name.length; i++) {
System.out.print(name[i] +'\n'); // i goes from 0 to length, Not correct
}
ArrayIndexOutOfBoundsException happens when you are trying to access
the name.length indexed element which does not exist (array index ends at length -1). just replacing <= with < would solve this problem.
for(int i = 0; i < name.length; i++) {
System.out.print(name[i] +'\n'); // i goes from 0 to length - 1, Correct
}
According to your Code :
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i<=name.length; i++) {
System.out.print(name[i] +'\n');
}
If You check
System.out.print(name.length);
you will get 3;
that mean your name length is 3
your loop is running from 0 to 3
which should be running either "0 to 2" or "1 to 3"
Answer
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i<name.length; i++) {
System.out.print(name[i] +'\n');
}
Each item in an array is called an element, and each element is accessed by its numerical index. As shown in the preceding illustration, numbering begins with 0. The 9th element, for example, would therefore be accessed at index 8.
IndexOutOfBoundsException is thrown to indicate that an index of some sort (such as to an array, to a string, or to a vector) is out of range.
Any array X, can be accessed from [0 to (X.length - 1)]
I see all the answers here explaining how to work with arrays and how to avoid the index out of bounds exceptions. I personally avoid arrays at all costs. I use the Collections classes, which avoids all the silliness of having to deal with array indices entirely. The looping constructs work beautifully with collections supporting code that is both easier to write, understand and maintain.
If you use an array's length to control iteration of a for loop, always remember that the index of the first item in an array is 0. So the index of the last element in an array is one less than the array's length.
ArrayIndexOutOfBoundsException name itself explains that If you trying to access the value at the index which is out of the scope of Array size then such kind of exception occur.
In your case, You can just remove equal sign from your for loop.
for(int i = 0; i<name.length; i++)
The better option is to iterate an array:
for(String i : name )
System.out.println(i);
This error is occurs at runs loop overlimit times.Let's consider simple example like this,
class demo{
public static void main(String a[]){
int[] numberArray={4,8,2,3,89,5};
int i;
for(i=0;i<numberArray.length;i++){
System.out.print(numberArray[i+1]+" ");
}
}
At first, I have initialized an array as 'numberArray'. then , some array elements are printed using for loop. When loop is running 'i' time , print the (numberArray[i+1] element..(when i value is 1, numberArray[i+1] element is printed.)..Suppose that, when i=(numberArray.length-2), last element of array is printed..When 'i' value goes to (numberArray.length-1) , no value for printing..In that point , 'ArrayIndexOutOfBoundsException' is occur.I hope to you could get idea.thank you !
You can use Optional in functional style to avoid NullPointerException and ArrayIndexOutOfBoundsException :
String[] array = new String[]{"aaa", null, "ccc"};
for (int i = 0; i < 4; i++) {
String result = Optional.ofNullable(array.length > i ? array[i] : null)
.map(x -> x.toUpperCase()) //some operation here
.orElse("NO_DATA");
System.out.println(result);
}
Output:
AAA
NO_DATA
CCC
NO_DATA
In most of the programming language indexes is start from 0.So you must have to write i<names.length or i<=names.length-1 instead of i<=names.length.
You could not iterate or store more data than the length of your array. In this case you could do like this:
for (int i = 0; i <= name.length - 1; i++) {
// ....
}
Or this:
for (int i = 0; i < name.length; i++) {
// ...
}
Can someone give me an algorithm to count distinct elements of an array of integers in one pass.
for example i can try to traverse through the array using a for loop
I will store the first element in another array.And the subsequent elements will be compared with those in the second array and if it is distinct then i will store it in that array and increment counter.
can someone give me a better algorithm than this.
Using c and c++
Supposing that your elements are integers and their values are between 0 and MAXVAL-1.
#include <stdio.h>
#include <string.h>
#define MAXVAL 50
unsigned int CountDistinctsElements(unsigned int* iArray, unsigned int iNbElem) {
unsigned int ret = 0;
//this array will contains the count of each value
//for example, c[3] will contain the count of the value 3 in your original array
unsigned int c[MAXVAL];
memset(c, 0, MAXVAL*sizeof(unsigned int));
for (unsigned int i=0; i<iNbElem; i++) {
unsigned int elem = iArray[i];
if (elem < MAXVAL && c[elem] == 0) {
ret++;
}
c[elem]++;
}
return ret;
}
int main() {
unsigned int myElements[10] = {0, 25, 42, 42, 1, 2, 42, 0, 24, 24};
printf("Distincts elements : %d\n", CountDistinctsElements(myElements, 10));
return 0;
}
Output : (Ideone link)
Distincts elements : 6
Maintain a array of structures.
structure should have a value and a counter of that value.
As soon as you pass an new element in an array being tested create a structure with value and increment the counter by 1.if you pass an existing element in the array then simply access the related structure and increment its counter by 1.
Finally after you do a one complete pass of the array, you will have the required result in the array of structures.
Edit: I wasn't aware you wanted just to count the elements. Updated code below.
int countUnique()
{
uniqueArray[numElements];
myArray[numElements];
int counter = 0;
int uniqueElements = 0;
for(int i = 0; i < numElements; i++)
{
element tempElem = myArray[i];
if(!doesUniqueContain(tempElem, counter, uniqueArray)//If it doesn't contain it
{
uniqueArray[counter] = tempElem;
uniqueElements++;
}
}
return uniqueElements;
}
bool doesUniqueContain(element oneElement, int counter, array *uniqueArray)
{
if(counter == 0)
{
return false; //No elements, so it doesn't contain this element.
}
for(int i = 0; i < counter; i++)
{
if(uniqueArray[i] == oneElement)
return true;
}
return false;
}
This is only so you can see the logic
How about using a hash table (in the Java HashMap or C# Dictionary sense) to count the elements? Basically you create an empty hash table with the array element type as the key type and the count as values. Then you iterate over your list. If the element is not yet in the hash table, you add it with count 1, otherwise you increment the count for that element.
This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
Finding a single number in a list
Given an array of numbers, except for one number all the others, occur
twice. What should be the algorithm to find that number which occurs only once in the
array?
Example
a[1..n] = [1,2,3,4,3,1,2]
should return 4
Let the number which occurs only once in the array be x
x <- a[1]
for i <- 2 to n
x <- x ^ a[i]
return x
Since a ^ a = 0 and a ^ 0 = a
Numbers which occur in pair cancel out and the result gets stored in x
Working code in C++
#include <iostream>
template<typename T, size_t N>
size_t size(T(&a)[N])
{
return N;
}
int main()
{
int a [] = {1,2,3,4,3,1,2};
int x = a[0];
for (size_t i = 1; i< size(a) ; ++i)
{
x = x ^ a[i];
}
std::cout << x;
}
Create new int i = 0
XOR each item with i
After all iterations there will be expected number in i
If you have quantities which cannot be reasonably xored (Big Integers or numbers represented as Strings, for example), an alternate approach which is also O(n) time, (but O(n) space rather than O(1) space) would be to simply use a hash table. The algorithm looks like:
Create a hash table of the same size as the list
For every item in the list:
If item is a key in hash table
then remove item from hash table
else add item to hash table with nominal value
At the end, there should be exactly one item in the hash table
I would do, C or C++ code, but neither of them have hash tables built in. (Don't ask me why C++ doesn't have a hash table in the STL, but does have a hash map based on a red-black tree, because I have no idea what they were thinking.) And, unfortunately, I don't have a C# compiler handy to test for syntax errors, so I'm giving you Java code. It's pretty similar, though.
import java.util.Hashtable;
import java.util.List;
class FindUnique {
public static <T> T findUnique(List<T> list) {
Hashtable<T,Character> ht = new Hashtable<T,Character>(list.size());
for (T item : list) {
if (ht.containsKey(item)) {
ht.remove(item);
} else {
ht.put(item,'x');
}
}
return ht.keys().nextElement();
}
}
Well i only know of the Brute force algo and it is to traverse whole array and check
Code will be like (in C#):
k=0;
for(int i=0 ; i < array.Length ; i++)
{
k ^= array[i];
}
return k;
zerkms' answer in C++
int a[] = { 1,2,3,4,3,1,2 };
int i = std::accumulate(a, a + 7, 0, std::bit_xor<int>());
You could sort the array and then find the first element that doesn't have a pair. That would require several loops for sorting and a loop for finding the single element.
But a simplier method would be setting the double keys to zero or a value that is not possible in the current format. Depends on the programming language, as well, as you cannot change key types in c++ unlike c#.