So i've tried interpreting this pseudocode a friend made and i wasn't exactly sure that my method returns the right result. Anyone who's able to help me out?
I've done some test cases where e.g. an array of [2,0,7] or [0,1,4] or [0, 8, 0] would return true, but not cases like: [1,7,7] or [2,6,0].
Array(list, d)
for j = 0 to d−1 do
for i = 0 to d−1 do
for k = 0 to d−1 do
if list[j] + list[ i] + list[k] = 0 then
return true
end if
end for
end for
end for
return false
And i've made this in java:
public class One{
public static boolean method1(ArrayList<String> A, int a){
for(int i = 0; i < a-1; i++){
for(int j = 0; j < a-1; j++){
for(int k = 0; k < a-1; k++){
if(Integer.parseInt(A.get(i)+A.get(j)+A.get(k)) == 0){
return true;
}
}
}
}
return false;
}
}
Thanks in advance
For a fix to your concrete problem, see my comment. A nicer way to write that code would be to actually use a list of Integer instead of String, because you will then want to convert the strings back to integers. So, your method looks better like this:
public static boolean method(List<Integer> A) {
for (Integer i : A)
for (Integer j : A)
for (Integer k : A)
if (i + j + k == 0)
return true;
return false;
}
See that you don't even need the size as parameter, since any List in Java embeds its own size.
Somehow offtopic
You're probably trying to solve the following problem: "Find if a list of integers contains 3 different ones that sum up to 0". The solution to this problem doesn't have to be O(n^3), like yours, it can be solved in O(n^2). See this post.
Ok, so here is what I believe the pseudo code is trying to do. It returns true if there is a zero in your list or if there are three numbers that add up to zero in your list. So it should return true for following test cases. (0,1,2,3,4,5), (1,2,3,4,-3). It will return false for (1,2,3,4,5). I just used d=5 as a random example. Your code is good for the most part - you just need to add the ith, jth and kth elements in the list to check if their sum equals zero for the true condition.
Related
can anyone pls tell me the time complexity of this func.
this is for generating all valid parenthesis , given the count of pairs
Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]
My code is working fine, but im not sure about time complexity of it.
pls help
class Solution {
public List generateParenthesis(int n) {
HashMap<String,Boolean> hm = new HashMap<>();
return generate(hm,n);
}
public static List<String> generate(HashMap<String,Boolean> hm, int n ){
if(n == 1){
hm.put("()",true);
List<String>temp = new ArrayList<>();
temp.add("()");
return temp;
}
List<String>temp = generate(hm,n-1);
List<String>ans = new ArrayList<>();
for(String pat : temp){
for(int i = 0; i < pat.length(); i++){
String newPat = pat.substring(0,i)+"()"+pat.substring(i);
if(!hm.containsKey(newPat)){
hm.put(newPat,true);
ans.add(newPat);
}
}
}
return ans;
}
}
You have two for loops, which each run over m and n elements, it can be written as O(m*n), and be contracted to O(n^2) because m can be equal to n.
Your function is recursively calling itself, making time complexity analysis a bit harder.
Think about it this way: You generate all valid parenthesis of length n. It turns out that the number of all valid parenthesis of length n (taking your definition of n) is equal to the nth catalan number. Each string has length 2*n, So the time complexity is not a polynomial, but O(n*C(n)) where C(n) is the nth catalan number.
Edit: It seems like this question is already answered here.
I'm referring to the leetcode question: Kth Smallest Element in a Sorted Matrix
There are two well-known solutions to the problem. One using Heap/PriorityQueue and other is using Binary Search. The Binary Search solution goes like this (top post):
public class Solution {
public int kthSmallest(int[][] matrix, int k) {
int lo = matrix[0][0], hi = matrix[matrix.length - 1][matrix[0].length - 1] + 1;//[lo, hi)
while(lo < hi) {
int mid = lo + (hi - lo) / 2;
int count = 0, j = matrix[0].length - 1;
for(int i = 0; i < matrix.length; i++) {
while(j >= 0 && matrix[i][j] > mid) j--;
count += (j + 1);
}
if(count < k) lo = mid + 1;
else hi = mid;
}
return lo;
}
}
While I understand how this works, I have trouble figuring out one issue.
How can we be sure that the returned lo is always in the matrix?
Since the search space is min and max value of the array, the mid need NOT be a value that is in the array. However, the returned lo always is.
Why is this happening?
For the sake of argument, we can move the calculation of count to a separate function like the following:
bool valid(int mid, int[][] matrix, int k) {
int count = 0, m = matrix.length;
for (int i = 0; i < m; i++) {
int j = 0;
while (j < m && matrix[i][j] <= mid) j++;
count += j;
}
return (count < k);
}
This predicate will do exactly same as your specified operation. Here, the loop invariant is that, the range [lo, hi] always contains the kth smallest number of the 2D array.
In other words, lo <= solution <= hi
Now, when the loop terminates, it is evident that lo >= hi
Merging those two properties, we get, lo = solution = hi, since solution is a member of array, it can be said that, lo is always in the array after loop termination and will rightly point to the kth smallest element.
Because We are finding the lower_bound using binary search and there cannot be any number smaller than the number(lo) in the array which could be the kth smallest element.
I'm trying to find the least common multiple of an array of integers, e.g. if there are 2 numbers given (7, 3) then my task is to find the LCM of the numbers 3 through 7 (3,4,5,6,7 in that case).
My solution would be to add the maximum number to a new variable (var common) until the remainders of all of the numbers in the array (common % numBetween[i]) equal 0. There are more efficient ways of doing this, for example applying the Euclidean Algorithm, but I wanted to solve this my way.
The code:
function smallestCommons(arr) {
var numBetween = [];
var max = Math.max.apply(Math, arr);
var min = Math.min.apply(Math, arr);
while (max - min !== -1) {
numBetween.push(min);
min += 1;
} //this loop creates the array of integers, 1 through 13 in this case
var common = max;
var modulus = [1]; //I start with 1, so that the first loop could begin
var modSum = modulus.reduce(function (a, b) {
return a + b;
}, 0);
while (modSum !== 0) {
modulus = [];
for (var i = 0; i < numBetween.length; i++) {
modulus.push(common % numBetween[i]);
}
if (modSum !== 0) {
common += max;
break; //without this, the loop is infinite
}
}
return common;
}
smallestCommons([1,13]);
Now, the loop is either infinite (without break in the if statement) so I guess the modSum never equals 0, because the modulus variable always contains integers other than 0. I wanted to solve this by "resetting" the modulus to an empty array right after the loop starts, with
modulus = [];
and if I include the break, the loop stops after 1 iteration (common = 26). I can't quite grasp why my code isn't working. All comments are appreciated.
Thanks in advance!
I may be false, but do you actually never change modSum within the while-loop? If so, this is your problem. You wanted to do this by using the function .reduce(), but this does not bind the given function, so you have to call the function each time again in the loop.
I have a quick question how can I loop over an NSMutable array starting from a certain index.
For Example I have these double loops I want k to start from the same index as l.
for (Line *l in L)
{
for (Line *k in L)
{
............
}
}
To elaborate further, lets say L has 10 object so l start from 0-10 and k from 0 -10. What I want is if l is equal 1 k should start from 1-10 rather than 0 - 10 and when l is equal 2 k should start from 2- 10 rather than 0. Any help is Appreciated
Objective-C is an extension of C, lookup the C for loop and you'll have your answer. HTH
Addendum
I was going to let you benefit from the learning experience of looking up the C for yourself, however at the time of writing all other answers since added give the code but it is not complete, so here is what you need to produce the l and k values in the order you wish:
for(NSInteger lIndex = 0; lIndex < L.count; lIndex++)
{
Line *l = L[lIndex]; // index into your array to get the element
for(NSInteger kIndex = lIndex; kIndex < L.count; kIndex++)
{
Line *k = L[kIndex];
// process your l and k
}
}
As you can see the for has three sub-parts which are the initialisation, condition, and increment. The initialisation is performed first, then the condition to determine whether to execute the for body, and the increment is executed after the statements in the body and before the condition is tested to determine if another iteration should be performed. A for loop is roughly (there are some differences that are unimportant here) to the while loop:
initialisation;
while(condition)
{
body statements;
increment;
}
You simply need to modify for-statement.
NSInteger indexYouNeed;
NSInteger iterationCount;
for (int i = indexYouNeed; i < iterationCount; i++) {
/* Your code here */
}
You may find this link helpfulll.
You have to use an indexed (ordinary) for loop instead of fast enumeration (for-in):
int l;
for (l=startValue; l<=endValue; l++)
{
int i;
for (int i=l; i<=endValue; i++)
{
…
}
}
I have a problem that I am unwilling to believe hasn't been solved before in Sage.
Given a pair of integers (d,n) as input, I'd like to receive a list (or set, or whatever) of all nondecreasing sequences of length d all of whose entries are no greater than n.
Similarly, I'd like another function which returns all strictly increasing sequences of length d whose entries are no greater than n.
For example, for d = 2 n=3, I'd receive the output:
[[1,2], [1,3], [2,3]]
or
[[1,1], [1,2], [1,3], [2,2], [2,3], [3,3]]
depending on whether I'm using increasing or nondecreasing.
Does anyone know of such a function?
Edit Of course, if there is such a method for nonincreasing or decreasing sequences, I can modify that to fit my purposes. Just something to iterate over sequences
I needed this algorithm too and I finally managed to write one today. I will share the code here, but I only started to learn coding last week, so it is not pretty.
Idea Input=(r,d). Step 1) Create a class "ListAndPosition" that has a list L of arrays Integer[r+1]'s, and an integer q between 0 and r. Step 2) Create a method that receives a ListAndPosition (L,q) and screens sequentially the arrays in L checking if the integer at position q is less than the one at position q+1, if so, it adds a new array at the bottom of the list with that entry ++. When done, the Method calls itself again with the new list and q-1 as input.
The code for Step 1)
import java.util.ArrayList;
public class ListAndPosition {
public static Integer r=5;
public final ArrayList<Integer[]> L;
public int q;
public ListAndPosition(ArrayList<Integer[]> L, int q) {
this.L = L;
this.q = q;
}
public ArrayList<Integer[]> getList(){
return L;
}
public int getPosition() {
return q;
}
public void decreasePosition() {
q--;
}
public void showList() {
for(int i=0;i<L.size();i++){
for(int j=0; j<r+1 ; j++){
System.out.print(""+L.get(i)[j]);
}
System.out.println("");
}
}
}
The code for Step 2)
import java.util.ArrayList;
public class NonDecreasingSeqs {
public static Integer r=5;
public static Integer d=3;
public static void main(String[] args) {
//Creating the first array
Integer[] firstArray;
firstArray = new Integer[r+1];
for(int i=0;i<r;i++){
firstArray[i] = 0;
}
firstArray[r] = d;
//Creating the starting listAndDim
ArrayList<Integer[]> L = new ArrayList<Integer[]>();
L.add(firstArray);
ListAndPosition Lq = new ListAndPosition(L,r-1);
System.out.println(""+nonDecSeqs(Lq).size());
}
public static ArrayList<Integer[]> nonDecSeqs(ListAndPosition Lq){
int iterations = r-1-Lq.getPosition();
System.out.println("How many arrays in the list after "+iterations+" iterations? "+Lq.getList().size());
System.out.print("Should we stop the iteration?");
if(0<Lq.getPosition()){
System.out.println(" No, position = "+Lq.getPosition());
for(int i=0;i<Lq.getList().size();i++){
//Showing particular array
System.out.println("Array of L #"+i+":");
for(int j=0;j<r+1;j++){
System.out.print(""+Lq.getList().get(i)[j]);
}
System.out.print("\nCan it be modified at position "+Lq.getPosition()+"?");
if(Lq.getList().get(i)[Lq.getPosition()]<Lq.getList().get(i)[Lq.getPosition()+1]){
System.out.println(" Yes, "+Lq.getList().get(i)[Lq.getPosition()]+"<"+Lq.getList().get(i)[Lq.getPosition()+1]);
{
Integer[] tempArray = new Integer[r+1];
for(int j=0;j<r+1;j++){
if(j==Lq.getPosition()){
tempArray[j] = new Integer(Lq.getList().get(i)[j])+1;
}
else{
tempArray[j] = new Integer(Lq.getList().get(i)[j]);
}
}
Lq.getList().add(tempArray);
}
System.out.println("New list");Lq.showList();
}
else{
System.out.println(" No, "+Lq.getList().get(i)[Lq.getPosition()]+"="+Lq.getList().get(i)[Lq.getPosition()+1]);
}
}
System.out.print("Old position = "+Lq.getPosition());
Lq.decreasePosition();
System.out.println(", new position = "+Lq.getPosition());
nonDecSeqs(Lq);
}
else{
System.out.println(" Yes, position = "+Lq.getPosition());
}
return Lq.getList();
}
}
Remark: I needed my sequences to start at 0 and end at d.
This is probably not a very good answer to your question. But you could, in principle, use Partitions and the max_slope=-1 argument. Messing around with filtering lists of IntegerVectors sounds equally inefficient and depressing for other reasons.
If this has a canonical name, it might be in the list of sage-combinat functionality, and there is even a base class you could perhaps use for integer lists, which is basically what you are asking about. Maybe you could actually get what you want using IntegerListsLex? Hope this proves helpful.
This question can be solved by using the class "UnorderedTuples" described here:
http://doc.sagemath.org/html/en/reference/combinat/sage/combinat/tuple.html
To return all all nondecreasing sequences with entries between 0 and n-1 of length d, you may type:
UnorderedTuples(range(n),d)
This returns the nondecreasing sequence as a list. I needed an immutable object (because the sequences would become keys of a dictionary). So I used the "tuple" method to turn the lists into tuples:
immutables = []
for s in UnorderedTuples(range(n),d):
immutables.append(tuple(s))
return immutables
And I also wrote a method which picks out only the increasing sequences:
def isIncreasing(list):
for i in range(len(list) - 1):
if list[i] >= list[i+1]:
return false
return true
The method that returns only strictly increasing sequences would look like
immutables = []
for s in UnorderedTuples(range(n),d):
if isIncreasing(s):
immutables.append(tuple(s))
return immutables