I am running code on leetcode with the question:
Given an array of integers, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
My answer is very straightforward:
class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
vector<int> result;
for(int i=0;i<numbers.size()-1;i++){
for(int j=1+i;j<numbers.size();j++){
if (numbers[i]+numbers[j]==target)
{ result.push_back(i+1);
result.push_back(j+1);
return result;}
}
}
return result;
}
};
By using double loops, its complexity should be (O**2). However, I got the error like "Running time limit exceeded". What is the problem please?
PS: I know the method using hash map can fasten it. I am just confused with the error. Thanks.
It means your program runs too long and was terminated.
Related
I am solving a problem on LeetCode:
Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence. You must write an algorithm that runs in O(n) time. So for nums = [100,4,200,1,3,2], the output is 4.
The Union Find solution to solve this is as below:
class Solution {
public:
vector<int> parent, sz;
int find(int i) {
if(parent[i]==i) return i;
return parent[i]=find(parent[i]);
}
void merge(int i, int j) {
int p1=find(i);
int p2=find(j);
if(p1==p2) return;
if(sz[p1]>sz[p2]) {
sz[p1]+=sz[p2];
parent[p2]=p1;
} else {
sz[p2]+=sz[p1];
parent[p1]=p2;
}
}
int longestConsecutive(vector<int>& nums) {
sz.resize(nums.size(),1);
parent.resize(nums.size(),0);
iota(begin(parent),end(parent),0);
unordered_map<int, int> m;
for(int i=0; i<nums.size(); i++) {
int n=nums[i];
if(m.count(n)) continue;
if(m.count(n-1)) merge(i,m[n-1]);
if(m.count(n+1)) merge(i,m[n+1]);
m[n]=i;
}
int res=0;
for(int i=0; i<parent.size(); i++) {
if(parent[i]==i && sz[i]>res) {
res=sz[i];
}
}
return res;
}
};
This gets accepted by the OJ (Runtime: 80 ms, faster than 76.03% of C++ online submissions for Longest Consecutive Sequence), but is this really O(n), as claimed by many answers, such as this one? My understanding is that Union Find is an O(NlogN) algorithm.
Are they right? Or, am I missing something?
They are right. A properly implemented Union Find with path compression and union by rank has linear run time complexity as a whole, while any individual operation has an amortized constant run time complexity. The exact complexity of m operations of any type is O(m * alpha(n)) where alpha is the inverse Ackerman function. For any possible n in the physical world, the inverse Ackerman function doesn't exceed 4. Thus, we can state that individual operations are constant and algorithm as a whole linear.
The key part for path compression in your code is here:
return parent[i]=find(parent[i])
vs the following that doesn't employ path compression:
return find(parent[i])
What this part of the code does is that it flattens the structure of the nodes in the hierarchy and links each node directly to the final root. Only in the first run of find will you traverse the whole structure. The next time you'll get a direct hit since you set the node's parent to its ultimate root. Notice that the second code snippet works perfectly fine, but it just does redundant work when you are not interested in the path itself and only in the final root.
Union by rank is evident here:
if(sz[p1]>sz[p2]) {...
It makes sure that the node with more children becomes the root of the node with less children. Therefore, less nodes need to be reassigned a new parent, hence less work.
Note: The above was updated and corrected based on feedback from #Matt-Timmermans and #kcsquared.
Problem Statement: You have n people sitting around in a circle. Each person has a skill value arr[i]. You want to assign people to these tasks. The people chosen for a specific task must have all been sitting consecutively and their skill gap (maximum to minimum) must not differ by more than K. Also, each person must be assigned exactly one task.
How many ways are there to assign tasks to all the people modulo 1e9+7
Two ways are considered different if person I and j are doing same task in different ways.
Example Input:
n=2
k=2
arr[]= {1,2}
Sample Output: 2
Explanation: there are 2 ways, either 1 and 2 do the same task or not. Both are valid statements.
I am stuck in this question. Earlier I tried to solve it like this:
//assuming necessary header files
int solve(int arr[], int n, int k){
map<int,vector<int>> mp;
for(int i=1;i<n;++i){
int temp= abs(arr[i]-arr[i-1]);
if(temp<=k)[
mp[temp].push_back(arr[i]);
mp[temp].push_back(arr[i-1]);
]
}
int ans=0;
for(auto it: mp){
vector<int> p= it.second;
int szz= p.size();
ans+= 2*(szz*(szz-1))/2;
}
return ans;
}
int main() {
int n,k;
cin>>n>>k
int arr[n];
for(int i=0;i<n;++i){
cin>>arr[i];
}
cout<<solve(arr,n,k)<<endl;
return 0;
}
My approach was to store all the consecutive people having skill gap less than k to be kept in a key value pair and then return twice of each key-value pair's size.
I got Wrong Answer verdict. Either I am unable to understand the question clearly or I have implemented it the wrong way. Can anyone please correct it and let me know the correct solution for the same?
Thank you
I'm working on a homework assignment for Comp Sci I, and I've tried literally everything I could find on the internet and in my textbook in an attempt to count the number of matching elements in two C++ arrays. I simply can't find a way to make my code count the number of matching elements in two arrays of the same size (same value, same position). For instance, if array1 has 5, 4, 3, 2, 1, and array2 has 1, 2, 3, 4, 5, there is one matching element, the 3. How can I get my code to perform this seemingly simple task?
Couldn't find an answer anywhere. My textbook only shows how to see if the arrays match exactly without counting the number of matches, and I haven't been able to find anyone else with this exact question for C++.
Here's my function so far:
int matchCounter(int lottery[5], int user[5], int matches, int SIZE)
{
int count = 0;
for (count < SIZE; count++;)
{
if (lottery[count] == user[count])
{
matches++;
}
}
return matches;
} // end matchCounter
SIZE is a constant that equals 5, by the way. It would seem that this should work, but every time I run the program, it displays 0 matches, even in there are some matches. If anyone could help me out, I would be forever in your debt. Seriously, there's got to be some incredibly simple thing I'm just not seeing. Thanks for you help in advance.
Your for statement is garbled. Try this:
int matchCounter(int lottery[5], int user[5], int matches, int SIZE)
{
for (int count = 0; count < SIZE; count++)
{
...
For reason why your program always fails: count++ is put in the position of the condition in the for loop, so every time this for loop is executed, count is evaluated and considered to be the condition, and then count is incremented by 1. However, every time count starts from 0 (as in int count = 0;), so the for loop actually never gets executed, and matches remain zero.
EDIT:
From your source code, you're not using your return value, which is also suggested by #jimhark. Change the function signature to:
int matchCounter(int lottery[5], int user[5], int SIZE)
And call with:
matches = matchCounter(lottery, user, 5);
Otherwise, your return value of the function matchCounter is ignored, while int matches only plays the role of a local variable in the function.
Read this about why the matches doesn't work: http://www.cs.utsa.edu/~wagner/CS2213/swap/swap.html (in a similar setting)
Since this is a homework, I will try to point the errors without giving the solution.
The form of a for loop is:
for (INIT; COND; STEP) {
BODY;
}
which is equivalent to:
INIT;
while (COND) {
BODY;
STEP;
}
The way you did, your INIT is count < SIZE, which doesn't do anything. Your "COND" is count++, which returns 0 on the first iteration. As you may know, 0 is false, in the context of a boolean. Therefore, your for loop body is not being executed at all.
Also, you should not pass matches as an argument. It should be a local variable initialized with 0.
Your may not be using your return value. Change sig to:
int matchCounter(int lottery[5], int user[5], int SIZE)
{
matches = 0;
And make sure your're calling with:
matches = matchCounter(lottery, user, 5);
If you need a running sum, use:
matches += matchCounter(lottery, user, 5);
In any case matchCounter doesn't care what running total is, so why pass it in?
Also you may need to sort the lists first.
Maybe this can help:
int matches=0;
for(int i=0;i<ARRAY.SIZE;i++) {
for(int j=0;j<ARRAY.SIZE;j++) {
if(arr1[i]==arr2[j]) {
matches++;
arr2[j]=NULL;
}
}
}
I was trying a problem on SPOJ,in which we have to simply find the Length of Longest Increasing Sub-sequence of the given Array A.
I had solved this problem using a dynamic programming O(n^2)algorithm and the solution got accepted..Here is the code,that got Accepted:
void LIS(int *A,int A_Length)
{
int Seq[MAX];
for(int i=0;i<A_Length;++i)
{
int maxima=0;
for(int j=0;j<i;++j)
{
if(A[i]>A[j])
{
maxima=max(Seq[j],maxima);
}
}
Seq[i]=maxima+1;
//cout<<Seq[i]<<endl;
}
cout<<*max_element(Seq,Seq+A_Length)<<endl;
}
But When i tried to solve it using the Second Method (LINK),which is ::
A simple way of finding the longest increasing subsequence is
to use the Longest Common Subsequence (Dynamic Programming) algorithm.
[1]Make a sorted copy of the sequence A, denoted as B. O(nlog(n)) time.
[2]Use Longest Common Subsequence on with A and B. O(n2) time.
,I got Wrong Answer .
This is my c++ code
//Global Variable
int A[100],B[100];
int DP[100][100];
//This function Finds the Longest common subsequce of Array A[1,2,3...,N] and B[1,2,3...,N]
void LIS(int N)
{
sort((B+1),(B+1)+N);//STL SORT sort from index 1 to N of Array B.
int i,j;
//Base Cases
for(i=0;i<=N;++i)
DP[i][0]=0;
for(j=0;j<=N;++j)
DP[0][j]=0;
for(i=1;i<=N;++i)
{
for(j=1;j<=N;++j)
{
if(A[i]==B[j])
DP[i][j]=DP[i-1][j-1]+1;
else
DP[i][j]=max(DP[i-1][j],DP[i][j-1]);
}
}
printf("%d\n",DP[N][N]);
}
int main()
{
int N,i;
scanf("%d",&N);
for(i=1;i<=N;++i)
{
scanf("%d",&A[i]);
B[i]=A[i];
}
LIS(N);
return 0;
}
I don't know why i am getting the Wrong Answer.Can You please Help me in Finding the Bug. Or the LIS by LCS Algorithm given in the site is incorrect??
The Second Method is correct, but can't be applied to this problem directly. That's because the numbers in the sequence are not guaranteed to be unique in this SPOJ problem, and the target is to find a strict increasing subsequence, while Your Second Method's output is non-decreasing subsequence here. Demonstrating on a simple test case [1,2,2,3] will help you find the difference.
This solution is also simple: just remove the duplicated elements after sorting.
as the title explains this is a program to find lcm of numbers between 1 to 20. i found an algorithm to do this, here's the link
http://www.cut-the-knot.org/Curriculum/Arithmetic/LCM.shtml
there is a java applet on the webpage that might explain the algorithm better
Problem: i wrote the code compiler shows no error but when i run the code the program goes berserk, i guess may be some infinite loopig but i can't figure it out for the life of me. i use turbo c++ 4.5 so basically if anyone can look at the code and help me out it would be great . thanks in advance
Algorithm:
say we need to find lcm of 2,6,8
first we find the least of the series and add to it the number above it, i.e the series become
4,6,8
now we find the least value again and add to it the intitial value in the column i.e 2
6,6,8
so the next iteration becomes
8,6,8
8,12,8
10,12,8
10,12,16
12,12,16
14,12,16
14,18,16
16,18,16
18,18,16
18,18,24
20,18,24
20,24,24
22,24,24
24,24,24
as you can see at one point all numbers become equal which is our lcm
#include<iostream.h>
/*function to check if all the elements of an array are equal*/
int equl(int a[20], int n)
{
int i=0;
while(n==1&&i<20)
{
if (a[i]==a[i+1])
n=1;
else
n=0;
i++;
}
return n;
}
/*function to calculate lcm and return that value to main function*/
int lcm()
{
int i,k,j,check=1,a[20],b[20];
/*loading both arrays with numbers from 1 to 20*/
for(i=0;i<20;i++)
{
a[i]=i+1;
b[i]=i+1;
}
check= equl(a,1);
/*actual implementation of the algorith*/
while(check==0)
{
k=a[0]; /*looks for the least value in the array*/
for(i=0;i<20;i++)
{
if(a[i+1]<k)
{
k=a[i+1]; /*find the least value*/
j=i+1; /*mark the position in array */
}
else
continue;
}
a[j]=k+b[j]; /*adding the least value with its corresponding number*/
check= equl(a,1);
}
return (a[0]);
/*at this point all numbers in the array must be same thus any value gives us the lcm*/
}
void main()
{
int l;
l=lcm();
cout<<l;
}
In this line:
a[j]=k+b[j];
You use j but it is unitialized so it's some huge value and you are outside of the array bounds and thus you get a segmentation fault.
You also have some weird things going on in your code. void main() and you use cout without either saying std::cout or using namespace std; or something similar. An odd practice.
Also don't you think you should pass the arrays as arguments if you're going to make lcm() a function? That is int lcm(int a[], int b[]);.
You might look into using a debugger also and improving your coding practices. I found this error within 30 seconds of pasting your code into the compiler with the help of the debugger.
Your loop condition is:
while(n==1&&i<20)
So your equl function will never return 1 because if n happens to be 1 then the loop will just keep going and never return a 1.
However, your program still does not appear to return the correct result. You can split the piece of your code that finds the minimum element and replace it with this for cleanliness:
int least(int a[], int size){
int minPos = 0;
for(int i=0; i<size ;i++){
if (a[i] < a[minPos] ){
minPos = i;
}
}
return minPos;
}
Then you can call it by saying j = least(a, 20);. I will leave further work on your program to you. Consider calling your variables something meaningful instead of i,j,k,a,b.
Your equl function is using array indices from 0-20, but the arrays only have 1-19
j in lcm() is uninitialized if the first element is the smallest. It should be set to 0 at the top of the while loop
In the following code, when i=19, you are accessing a[20], which is out of the bounds of the array. Should be for(i=0;i<19;i++)
for(i=0;i<20;i++) {
if(a[i+1]<k)
You are not actually using the std namespace for the cout. this should be std::cout<<l
Your are including iostream.h. The standard is iostream without the .h, this may not work on such an old compiler tho
instead of hard-coding 20 everywhere, you should use a #define. This is not an error, just a style thing.
The following code does nothing. This is the default behavior
else
continue;