My implementation of selection sort does not work in case of j < n-2 or n-1 or n. What am I doing wrong?
Is there an online IDE that lets us put a watch for the control loops?
#include <stdio.h>
#define n 4
int main(void) {
int a[n]={4,3,2,1};
int j,min;
for(int i=0;i<n;i++){
min=i;
for(j=i+1;j<n-3;j++)
if(a[j]>a[j+1])
min=j+1;
if(min!=i){
int t=a[min];
a[min]=a[i];
a[i]=a[t];
}
}
for(int i=0;i<n;i++)
printf("%d",a[i]);
return 0;
}
I tried it here
Your code has indeed a strange limit on n-3, but it has also some other flaws:
To find a minimum you should compare with the current minimum (a[min]), not the next/previous element in the array
The code to swap is not correct: the last assignment should not be from a[t], but t itself.
Here is the corrected code:
int main(void) {
int a[n]={4,3,2,1};
int j,min;
for(int i=0;i<n;i++){
min=i;
for(j=i+1;j<n;j++)
if(a[min]>a[j])
min=j;
if(min!=i){
int t=a[min];
a[min]=a[i];
a[i]=t;
}
}
for(int i=0;i<n;i++)
printf("%d",a[i]);
return 0;
}
https://ideone.com/AGJDPS
NB: To see intermediate results in an online IDE, why not add printf calls inside the loop? Of course, for larger code projects you'd better use a locally installed IDE with all the debugging features, and step through the code.
Related
i am getting 0xc0000005 error(access violation error), where am i wrong in this code?
i couldnt debug this error. please help me.
question is this
Formally, given a wall of infinite height, initially unpainted. There occur N operations, and in ith operation, the wall is painted upto height Hi with color Ci. Suppose in jth operation (j>i) wall is painted upto height Hj with color Cj such that Hj >= Hi, the Cith color on the wall is hidden. At the end of N operations, you have to find the number of distinct colors(>=1) visible on the wall.
#include<iostream>
#include <bits/stdc++.h>
#include <algorithm>
using namespace std;
int main()
{
int t;
cin>>t;
for(int tt= 0;tt<t;tt++)
{
int h,c;
int temp = 0;
cin>>h>>c;
int A[h], B[c];
vector<int> fc;
for(int i = 0;i<h;i++)
{
cin>>A[i];
}
for(int j =0;j<h;j++)
{
cin>>B[j];
}
if(is_sorted(A,A+h))
{
return 1;
}
if(count(A,A+h,B[0]) == h)
{
return 1;
}
for(int i = 0;i<h;i++)
{
if(A[i]>=temp)
{
temp = A[i];
}
else
{
if(temp == fc[fc.size()-1])
{
fc[fc.size()-1] = B[i];
}
else
{
fc.push_back(B[i]);
}
}
}
}
}
There are several issues.
When reading values into B, your loop check is j<h. How many elements are in B?
You later look at fc[fc.size()-1]. This is Undefined Behavior if fc is empty, and is the likely source of your problem.
Other issues:
Don't use #include <bits/stdc++.h>
Avoid using namespace std;
Variable declarations like int A[h], where h is a variable, are not standard C++. Some compilers support them as an extension.
Which is more efficient in terms of memory and time complexity hashing using int array or unordered_map in STL?
By hashing I mean storing elements formed by the combination of a key value and a mapped value, and fast retrieval of individual elements based on their keys.
Actually I was trying to solve this question.
Here's my solution:-
#include <bits/stdc++.h>
#define MAX 15000005
using namespace std;
/*
* author: vivekcrux
*/
int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
int c[MAX];
int n;
int sieve()
{
bitset<MAX> m;
m.set();
int ans = 0;
for(int i=2;i<MAX;i++)
{
if(m[i])
{
int mans = 0;
for(int j=i;j<MAX;j+=i)
{
m[j]=0;
mans += c[j];
}
if(mans<n)
ans = max(ans,mans);
}
}
return ans;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int i,j;
cin>>n;
int a[n+1];
for(i=0;i<n;i++)
{
cin>>a[i];
}
int g = a[0];
for(i=1;i<n;i++)
{
g = gcd(g,a[i]);
}
for(i=0;i<n;i++)
{
a[i] /= g;
if(a[i]!=1) c[a[i]]++;
}
int m = sieve();
if(m==0)
cout<<"-1";
else
cout<<n - m<<endl;
return 0;
}
In this code if I use
unordered_map<int,int> c;
instead of
int c[MAX];
I get a Memory limit exceeded verdict.I have found here that unordered_map has a constant average time complexity on average, but no details about space complexity is mentioned here.I wonder why am I getting MLE with unordered_map.
unordered_map uses bucket to store values. A bucket is a slot in the container's internal hash table to which elements are assigned based on the hash value of their key. Lets see the following code in C++17.
#include <bits/stdc++.h>
using namespace std;
int main() {
unordered_map<int,int> mp;
mp[4] = 1;
mp[41] = 5;
mp[67] = 6;
cout<<mp.bucket_count();
}
The output comes out be 7 (depends on compiler). This is the number of buckets used in the above code. But if we use an array of size 67, it will obviously take more memory. Another case would be that if we would had numbers 1, 2 and 3 instead of 4, 41 and 67, the output would have been 7. Here using array was the way to go for saving space. So it depends on the keys you are storing in the hash table. For time complexity, both performs equally same. There is a collision condition in unordered_map which would blow the overall time complexity of the code. Here is the codeforces link of the blog.
When i try to run this code, it compiles without any error, but i wanted that it should display the string and int array that it takes as input. Instead, after giving one input, I am presented with a list of zeroes and program terminates.
however,when i removed the line containing k[j] from both for loops, it worked.
What am I doing wrong? Pardon if asked something stupid as I'm a novice, please help...
#include <iostream>
#include <string>
using namespace std;
int main ()
{
int t,count = 0;
string state[t];
int k[t];
cin>>t;
for (int j=0; j<t; j++)
{
getline(cin, state[j]);
cin>>k[j];
}
for (int j=0; j<t; j++)
{ cout<< state [j]<<'\t'<<k[j];
cout<<endl;
}
return 0
}
You declare both arrays k[] and state[] with t wich is undefined at this moment.
t should be initiated with a valid value beforehand!
Please tell me why this code is giving garbage values
Compiles well, implemented this based on the Cormen algorithm for mergesorting
Basically taking given numbers in a dynamic array. two void functions are taken.One is to merge the two sub arrays via merge sort and the other to recursively split the array to sub arrays
#include<iostream>
using namespace std;
void merge(int *A,int p, int q, int r)// a function to merge two //sub arrays
{
int n1= q-p+1;
int n2=r-q;
int L[n1];
int R[n2];
for (int i=0;i<n1;i++)
{
L[i]=A[p+i];
}
int m=1;
for(int j=0; j<n2 ;j++)
{
R[j]=A[q+m];
m=m+1;
}
int i=0;
int j=0;
for(int k=0;k<r;k++)
{
if (L[i]<=R[j])
{
A[k]=L[i];
i=i+1;
}
else
{
A[k]=R[j];
j=j+1;
}
}
}
void mergesort(int *A,int p,int r)// dividng the sequence to sub arrays
{
if (p<r)
{
int q;
q=(p+r)/2;
mergesort(A,p,q);
mergesort(A,(q+1),r);
merge(A,p,q,r);
}
}
int main()
{
int n;
cout<<"Enter the number of numbers to be sorted by merge sort"<<endl;
cin>>n;
int* a=NULL;
a=new int[n];
int temp;
cout<<"Enter the numbers"<<endl;
for(int i=0;i<n;i++)
{
cin>>temp;
*(a+i)=temp;// inputting the given numbers into a dynamic array
}
cout<<"The given numbers are:"<<endl;
for(int j=0;j<n;j++)
cout<<*(a+j)<<" ";
mergesort(a,0,n-1);
cout<<"The merged sorted numbers are:"<<endl;
for(int s=0;s<n;s++)
cout<<*(a+s)<<" ";
delete [] a;
system("pause");
return 0;
}
You are getting your intervals wrong pretty much everywhere in your code. For example:
Based on your usage in main, mergesort is supposed to sort the sublist of indices [0,n-1].
With this meaning, your recursion in mergesort says in order to sort the indices [p,r-1], you should first sort [p,q-1] then sort [q+1,r-1]: you completely ignore index q.
Similarly, merge is confused: once you fix the typo when coping into L (A[i] should be A[p+i]), it takes [p,q] as one list, and [q,r] as the other list: note you copy entry q twice, and you also copy r when you probably shouldn't be.
To fix your code, you need to straighten out exactly what intervals everything is supposed to be working on. This isn't a hard problem, you just have to bring yourself to write down explicitly exactly what all of your functions and loops and stuff are supposed to be doing.
The typical convention these days is half-open intervals: you should generally think of taking indices [p,q) from a list. ([p,q) is the same as [p,q-1]) Here are several examples of why this is preferred:
The number of entries of [p,r) is simply r-p
A for loop iterating through [p,r) is the usual for(i=p; i<r; ++i) (not <=)
Splitting the interval [p,r) into parts gives you intervals [p,q) and [q,r) -- there is no worry about remembering to add 1 in places.
e.g. merge would normally be designed to take the first list comes from indices [p,q) and the second list from indices [q,r).
Halo I just write code to perform Matrix chain multiplication, which can be solved by Dynamic Programming
http://en.wikipedia.org/wiki/Matrix_chain_multiplication#A_Dynamic_Programming_Algorithm
Here is the code I wrote, which I think is simpler than the one provided by wikipedia. So I doubt am i doing dynamic programming or not?
and I can't figure out the time complexity of my program. Can someone help me to figure the time complexity of this program?
Here's my guess..
the for loop will run n times for each call? if mem is not used..
for each loop, it will then expand into two
if mem is used, it prevent recalculation...
ahhh I can't figure it out, hope someone can help me :-)
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <climits>
using namespace std;
int mem[10][10];
int row[10];
int col[10];
int m[10];
#define NUM 4
int DP(int c, int r){
if(mem[c][r] != INT_MAX) return mem[c][r];
if(c == r) return 0;
int min_cost;
for(int j=c; j<r; j++){
min_cost = DP(c, j) + DP(j+1, r) + m[c-1]*m[j]*m[r];
if(min_cost < mem[c][r])
mem[c][r] = min_cost;
}
return mem[c][r];
}
int main(){
for(int i=0; i< 10;i++){
for(int j=0; j<10;j++){
mem[i][j] = INT_MAX;
}
}
int n = NUM; // MAX 4 matrix
int a,b;
for(int i=0; i< NUM+1; i++){
cin >> a;
m[i] = a;
}
cout << "Lowest Cost for matrix multiplicatoin " << DP(1,NUM);
}
The technique you have used is called memoization. Most of the time, you may solve DP problems using memoization with little (or no) overhead.
The complexity of your implementation is just like the original DP solution: O(n^3) (Note: Every cell of mem array should be computed at least once, and each cell takes O(n) time to be computed. Further computation of a cell, does not involve any loop, since it would be a simple lookup.)
See also http://en.wikipedia.org/wiki/Memoization