I am learning container in C++ and trying to insert and print element of list with the help of iterator. I am getting a different output then my expectations.
#include<iostream>
#include<list>
#include<vector>
#include<deque>
#include<string>
using namespace std;
int main()
{
list<int> ilist(10);
cout<<ilist.size()<<endl;
list<int>::iterator litr = ilist.begin();
int i =0;
for(litr = ilist.begin();litr != ilist.end();litr++,i++)
{
cout<<"i:"<<i<<" ";
//ilist.push_back(i);
*litr=i;
litr++;
}
litr = ilist.begin();
cout<<endl;
cout<<ilist.size();
while(litr != ilist.end())
{
i=*litr;
cout<<i<<endl;
litr++;
}
return 0;
}
output :
10
i:0 i:1 i:2 i:3 i:4
100
0
1
0
2
0
3
0
4
0
why size changed after Inserting element and why elements are not properly inserted ?
Thanks in advance.
In the first loop, you are incrementing the iterator twice per iteration (once in the for statement and once more at the bottom of the loop).
This causes the iterator to skip every other element.
The size of the list remains unchanged: the 100 is 10 immediately followed by 0. The 10 is the size of the list and the 0 is the first element of the list. Print a space or a newline after printing out the size and you'll see for yourself.
Size did not change. Before second loop you display size of the list and then it's contents without any separator. Since the first element in the list is equal to 0 it looks as if the list grew to the length of 100.
Also, like NPE has mentioned, in the first loop you increment the iterator twice.
In this loop
for(litr = ilist.begin();litr != ilist.end();litr++,i++)
{
cout<<"i:"<<i<<" ";
//ilist.push_back(i);
*litr=i;
litr++;
}
you are increasing iterator litr twice: in the control statement and within the body of the loop.
It could be written simpler
list<int>::iterator litr;
int i = 0;
for ( litr = ilist.begin(); litr != ilist.end(); litr++, i++ )
{
cout<<"i:"<<i<<" ";
*litr = i;
}
Or
list<int>::iterator litr;
int i = 0;
for ( litr = ilist.begin(); litr != ilist.end(); litr++ )
{
cout<<"i:"<<i<<" ";
*litr = i++;
}
As for the output then that it would be more clear you need to insert std::endl after printing the size the second time. For example
cout<<ilist.size() << endl;
// ^^^^^^^
while(litr != ilist.end())
{
i=*litr;
cout<<i<<endl;
litr++;
}
Then the output will look like
10
i:0 i:1 i:2 i:3 i:4
10
0 0 1 0 2 0 3 0 4 0
Take into account that there are several standard algorithms declared in header <algorithm> that can do the same task. For example you could use algorithm std::generate.
#include <algorithm>
//...
int i = 0;
std::generate( ilist.begin(), ilist.end(), [&] { return i++; } );
Related
I have a code for "Minimum number of jumps to reach end of the array with its sequence using recursion". But I am not able to print the sequence. ( There is nothing in vector vec to print )
Any help will be appreciated.
Explanation :
I want to reach from 1st element ( i.e. 2) to
last element ( i.e. 4) of the array in minimum Jump.
How Jump will be :
1st element is 2. It means I can make upto 2 jumps in array. If I take 1st jump then I can reach 2nd element ( i.e. 3) or if I take
2nd jump then I can reach 3rd element (i.e. 1)
2nd element is 3 ,so I can make maximum 3 jumps. In 1st jump I can reach to 1 , in 2nd jump I can reach to 0 and in 3rd jump I can
reach to 4
In this way I want to reach from 1st element to last element of the array in minimum number of jumps.
So output will be like , from 1st element 2, I will jump to 3. Then from 3 I will jump to 4 (last element). So 2 Jumps. ( 2 - 3 - 4 )
#include<iostream>
#include<vector>
#include<climits>
using namespace std;
int jump(int arr[], int n, int start, vector<int> &vec)
{
if(start == n-1) // if start is the last element in array
return 0;
if( arr[start] == 0) // if array element is 0
return 0;
vector<int> vec1 = vec;
vector<int> vec2 = vec;
int minimum = INT_MAX;
for( int i = 1 ; i <= arr[start]; i++ )
{
vec1.push_back(start);
int _jump = 1 + jump( arr, n, start+i, vec1); // considering every jump
vec = (_jump < minimum) ? vec1 : vec2;
minimum = min(minimum, _jump);
}
return minimum;
}
int main()
{
int arr[] = { 2, 3, 1, 0, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
vector<int> vec;
cout << "Number of jumps " << jump(arr, n, 0, vec) << endl;
cout<<"Sequence is "<<endl;
for( auto x : vec)
cout << x <<" ";
return 0;
}
output
Number of jumps 2
Sequence is
Expected output
Number of jumps 2
Sequence is 2 3 4
Here is an example that will set a vector where each index stores the correct next step in the sequence after visiting that index. I leave it to you to code following the sequence from the first element to the end, using the result vector. I also corrected this condition if( arr[start] == 0) to return "infinity" since if we visit this element, we cannot complete the sequence.
#include<iostream>
#include<vector>
#include<climits>
using namespace std;
int jump(int arr[], int n, int start, vector<int> &vec)
{
if(start == n-1) // if start is the last element in array
return 0;
if( arr[start] == 0) // if array element is 0
return INT_MAX - n;
int minimum = INT_MAX;
int step;
for( int i = 1 ; i <= arr[start]; i++ )
{
int _jump = 1 + jump( arr, n, start+i, vec); // considering every jump
if (_jump < minimum){
minimum = _jump;
step = start + i;
}
}
vec.at(start) = step;
return minimum;
}
int main()
{
int arr[] = { 2, 3, 1, 0, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
vector<int> vec(n, -1);
cout << "Number of jumps " << jump(arr, n, 0, vec) << endl;
cout<<"Vector: "<<endl;
for( auto x : vec)
cout << x <<" ";
return 0;
}
Essentially, this is the minimal fix so that the sample data would works. I have not check all edge cases. For example, one might want to print something else than the value of INT_MAX is the end is not reachable.
Problem 1
You want to output values (i.e. 2, 3, 4 in your example) and not index (0, 1, 4). Thus you must push values instead of indexes.
vec1.push_back(arr[start]);
Problem 2
if(start == n-1) // if start is the last element in array
return 0;
This will not add the final value when the end is reached. You must add last value with:
vec.push_back(arr[start]);
Problem 3
if( arr[start] == 0) // if array element is 0
return 0;
A sequence that does not reach the end, would be considered to be very good. You should return a large value. Since _jump is 1 + return value of jump, the return value should be INT_MAX - 1 and minimum should also be initialized to that value for same reason.
Alternatively, you could return other values like n too instead.
Problem 4
Finally, the following condition is incorrect:
vec = (_jump < minimum) ? vec1 : vec2;
When the condition is not verified, it is vect2 that need to be copied in vec1 since the loop uses vect1.
Can somebody tell me what is the problem with this program?
The idea is to display "Yes" if the vector array satisfies all these criteria:
The array elements are not sorted in ascending order.
The array contains distinct elements.
All the array elements should have a value between 1 to n inclusive.
Otherwise "No".
The program aborts when it reaches the line at if(bSort).
Is there anything wrong with the iterator increment?
#include <iostream>
#include <string>
#include <bits/stdc++.h>
using namespace std;
std::string solve(vector <int> &a, int n) {
vector<int> visited (n);
int i=0;
for(std::vector<int>::iterator it = a.begin(); it != a.end(); ++it) {
i++;
if((it+1)!=a.end() && (*it > *(it+1)))
{
bSort = false;
}
if(std::find(visited.begin(), visited.end(), *it)!=visited.end())
{
return "No";
}
else
{
visited[i] = *it;
}
if(*it <= 0 || *it > n)
{
return "No";
}
}
if(bSort)
return "No";
else
return "Yes";
}
int main() {
int q;
cin >> q;
for(int a0 = 0; a0 < q; a0++){
int n;
cin >> n;
vector<int> a(n);
for(int a_i = 0; a_i < n; a_i++){
cin >> a[a_i];
}
std::string result = solve(a,n);
cout << result << endl;
}
return 0;
}
The issue appears to be happening only with the following input:
1
30
18 8 24 20 7 17 5 9 26 21 25 12 11 15 30 13 19 16 22 10 14 1 3 29 23 2 6 28 4 27
I'm not sure the problem has to do with iterators specifically.
At the very beginning of the loop, variable i is incremented before it's used, which means the set of numbers that i will range between is [1, vector.size()]. This means that at some point, you'll access vector[vector.size()], which is undefined behavior and can crash your program.
In your program specifically, given the input you provided, because none of the numbers in your example code are duplicated, the else branch of the std::find(...) conditional statement is always executed, which means you end up calling visited[30] at some point, which again, is out of bounds and undefined behavior, potentially causing a crash.
I have a code where I am reverse iterating a map in C++. For n pairs in map, sometimes the map prints out n pairs, sometimes n-1 pairs.
for test case
4 2
2 1
3 5
2 3
1 5
it prints 4 hi's, for the next case 3 hi's
4 1
0 2
0 3
3 5
0 6
My code:
include cmath, cstdio, vector, iostream, map, algorithm, using namespace std;
int main() {
int n,f;
cin >> n >> f;
int k[n],l[n],dif[n];
map<int,int> m;
map<int,int>::reverse_iterator it;
for (int i=0;i<n;i++){
scanf("%d %d\n",&k[i],&l[i]);
dif[i] = l[i]-k[i];
m.insert(make_pair(dif[i],l[i]));
}
int a = 0;
int sum = 0;
for(it = m.rbegin(); it != m.rend(); ++it ) {
cout << "hi" << endl;
int article = it->second - it->first;
int people = it->second;
if (a < f and article!=0){
a++;
if (2*article > people){
sum = sum + people;
} else {
sum = sum + (2*article);
}
} else {
if (article > people){
sum = sum + people;
} else {
sum = sum + article;
}
}
}
cout << sum;
return 0;
}
This has nothing to do with reverse iteration, it would be the same with forward iteration. A map has a pair of a key and a value. Each key can only be included in the map once. When you insert the same key again the previous value is overwritten. In your second test case you are setting the key 0 three times:
4 1
0 2 <--
0 3 <--
3 5
0 6 <--
So your map actually only has three values, so the three "Hi" are correct. In your first example you overwrite key 2 only twice, so the four "Hi". Maybe a map is not the correct container you are looking for.
This is an spoj problem. It works, but It's too slow.
Here is the question:
Iterated Difference
You are given a list of N non-negative integers a(1), a(2), ... ,
a(N). You replace the given list by a new list: the k-th entry of the
new list is the absolute value of a(k) - a(k+1), wrapping around at
the end of the list (the k-th entry of the new list is the absolute
value of a(N) - a(1)). How many iterations of this replacement are
needed to arrive at a list in which every entry is the same integer?
For example, let N = 4 and start with the list (0 2 5 11). The successive iterations are:
2 3 6 11
1 3 5 9
2 2 4 8
0 2 4 6
2 2 2 6
0 0 4 4
0 4 0 4
4 4 4 4
Thus, 8 iterations are needed in this example.
Input
The input will contain data for a number of test cases. For each case,
there will be two lines of input. The first line will contain the
integer N (2 <= N <= 20), the number of entries in the list. The
second line will contain the list of integers, separated by one blank
space. End of input will be indicated by N = 0.
Output
For each case, there will be one line of output, specifying the case
number and the number of iterations, in the format shown in the sample
output. If the list does not attain the desired form after 1000
iterations, print 'not attained'.
Sample Input
4
0 2 5 11
5
0 2 5 11 3
4
300 8600 9000 4000
16
12 20 3 7 8 10 44 50 12 200 300 7 8 10 44 50
3
1 1 1
4
0 4 0 4
0
Sample Output
Case 1: 8 iterations
Case 2: not attained
Case 3: 3 iterations
Case 4: 50 iterations
Case 5: 0 iterations
Case 6: 1 iterations
I'm not sure of what to do to make it faster. I tried using arrays, but I get all sorts of problems trying to allocate the memory and set one array to another.
How can I make it faster? Here's my code:
#include <iostream>
#include <vector>
#include <cmath>
#include <sstream>
#include <string>
using namespace std;
bool checker(vector<int>& nums2) {
int n = nums2[0];
for (int i = 1; i < nums2.size(); i++)
{
if (n != nums2[i])
return false;
}
return true;
}
vector<int> iterate(vector<int>& nums, int& iter, bool& attained) {
if (iter == 1000) {
attained = false;
return nums;
}
vector<int> nums2;
for (int i = 0; i < nums.size(); i++) {
if (i == nums.size() - 1)
nums2.push_back((int)abs((double)nums[i] - (double)nums[0]));
else
nums2.push_back((int)abs((double)nums[i] - (double)nums[i + 1]));
}
iter++;
return nums2;
}
int main()
{
int N = -1, count = 1;
while (1) {
int num = 0;
vector<int> nums;
string List = "";
stringstream ss;
cin >> N;
if (N == 0)
break;
cin.ignore();
cin.clear();
getline(cin, List);
ss << List;
while (ss >> num) {
nums.push_back(num);
}
int iterations = 0;
bool attained = true;
while (!checker(nums)) {
nums = iterate(nums, iterations, attained);
}
if (!attained)
cout << "case " << count << ": not attained";
else
cout << "case " << count << ": " << iterations << " iterations" << endl;
count++;
}
}
I fixed it. It was a problem with the while loop in the main function. The condition was:
while (!checker(nums)) { ... }
It would stay in the loop and repeatedly call the iterate function because if it is not attainable, then the checker will always be false. So changing the condition to:
while (!checker(nums) && attained) { ... }
would break the loop if it was not attainable.
Basically, it was just getting stuck on doing the same thing over and over; it wasn't actually slow.
Thanks, xan, for your answer.
If you want it to be a little faster you should work on debugging your array variation to avoid the vector allocations. If you want it to be a lot faster you need to do some analysis of the problem to find a better algorithm. For instance, if you see the same list twice you're in a loop and will exceed 1000 iterations. And the result will be the same if you rotate the list, which you can consider when checking for a repeated list.
Your implementation executes 1000 iterations in 25ms on my mainstream lapton. Fixed one, because there's a bug and case 2 will execute forever.
To do faster you can reuse the same vector and modify it in place, your iterator() function signature would look like:
void iterate(vector<int>& nums);
This version takes 7ms on my machine, because it doesn't allocate memory in loop.
I built a program for converting a multigraph into undirected graph with multiple edges and self loops removed using adjacency-list as Graph representation.
`
#include<iostream>
#include<istream>
#include<algorithm>
#include<list>
using namespace std;
int main()
{
list<int> adj[3];
list<int> auxArray[3];
list<int> adjnew[3];
cout<<adjnew[2].back()<<endl; // Gives output 0, whereas it should have some garbage
//value
for(int i = 0;i<3;i++){
int x;
while(true){ // reading a line of integers until new line is encountered , peek()
returns the next input character without extracting it.
cin>>x;
adj[i].push_back(x);
auxArray[i].push_back(x);
if(cin.peek() == '\n') break;
}
}
//flatten the adj-list
for(int i = 0;i<3;i++){
list<int>::iterator it = adj[i].begin();
while(it != adj[i].end()){
auxArray[*it].push_back(i);
it++;
}
}
for(int i = 0;i<3;i++){
list<int>::iterator it = auxArray[i].begin();
while(it != auxArray[i].end()){
//cout<<*it<<" "<<adjNew[*it].back()<<endl;
if((*it != i) && ((adjnew[*it].back()) != i)){
// cout<<*it<<" -> "<<i<<endl;
adjnew[*it].push_back(i);
}
it++;
}
}
for(int i = 0;i<3;i++){
list<int>::iterator it = adjnew[i].begin();
while(it != adjnew[i].end()){
cout<<*it<<" ";
it++;
}
cout<<endl;
}
return 0;
}
`
But it shows St9bad_alloc error whereas my list has size of just 3.
Also, adjnew[2].back() is assigned to "0" without being initialized, whereas it should have some garbage value.
'
Input:
1 2 1
0
1 1
Output of Program(Incorrect because of 0 as back element in adjnew[2]):
1 2
0 2
1
Correct Output:
1 2
0 2
0 1
'
All suggestions are welcomed!
The
cout<<adjnew[2].back()<<endl;
at begin is plain undefined behavior on an empty container.
valgrind gives
Conditional jump or move depends on uninitialised value(s)
for this line:
if ((*it != i) && ((adjnew[*it].back()) != i))
Again an undefined behavior on an empty container.
Hint: You could use container.at() instead of operator [] to have a range check.