Related
I don't think you'll need to know the context of the problem to answer this question, but I'll give it just in case.
-In the past N weeks, we've measured the amount of rainfall every day, and noted it down for each day of the week. Return the number of the first week of the two week period where there were the most days without rain.
The code gives no warnings or errors, and if I try to print dryestweeks inside the second for loop, then it returns the correct answer. However, all of the code after the second for loop seems to be getting ignored, and I'm getting Process returned -1073741819 (0xC0000005). The issue has to lie in the 2nd for loop, because if I comment it out then both "test2" and dryestweeks get printed, and the program returns 0.
#include <iostream>
#include <vector>
#include <bits/stdc++.h>
using namespace std;
int main() {
int weeks;
cin >> weeks;
vector<int> v[weeks];
for (int i = 0;i < weeks; i++) {
int a, b, c, d, e, f, g;
cin >> a >> b >> c >> d >> e >> f >> g;
v[i].push_back(a);
v[i].push_back(b);
v[i].push_back(c);
v[i].push_back(d);
v[i].push_back(e);
v[i].push_back(f);
v[i].push_back(g);
}
int mostdrydays = 0;
int dryestweeks = 0;
for (int i = 0; i < weeks; i++) {
int weeklydrydays = count(v[i].begin(), v[i].end(), 0);
int nextweekdrydays = count(v[i+1].begin(), v[i+1].end(), 0);
int biweeklydrydays=weeklydrydays+nextweekdrydays;
if (biweeklydrydays > mostdrydays) {
mostdrydays = biweeklydrydays;
dryestweeks = i + 1;
}
}
cout << "test2" << endl;
cout << dryestweeks << endl;
return 0;
}
An example of an input would be:
6
5 10 15 20 25 30 35
0 2 0 0 0 0 0
0 0 0 1 0 3 0
0 1 2 3 4 5 6
5 1 0 0 2 1 0
0 0 0 0 0 0 0
The program should print "2" with the above input.
The second loop has an overflow.
You first defined v[weeks] and then the second loop goes from [0, weeks[ but you are retrieving the next week with v[i + 1]. I don't know exactly what are you are trying to achieve, but if you do
for(int i = 0; i < weeks - 1; i++)
{
...
}
it executes properly.
For the given example of input, in the last iteration (i = 5) of the second loop, index i + 1(=6) will be out of the bound for v[i + 1] (legal indices for v will be from 0 to 5).
The second loop is iterating one more time than required.
Smaller Greater Equal Numbers
PrepBuddy has N baskets containing one fruit each with some quality factor(ith basket have Ai quality factor) and Tina has one single basket with one fruit having quality factor K. She wants to know how many PrepBuddy's baskets have quality factor less(L) than K, how many baskets have quality factor more(M) than K and how many baskets have quality factor equal(E) to K.
Input format
The first line contains an integer T, representing the number of test cases.T test cases follow,First linecontains two space-separated integers N and K.The second line contains N space-separated integers representing the quality factor of the basket.
Output format
For each test case on a new line, print three space-separated integers representing the values of L, M,and E.
Constraints
1<=T<=100
1<=N,K<=10^5
−10^6<=A[i]<=10^6
Sum of all N over any test case file doesn't exceed 5∗10^6
Time Limit
1 second
Example
Input
2
5 2
-1 0 -3 1 2
5 3
1 -1 -5 2 4
Output
4 0 1
4 1 0
Sample test case explanation
In the first test case,
K=2, the baskets with quality factor smaller than K are [1,2,3,4], there is no basket which has quality factor more than K and there is one basket [5] which have quality factor equal to K.
My solution
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
ll n, k;
cin >> n >> k;
ll arr[n];
for (ll i = 0; i < n; i++) {
cin >> arr[i];
}
int less = 0, more = 0, equal = 0;
for (ll i = 0; i < n; i++) {
if (arr[i] < k) {
less++;
} else if (arr[i] > k) {
more++;
} else {
equal++;
}
cout << less << " " << more << " " << equal << " ";
}
cout << endl;
}
return 0;
}
Input
2
5 2
-1 0 -3 1 2
5 3
1 -1 -5 2 4
Output
1 0 0 2 0 0 3 0 0 4 0 0 4 0 1
1 0 0 2 0 0 3 0 0 4 0 0 4 1 0
Why am I getting additional numbers like 1 0 0 2 0 0 3 0 0 4 0 0 along with my answer.How to correct this?? Please help
The error in your code was quiet simple, and easy to debug. You were printing the output every run through the array, thus, getting all these extra prints.
How does it become easy to debug? Reorganizing the code so it will be more readable, made it quiet possible. Actually, the fix was moving the line:
cout<<less<<" "<<more<<" "<<equal<<" ";
Two line lower than it was.
In order to demonstrate it, here is the code fixed and organized:
#include <iostream>
#include <vector>
#include <cstdint>
int main()
{
int t;
std::cin >> t;
while(t--)
{
std::int64_t n,k;
std::cin >> n >> k;
std::vector<int> vec{n};
for(std::size_t i = 0; i < n; ++i)
{
std::cin >> vec[i];
}
int less=0, more=0, equal=0;
for (std::size_t i = 0; i < n; i++)
{
if (vec[i] < k)
{
less++;
}
else if (vec[i] > k)
{
more++;
}
else
{
equal++;
}
// The next output line was here, under the for loop!!!
}
std::cout << less << " " << more<< " " << equal << " "; // this is its place!
std::cout << std::endl;
}
return 0;
}
I have made only 3 changes:
Removed the using namespace std; and #include <bit/stdc++.h>.
Realigned the code according to its logical order - each new scope has its own indentation level, showing which command runs in which scope, revealing the mistake.
Used proper types to each thing: In C++ there are no dynamic arrays in the format arr[n], thus, you need to use std::vector. Also, there are fixed size lengths in cstdint, and you should use those to ensure you are using the right type. Also, prefer using unsigned values when possible, for example, when indexing, use either std::size_t, std::uint32_t or std::uint64_t.
This code worked fine for me
#include <bits/stdc++.h>
using namespace std;
int main()
{
int T,N,K,arr[N];
cin>>T;
while(T--)
{
cin>>N;
cin>>K;
for(int i=0;i<N;i++)
{
cin>>arr[i];
}
int L=0,M=0,E=0;
for(int i=0;i<N;i++)
{
if(arr[i]<K){
L++;
}
else if(arr[i]>K)
{
M++;
}
else{
E++;
}
}
cout<<L<<" "<<M<<" "<<E<<endl;
}
return 0;
}
I am a beginner programmer and I need some assistance.
I need to write a program that reads an array of 10 numbers from a user, then scans it and figures out the most common number/s in the array itself and prints them. If there is only one number that is common in the array, only print that number. But, if there's more than one number that appears more than once, print them also in the order they appear in in the array.
For example- 1 2 3 3 4 5 6 7 8 9 - output would be 3
For- 1 2 3 4 1 2 3 4 5 6 - output would be 1 2 3 4
for- 1 1 1 1 2 2 2 3 3 4 - output would be 1 2 3
Now, the problem I've been running into, is that whenever I have a number that repeats more than twice (see third example above), the output I'm getting is the number of iterations of the loop for that number and not only that number once.
Any assistance would be welcome.
Code's attached below-
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
int array [10], index, checker, common;
main ()
{
for (index=0; index<10; index++)
{
cin >> array [index];
}
for (index=0; index<10; index++)
{
int tempcount=0;
for (checker=(index+1);checker<10;checker++)
{
if (array[index]==array[checker])
tempcount++;
}
if (tempcount>=1)
cout << array[index]<<" ";
}
return 0;
}
Use appropriate data structures for the task.
Create a std::unordered_map that maps value to number_of_occurrences, and make a single pass over the input data.
Then create another map from number_of_occurrences to value. Sort it, in descending order. Report the first value, plus any additional ones that occurred as many times as the first did.
The reason you are having problems is that anytime a number appears two times or more it will print out. A solution is that you create another variable maxCount, then find the maximum times a number appears. Then loop through the array and print out all the numbers that appears the maximum amount of times.
Hope this helps.
Jake
Rather than writing you a solution, I will try to give you some hints that you can hopefully use to correct your code. Try to keep track of the following things:
Remember the position of the first occurrence of each distinct number in the array.
Count the number of times each number appears
and combine the two to get your solution.
EDIT:
int array[] = {1, 2, 3, 4, 1, 2, 3, 4, 5, 6};
int first [11], cnt[11];
for(int i = 0; i < 11; i++){
first[i] = -1;
cnt[i] = 0;
}
int max = 0;
for(int i = 0; i < 10; i++){
cnt[array[i]]++;
if(max < array[i]) max = array[i];
}
for(int i = 0; i <= max; i++){
if(cnt[i] > 1 && first[i] == -1) {
printf(" %d", i);
first[i] = i;
}
}
You could do something like this. At any index in the array look for previous occurences of that element. If you find that that it is the first occurence of that element, you only need to look if there is an occurence of that element ahead in the array.
Lastly display the element whose frequency(here num) would be greater than 1.
for (int i = 0; i < 10; i++)
{
int presentBefore = 0;
for (int j = 0; j < i; j++) //if any previous occurence of element
{
if (array[i] == array[j]) presentBefore++;
}
if (presentBefore == 0)//if first occurence of the element
{
int num = 1;
for (int j = i + 1; j < 8; j++)// if occurences ahead in the array
{
if (array[i] == array[j]) num++;
}
if(num>1)cout<<array[i]<<" ";
}
}
Here is another solution using STL and std::set.
#include <iostream>
#include <algorithm>
#include <set>
#include <iterator>
int main()
{
int array[12] = { 1, 2, 3, 1, 2, 4, 5, 6, 3, 4, 1, 2 };
std::set<int> dupes;
for (auto it = std::begin(array), end = std::end(array); it != end; ++it)
{
if (std::count(it, end, *it) > 1 && dupes.insert(*it).second)
std::cout << *it << " ";
}
return 0;
}
Prints:
1 2 3 4
I will try to explain how this works:
The original array is iterated from start to finish (BTW as you can see it can be any length, not just 10, as it uses iterators of beginning and end)
We are going to store duplicates which we find with std::count in std::set
We count from current iterator until the end of the array for efficiency
When count > 1, this means we have a duplicate so we store it in set for reference.
std::set has unique keys, so trying to store another number that already exists in set will result in insert .second returning false.
Hence, we print only unique insertions, which appear to be in the order of elements appearing in the array.
In your case you can use class std::vector which allows you to Erase elements, resize the array...
Here is an example I provide which produces what you wanted:
1: Push the values into a vector.
2: Use 2 loops and compare the elements array[i] and array[j] and if they are identical push the the element j into a new vector. Index j is always equal to i + 1 in order to avoid comparing the value with itself.
3- Now you get a vector of the repeated values in the temporary vector; You use 2 loops and search for the repeated values and erase them from the vector.
4- Print the output.
NB: I overloaded the insertion operator "<<" to print a vector to avoid each time using a loop to print a vector's elements.
The code could look like :
#include <iostream>
#include <vector>
std::ostream& operator << (std::ostream& out, std::vector<int> vecInt){
for(int i(0); i < vecInt.size(); i++)
out << vecInt[i] << ", ";
return out;
}
int main() {
std::vector< int > vecInt;
//1 1 1 1 2 2 2 3 3 4
vecInt.push_back(1);
vecInt.push_back(1);
vecInt.push_back(1);
vecInt.push_back(1);
vecInt.push_back(2);
vecInt.push_back(2);
vecInt.push_back(2);
vecInt.push_back(3);
vecInt.push_back(3);
vecInt.push_back(4);
std::vector<int> vecUniq;
for(int i(0); i < vecInt.size(); i++)
for(int j(i + 1); j < vecInt.size(); j++)
if(vecInt[i] == vecInt[j])
vecUniq.push_back(vecInt[j]);
std::cout << vecUniq << std::endl;
for(int i = 0; i < vecUniq.size(); i++)
for(int j = vecUniq.size() - 1 ; j >= 0 && j > i; j--)
if(vecUniq[i] == vecUniq[j])
vecUniq.erase(&vecUniq[j]);
std::cout << vecUniq << std::endl;
std::cout << std::endl;
return 0;
}
The input: 1 2 3 3 4 5 6 7 8 9
The output: 3
The input: 1 2 3 4 1 2 3 4 5 6
The output: 1 2 3 4
The input: 1 1 1 1 2 2 2 3 3 4
The output: 1 2 3
For this problem, you can use a marking array that will count the number of times you a digit is visited by you, it's just like counting sort. let's first see the program :
#include <iostream>
using namespace std;
int print(int a[],int b[])
{
cout<<"b :: ";
for (int index=0;index<10;index++)
{
cout<<b[index]<<" ";
}
cout<<endl;
}
int main ()
{
int a[10],b[11], index, checker, common;
for (index=0; index<10; index++)
{
cin >> a [index];
b[index] = 0;
}
b[10] =0;
for (index=0;index<10;index++)
{
b[a[index]]++;
if (b[a[index]] == 2)
cout<<a[index];
//print(a,b);
}
return 0;
}
As you can see that I have used array b as marking array which counts the time a number is visited.
The size of array b depends upon what is the largest number you are going to enter, I have set the size of array b to be of length 10 that b[11] as your largest number is 10. Index 0 is of no use but you need not worry about it as it will be not pointed until your input has 0.
Intially all elements in array in b is set 0.
Now assume your input to be :: 1 2 3 4 1 2 3 4 5 6
Now value of b can be checked after each iteration by uncommenting the print function line::
b :: 0 1 0 0 0 0 0 0 0 0 ....1
b :: 0 1 1 0 0 0 0 0 0 0 ....2
b :: 0 1 1 1 0 0 0 0 0 0 ....3
b :: 0 1 1 1 1 0 0 0 0 0 ....4
b :: 0 2 1 1 1 0 0 0 0 0 ....5
b :: 0 2 2 1 1 0 0 0 0 0 ....6
b :: 0 2 2 2 1 0 0 0 0 0 ....7
b :: 0 2 2 2 2 0 0 0 0 0 ....8
b :: 0 2 2 2 2 1 0 0 0 0 ....9
b :: 0 2 2 2 2 1 1 0 0 0 ....10
In line 5 you can b's at index 1 has value 2 so it will print 1 that is a[index].
And array a's element will be printed only when it is repeated first time due to this line if(b[a[index]] == 2) .
This program uses the idea of counting sort so if you want you can check counting sort.
So I N - numbers I have to input, and I got M - numbers of places for those numbers and I need to find all combinations with repetition of given numbers.
Here is example:
Let's say that N is 3(I Have to input 3 numbers), and M is 4.
For example let's input numbers: 6 11 and 533.
This should be result
6,6,6,6
6,6,6,11
6,6,6,533
6,6,11,6
...
533,533,533,533
I know how to do that manualy when I know how much is N and M:
In example where N is 3 and M is 4:
int main()
{
int N = 3;
int M = 4;
int *numbers = new int[N + 1];
for (int i = 0; i < N; i++)
cin >> numbers[i];
for (int a = 0; a < N; a++)
for (int b = 0; b < N; b++)
for (int c = 0; c < N; c++)
for (int d = 0; d < N; d++)
{
cout << numbers[a] << " " << numbers[b] << " " << numbers[c] << " " << numbers[d] << endl;
}
return 0;
}
But how can I make algorithm so I can enter N and M via std::cin and I get correct resut?
Thanks.
First one short tip: don't use "new" or C-style arrays in C++ when we have RAII and much faster data structures.
For the solution to your problem I would suggest making separate function with recursion. You said you know how to do it manually so the first step in making it into algorithm is to tear down you manual solution step by step. For this problem when you solve it by hand you basically start with array of all first numbers and then for last position you just loop through available numbers. Then you go to the second last position and again loop through available numbers just now with the difference that for every number there you must also repeat the last spot number loop. Here is the recursion. For every "n"th position you must loop through available numbers and for every call the same function for "n+1"th number.
Here is a simplified solution, leaving out the input handling and exact print to keep code shorter and more focused on the problem:
#include <vector>
#include <iostream>
void printCombinations(const std::vector<int>& numbers, unsigned size, std::vector<int>& line) {
for (unsigned i = 0; i < numbers.size(); i++) {
line.push_back(numbers[i]);
if (size <= 1) { // Condition that prevents infinite loop in recursion
for (const auto& j : line)
std::cout << j << ","; // Simplified print to keep code shorter
std::cout << std::endl;
line.erase(line.end() - 1);
} else {
printCombinations(numbers, size - 1, line); // Recursion happens here
line.erase(line.end() - 1);
}
}
}
int main() {
std::vector<int> numbers = {6, 11, 533};
unsigned size = 4;
std::vector<int> line;
printCombinations(numbers, size, line);
return 0;
}
If you have any questions feel free to ask.
Totally there is no need for recursion here. This is a typical job for dynamic programming. Just get the first solution right for n = 1 (1 slot is available) which means the answer is [[6],[11],[533]] and then move on one by one by relying on the one previously memoized solution.
Sorry that i am not fluent in C, yet in JS this is the solution. I hope it helps.
function combosOfN(a,n){
var res = {};
for(var i = 1; i <= n; i++) res[i] = res[i-1] ? res[i-1].reduce((r,e) => r.concat(a.map(n => e.concat(n))),[])
: a.map(e => [e]);
return res[n];
}
var arr = [6,11,533],
n = 4;
console.log(JSON.stringify(combosOfN(arr,n)));
Normally the easiest way to do dynamic nested for loops is to create your own stack and use recursion.
#include <iostream>
#include <vector>
void printCombinations(int sampleCount, const std::vector<int>& options, std::vector<int>& numbersToPrint) {
if (numbersToPrint.size() == sampleCount) {
// got all the numbers we need, print them.
for (int number : numbersToPrint) {
std::cout << number << " ";
}
std::cout << "\n";
}
else {
// Add a new number, iterate over all possibilities
numbersToPrint.push_back(0);
for (int number : options) {
numbersToPrint.back() = number;
printCombinations(sampleCount, options, numbersToPrint);
}
numbersToPrint.pop_back();
}
}
void printCombinations(int sampleCount, const std::vector<int>& options) {
std::vector<int> stack;
printCombinations(sampleCount, options, stack);
}
int main()
{
printCombinations(3, {1,2,3});
}
output
1 1 1
1 1 2
1 1 3
1 2 1
1 2 2
1 2 3
1 3 1
1 3 2
1 3 3
2 1 1
2 1 2
2 1 3
2 2 1
2 2 2
2 2 3
2 3 1
2 3 2
2 3 3
3 1 1
3 1 2
3 1 3
3 2 1
3 2 2
3 2 3
3 3 1
3 3 2
3 3 3
Here is an algorithm to solve this, that does't use recursion.
Let's say n=2 and m=3. Consider the following sequence that corresponds to these values:
000
001
010
011
100
101
110
111
The meaning of this is that when you see a 0 you take the first number, and when you see a 1 you take the second number. So given the input numbers [5, 7], then 000 = 555, 001=557, 010=575 etc.
The sequence above looks identical to representing numbers from 0 to 7 in base 2. Basically, if you go from 0 to 7 and represent the numbers in base 2, you have the sequence above.
If you take n=3, m=4 then you need to work in base 3:
0000
0001
0002
0010
0011
0012
....
So you go over all the numbers from 0 to 63 (4^3-1), represent them in base 3 and follow the coding: 0 = first number, 1 = second number, 2 = third number and 3 = fourth number.
For the general case, you go from 0 to M^N-1, represent each number in base N, and apply the coding 0 = first number, etc.
Here is some sample code:
#include <stdio.h>
#include <math.h>
void convert_to_base(int number, char result[], int base, int number_of_digits) {
for (int i = number_of_digits - 1; i >= 0; i--) {
int remainder = number % base;
number = number / base;
result[i] = '0' + remainder;
}
}
int main() {
int n = 2, m = 3;
int num = pow(n, m) - 1;
for (int i = 0; i <= num; i++) {
char str[33];
convert_to_base(i, str, n, m);
printf("%s\n", str);
}
return 0;
}
Output:
000
001
010
011
100
101
110
111
I have got this code:
int n,m; // n - rows count , m - columns in each row count
cin >> n >> m;
string wiersze[n];
int wynik=0;
for(int i = 0; i < n;i++)
{
cin >> wiersze[i];
}
for(int i = 0; i < n;i++)
{
for(int j =0; j<m-1;j++)
{
cout << j << " " << j+1<<endl; // debbuging line
cout<<wiersze[i].at(j)<<wiersze[i].at(j+1)<<endl; // debbuging line
if(wiersze[i].at(j) == wiersze[i].at(j+1))wynik++;
}
}
When there are two the letters in the string, I should increase variable wynik.
I do not know why, but running that code on this example:
4 4
aaba
aaab
aaba
bbaa
causes std::out_of_range error on string::at in row if(wiersze[i].at(j) == wiersze[i].at(j+1))wynik++;.
I tried debbuging and added two lines in loop to see what is happeing. Results:
0 1
aa
1 2 ab
2 3
ba
0 1
aa
1 2
aa
2 3
ab
0 1
aa
1 2
ab
2 3
ba
0 1
bb
1 2
ba
2 3
aa
Please look at two last lines - it shows exactly that what by computation next line does not exists! I, of course, tried to look for "stupid" mistakes and small debugging, but without any result exepct that, you can see above. Please help me in my case.
First of all you CAN NOT allocate memory for array as you did. In wiersze[n] variable n must be known at compile time, but here it is not. You have to initialize the array dynamically:
string* wiersze = new string[n];