i made about 90% from the problem but the issue is when there is no mode
that what i did , i if there is no mode to print -1 i tried a lot but always fail i tast case
this the function I wrote
void mode(int arr[], int size)
{
int max_count = 0; // for the most numbers occuer
cout << endl;
cout << "Mode is: ";
for (int i = 0; i < size; i++) {
int count = 1;
/*to count how many times num occuer to compare it with max count
it is inside the loop to start each time from 1*/
for (int j = i + 1; j < size; j++)
if (arr[i] == arr[j])
count++;
if (count > max_count)
max_count = count;
}
for (int i = 0; i < size; i++) {
int count = 1;
for (int j = i + 1; j < size; j++)
if (arr[i] == arr[j])
count++;
if (count == max_count)
cout << arr[i] << " ";
}
}
Related
I have this task:
A user inputs a number N and you have to output this pyramid:
0
101
21012
.......
N.21012.N
For N=5 it will be :
0
101
21012
3210123
432101234
54321012345
I managed to only get it working for N<10 with this code:
int n;
cin >> n;
for (int i = 0; i < n + 1; i++) {
for (int j = 0; j < n - i; j++)
cout << " ";
int dir = -1;
for (int k = i; k <= i; k += dir) {
cout << k;
if (k == 0)
dir = 1;
}
cout << endl;
}
For N=10 it will look like this :
0
101
21012
3210123
432101234
54321012345
6543210123456
765432101234567
87654321012345678
9876543210123456789
10987654321012345678910
After the answers I settled on this :
#include <iostream>
#include <cmath>
#include <string>
using namespace std;
int main()
{
int n, spaces;
string number;
cin >> n;
if (n < 10)
spaces = n;
else
{
spaces = 9;
int pwr = 0, k = n;
while (k > 9)
{
pwr++;
k /= 10;
}
for (int i = 1; i < pwr; i++)
{
spaces += pow(10, i) * 9 * (i + 1);
}
spaces += (n - pow(10, pwr) + 1) * (pwr + 1);
}
// cout << spaces << endl;
for (int i = 0; i < n + 1; i++)
{
for (int j = i; j > -1; j--)
number += to_string(j);
int len = number.length() - 1;
for (int j = 0; j < spaces - len; j++)
cout << " ";
for (int j = 1; j <= i; j++)
number += to_string(j);
cout << number << endl;
number.clear();
}
cout << endl;
return 0;
}
int padding(int n) {
constexpr auto singleDigitNumbersCount = 9;
constexpr auto doubleDigitNumbersCount = 90; // from 10 to 99
if (n < 10) return n;
if (n < 100) return 2*n - singleDigitNumbersCount;
return 3*n - doubleDigitNumbersCount - 2*singleDigitNumbersCount;
}
int main() {
int n;
cin >> n;
for (int i = 0; i < n + 1; i++) {
std::cout << std::string(padding(n) - padding(i), ' ');
for (int k = i; k >= 0; k--) {
cout << k;
}
for (int k = 1; k <= i; k++) {
cout << k;
}
cout << '\n';
}
return 0;
}
https://godbolt.org/z/EEaeWEvf4
I made this a bit ago Compiler Explorer
Not sure if that'd help 🤔
Here is the working code:
#include <string>
#include <iostream>
using namespace std;
#define MAX_SPACE 50
int main()
{
int n;
cin >> n;
string output = "";
for (int i = 0; i < n + 1; i++)
{
for (int k = i; k >= 0; k--) {
output += to_string(k);
}
for (int k = 1; k <= i; k++) {
output += to_string(k);
}
for (uint8_t i = 0, max = MAX_SPACE - output.length() / 2.00; i < max; i++) // Print max spaces minus the integer length divided by 2
{
cout << " ";
}
cout << output << endl; // Print number
output = "";
}
return 0;
}
So I have array A and B, two of them contain random numbers and I need to write in the C array initially even numbers of A and B and then odd. I have made this wtih vector but I wonder if there is other way to do it like in Javascript there are methods like .unshift(), .push() etc
#include<iostream>
#include<vector>
using namespace std;
int main() {
const int n = 4;
int A[n];
int B[n];
vector<int>C;
for (int i = 0; i < n; i++)
{
A[i] = rand() % 10;
cout << A[i] << " ";
}
cout << endl;
for (int i = 0; i < n; i++)
{
B[i] = rand() % 30;
cout << B[i] << " ";
}
for (int i = 0; i < n; i += 1)
{
if (A[i] % 2 == 0)
{
C.push_back(A[i]);
}
if (B[i] % 2 == 0)
{
C.push_back(B[i]);
}
}
for (int i = 0; i < n; i++)
{
if (A[i] % 2 != 0)
{
C.push_back(A[i]);
}
if (B[i] % 2 != 0)
{
C.push_back(B[i]);
}
}
cout << endl;
for (int i = 0; i < C.size(); i++)
cout << C[i] << " ";
}
I would suggest interleaving A and B initially:
for (int i = 0; i < n; i += 1)
{
C.push_back(A[i]);
C.push_back(B[i]);
}
And then partitioning C into even and odd elements:
std::stable_partition(C.begin(), C.end(), [](int i) { return i % 2 == 0; });
vector::push_back is the simplest way to have a collection that grows as you add things to the end.
Since you have fixed size for A and B, you could make them primitive arrays instead, which is what you have done. But for C you don't know how long it will be, so a collection that has a changeable size is appropriate.
You can use std::array, if you know the size you need in compile time. You can then add using an iterator.
#include<vector>
using namespace std;
int main() {
const int n = 4;
int A[n];
int B[n];
std::array<int, n+n>C; // <-- here
auto C_it = C.begin(); // <-- here
for (int i = 0; i < n; i++)
{
A[i] = rand() % 10;
cout << A[i] << " ";
}
cout << endl;
for (int i = 0; i < n; i++)
{
B[i] = rand() % 30;
cout << B[i] << " ";
}
for (int i = 0; i < n; i += 1)
{
if (A[i] % 2 == 0)
{
*C_it++ = A[i]; // <-- here
}
if (B[i] % 2 == 0)
{
*C_it++ = B[i];
}
}
for (int i = 0; i < n; i++)
{
if (A[i] % 2 != 0)
{
*C_it++ = A[i];
}
if (B[i] % 2 != 0)
{
*C_it++ = B[i];
}
}
cout << endl;
for (int i = 0; i < C.size(); i++)
cout << C[i] << " ";
}
Alternatively if you want to be more safe you can hold the next unwritten index and access elements with C.at(last++) = A[i], which checks for out-of-bounds and throws an exception instead of UB.
well you don't to change much.
first of declare C array as int C[n+n]; and declare a variable for incrementing through c array as int j=0;
and in if statements of loops do this C[j]=A[i]; j++; for first if and C[j]=B[i]; j++; for the second if statements
int main() {
const int n = 4;
int A[n];
int B[n];
int C[n+n];
int j=0;
for (int i = 0; i < n; i++)
{
A[i] = rand() % 10;
cout << A[i] << " ";
}
cout << endl;
for (int i = 0; i < n; i++)
{
B[i] = rand() % 30;
cout << B[i] << " ";
}
for (int i = 0; i < n; i += 1)
{
if (A[i] % 2 == 0)
{
C[j]=A[i];
j++;
}
if(B[i]%2==0){
C[j]=B[i];
j++;
}
}
for (int i = 0; i < n; i++)
{
if (A[i] % 2 != 0)
{
C[j]=A[i];
j++;
}
if (B[i] % 2 != 0)
{
C[j]=B[i];
j++;
}
}
j=0;
cout << endl;
for (int i = 0; i < C[].lenght(); i++)
cout << C[i] << " ";
}
Can't use libraries and other methods.
As you can see my program finds the repeated numbers and print it but I need to print the numbers just once.
As example if entered:
7 1 1 2 1 2 2 9
It should print
1 2
In case there is no any repeated number:
7 1 2 3 4 5 6 7
There should not be any output!
Also note, that the first number is the length of array:
#include <iostream>
int main()
{
unsigned size;
std::cin >> size;
int* myArray = new int[size];
for (int i = 0; i < size; i++) {
std::cin >> myArray[i];
}
for (int i = 0; i < size; i++) {
bool found = false;
for (int j = 0; j < i && !found; j++) {
found = (myArray[i] == myArray[j]);
}
if (!found) {
std::cout << myArray[i] << " ";
}
}
delete []myArray;
}
The easiest approach would probably be to use a set, but I'm not sure if that's allowed under the "can't use other libraries" rule.
Using just arrays, for each item you could iterate over all the items before it, and only print it if it wasn't found there:
for (int i = 0; i < size; i++) {
bool found = false;
for (int j = 0; j < i && !found; j++) {
found = (myArray[i] == myArray[j]);
}
if (!found) {
cout << myArray[i] << " ";
}
}
The first occurrence of a repeated number has no occurrences before it and at least one after it.
This is reasonably easy to detect:
for (int i = 0; i < size; i++) {
bool before = false;
for (int j = 0; j < i && !before; j++) {
before = myArray[i] == myArray[j];
}
if (!before) {
bool after = false;
for (int j = i + 1; j < size && !after; j++) {
after = myArray[i] == myArray[j];
}
if (after)
{
cout << myArray[i] << " ";
}
}
}
Instead of two bools as the other user suggested I would use the counter basically doing the same thing but with one variable. The trick is to check if you have already had the number you are checking right now before so that you wouldn't print it again.
And then to check the rest of the list for duplicates.
#include <iostream>
int main()
{
unsigned size;
std::cin >> size;
int* myArray = new int[size];
for (int i = 0; i < size; i++) {
std::cin >> myArray[i];
}
for (int i = 0; i < size; i++) {
int count = 0;
//check if the number has already been found earlier
for (int j = 0; j < i && !count; j++) {
if(myArray[i] == myArray[j]) count++;
}
//check the rest of the array for the repeated number
if (!count) {
for (int j = i; j < size; j++) {
if(myArray[i] == myArray[j]) count++;
}
}
//print if repeated
if (count > 1) {
std::cout << myArray[i] << " ";
}
}
delete []myArray;
}
Getting error in this code, even though it passed the basic test cases. But still, it gives the wrong answer.
Cannot find the test case where it fails, Solution to Codechef Count of maximum problem. I think a part of the code is making it fail for a certain test case(s).
Can anyone help me find the error in this code, please?
#include <bits/stdc++.h>
using namespace std;
int main()
{
int k;
cin >> k;
for (int j = 0; j < k; j++)
{
int n;
cin >> n;
int a[n];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int maxCount = 0;
int number;
int index = 0;
for (int i = 0; i < n; i++)
{
int count = 0;
for (int l = 0; l < n; l++)
{
if (a[i] == a[l])
{
count++;
if (count > maxCount)
{
maxCount = count;
index = i;
}
if (count == maxCount) {
(a[i] > a[index]) ? number = a[index] : number = a[i];
}
}
}
}
cout << number << " " << maxCount << endl;
}
}
Your number variable is redundant. You need to track theindex of the elements in the array.
That means, change this line
(a[i] > a[index]) ? number = a[index] : number = a[i];
to
(a[i] > a[index]) ? index = index : index = i;
and print
std::cout << a[index] << " " << maxCount << std::endl;
How can I write this same code with the while loop instead of the for loop?
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
for (int j = n; j >= i; j--) {
cout << j;
}
cout << endl;
}
This is my attempt, but it does not achieve the same affect. I'm not sure why.
int n;
cin >> n;
int i = 1;
int j = n;
while (i <= n) {
while (j >= i) {
cout << j;
j--;
}
i++;
cout << endl;
}
You have to reset j before the while(j >= i) loop.
while (i <= n) {
j = n; //<<<<<<<< Reset j to the starting value
while (j >= i) {
cout << j;
j--;
}
i++;
cout << endl;
}