I am trying to write a function that inputs an array_string and finds unique words, then copies them over to unique_array. This is a bit over my head I feel like and need some advice on where to go from here..
void unique140(string str_array[], int array_size, string unique_array[], int&unique_size)
{
int i = 0;
int j = 0;
for(i = 0; i < array_size; i++)
{
for(j = 0; j < unique_size; j++)
if (str_array[i] != unique_array[j])
{unique_array[i];}
if(str_array[i] == unique_array[j]
break;
}
Try something like:
void unique140(string str_array[], int array_size,
string unique_array[], int&unique_size)
{
unordered_map<string,int> count;
for(int i=0; i<array_size; i++) {
count[str_array[i]]++;
}
unique_size = 0;
for(auto it=count.begin(); it!=count.end(); ++it) {
if(it->second==1) {
unique_array[unique_size++] = it->first;
}
}
}
use a std::map if you dont have a std::unordered_map.
You're not a million miles away, the loops look good. Now you have to use those loops to make a decision, is the word str_array[i] present in the unique_array or not? Now just a minutes thought will tell you that str_array[i] is present if any one of the equality tests str_array[i] == unique_array[j] is true. Here's how to code that
for(i = 0; i < array_size; i++)
{
bool in_unique_array = false;
for(j = 0; j < unique_size; j++)
if (str_array[i] == unique_array[j])
in_unique_array = true;
...
}
Next we have to add str_array[i] to unique_array if it's not already there. So that's
for(i = 0; i < array_size; i++)
{
bool in_unique_array = false;
for(j = 0; j < unique_size; j++)
if (str_array[i] == unique_array[j])
in_unique_array = true;
if (!in_unique_array)
{
...
}
}
Finally having decided to add the string to unique_array we have to do that
for(i = 0; i < array_size; i++)
{
bool in_unique_array = false;
for(j = 0; j < unique_size; j++)
if (str_array[i] == unique_array[j])
in_unique_array = true;
if (!in_unique_array)
{
unique_array[unique_size] = str_array[i];
unique_size++;
}
}
Related
#include<iostream>
using namespace std;
int main()
{
int arr[] = {40,20,14,20,55,14,22,45,22,447,441,224,421,2,14,1,9};
int size = sizeof(arr) / sizeof(int);
for (int i = 0; i < size; i++)
{
cout<<arr[i]<<" ";
}
cout<<endl;
for (int i = 0; i < size; i++)
{
int Index_of_Min = i;
for (int j = i+1; j < size; j++)
{
if (arr[j] < arr[Index_of_Min])
{
Index_of_Min = j;
}
swap(arr[Index_of_Min], arr[i] );
}
}
for (int i = 0; i < size; i++)
{
cout<<arr[i]<<" ";
}
cout<<endl;
return 0;
}
in the above program when we run it the number 9, 224 are not getting sorted
i want to understand why the program is not working anyone who knows the solution then your help is much appreciated
and please explain me what was my mistake so i will not repeat it again.
Thank You
Saw the exact same error only a few days ago. Your loop is wrong, specifically the swap is in the wrong place. This is how it should look
for (int i = 0; i < size; i++)
{
int Index_of_Min = i;
for (int j = i+1; j < size; j++)
{
if (arr[j] < arr[Index_of_Min])
{
Index_of_Min = j;
}
}
swap(arr[Index_of_Min], arr[i] );
}
You do the swap after the inner loop has found the index of the minimum element, not while it is finding that index.
#include<bits/stdc++.h>
using namespace std;
int
main ()
{
int a[3] = { 3, 2, 1 }, i, j, t = 0, k = 0;
int m;
for (int i = 0; i < 2; i++)
{
m = a[i];
for (int j = i++; j < 3; j++)
{
if (m > a[j])
{
m = a[j];
k = j;
}
}
t = a[i];
a[i] = m;
a[k] = t;
}
for (int i = 0; i < 3; i++)
{
cout << a[i];
}
return 0;
}
You seem to want to implement an insertion sort algorithm by hand. For this task, you should consider using std::sort. I refactored your code such that it basically does the same thing as you wanted and included some tips to make the code more readable and easier to debug for you and others:
#include <algorithm> //only include necessary headers
#include <iostream>
//do not use "using namespace std"
int main ()
{
int a[3] = { 3, 2, 1 };
/*
Declare local variables at the first point they are used an as local as possible.
It is much easier to read, if a no longer necessary variables leaves scope.
*/
for (int i = 0; i < 2; i++) {
int min = a[i]; //use names that tell something about what you are doing
int bestIndex = i;
for (int j = i+1; j<3; j++) { //here was your major bug: i++ also increcments i
if (a[j] < min){
min = a[j];
bestIndex = j;
}
}
std::swap(a[i], a[bestIndex]);
//try to use standard algorithms as often as possible. They document your code, are optimized and easier to read.
}
for (int i = 0; i < 3; i++){
std::cout << a[i];
}
return 0;
}
Look at this:
for (int i = 0; i < 2; i++)
{
for (int j = i++; j < 3; j++)
{
}
}
On the first iteration of the outer loop:
i is 0.
Then the inner loop increments it to 1.
Then the outer loop increments it to 2.
Then the outer loop is done, after just one iteration.
You want i+1 in the inner loop, not i++.
#include<bits/stdc++.h>
using namespace std;
int main ()
{
int a[3] = { 3,2,1}, t = 0, k = 0;
int m=0;
bool flag_change =false;
for (int i = 0; i < 2; i++)
{
m = a[i];
for (int j = i+1; j < 3; j++)
{
if (m < a[j]) //changing sing (<>) sort by ascending/sort by descending order
{
m = a[j];
k = j;
flag_change = true;
}
}
if(flag_change)
{
t = a[i];
a[i] = m;
a[k] = t;
flag_change = false;
}
}
for (int i = 0; i < 3; i++)
{
cout << a[i];
}
return 0;
}
I'm done correct you code. Code is working.But it is not good implementation.
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;
}
I was trying to understand bubble sort. While implementing my own version of it for practicing, I implemented it in such a way:
public int[] sort(int[] nums) {
int x = 0;
for (int i = 0; i < nums.length; i++) {
for (int j = 0; j < i; j++) {
System.out.println(++x);
if (nums[j] > nums[i]) {
int temp = nums[j];
nums[j] = nums[i];
nums[i] = temp;
}
}
}
return nums;
}
Notice how the inner loop goes from 0 to i.
When I googled this algorithm to find the conventional implementation, I found this implementation:
public int[] sortnew(int[] nums) {
int x = 0;
for (int i = 0; i < nums.length - 1; i++) {
for (int j = 0; j < nums.length - i - 1; j++) {
System.out.println(++x);
if (nums[j] > nums[j+1]) {
int temp = nums[j];
nums[j] = nums[j+1];
nums[j+1] = temp;
}
}
}
return nums;
}
Here i goes from 0 to array length - i - 1.
In both the cases the number of prints of 'x' are the same. This I believe indicates that the number of computations are same. Then what exactly is the difference between the two practically?
Thank you!
Say that I have a sequence:
int seq[4][4];
Then, lets say seq[1][2]=8;
No other values of the sequence yields 8.
If I want to find the values of a sequence and print out which one it is, (e.g. 1,2 and make x=1 and y=2) how can I do that? What
int x,j;
for (int i = 0; i < 4; i++) // looping through row
{
for(int j = 0; j < 4; j++) //looping through column
{
if (seq[i][j] == 8) //if value matches
{
x = i; y = j; //set value
i = 4; //set i to 4 to exit outer for loop
break; //exit inner for loop
}
}
}
int numberBeingSearchedFor = *Any Value Here*;
int array[*numRows*][*numColumns*];
int firstOccuranceRow = -1, firstOccuranceColumn = -1;
for(int i = 0; i < numRows; ++i)
{
for(int j = 0; j < numColumns; ++j)
{
if(array[i][j] == numberBeingSearchedFor)
{
firstOccuranceRow = i;
firstOccuranceColumn = j;
i = numRows; //Credit to other answer, I've never seen that :) It's cool
break;
}
}
}
if(firstOccuranceRow == -1 || firstOccuranceColumn == -1)
{
//Item was not in the array
}