Adding two polynomials in C, Abort trap: 6 - c
I am trying to add 2 polynomials using int arrays. For example,
co1[ ] = {5, 3, 2, 0, 0, 0, ...} & ex1[ ] = {6, 2, 1, 0, 0, 0, ...} represents the polynomial 5x^6 + 3x^2 + 2x. I tried doing this in different ways but I keep getting the message "Abort trap: 6" whenever I try to use this function. Any help would be appreciated. Thank you.
void add_polynom( int co1[ ], int ex1[ ], int co2[ ], int ex2[ ] )
{
int tmpC[ASIZE]; //ASIZE = 50
int tmpE[ASIZE];
int i, j, k, b, x = 0;
init_polynom(tmpC, tmpE); //sets all values within tmpC and tmpE to 0
for(i=0;i<ASIZE;i++)
{
for(j=0;j<ASIZE;j++)
{
if(ex1[i] > ex2[j])
{
tmpC[x] = co1[i];
tmpE[x] = ex1[i];
x++;
}
else if(ex1[i] == ex2[j])
{
if((ex1[i] == 0) && (ex2[j] == 0))
{
if(k == 0)
{
tmpC[x] = co1[i] + co2[j];
tmpE[x] = 0;
x++;
k = 1;
}
else
{
goto jump;
}
}
else
{
tmpC[x] = co1[i] + co2[j];
tmpE[x] = ex1[i];
x++;
}
}
else if(ex1[i] < ex2[j])
{
tmpC[x] = co2[j];
tmpE[x] = ex2[j];
x++;
}
}
}
jump:
for(b=0;b<ASIZE;b++)
{
co1[b] = tmpC[b];
ex1[b] = tmpE[b];
}
}
try to give k a value before entering the loop
In your inner loop you're incrementing x beyond ASIZE yet you use it to index into arrays you created with a length of ASIZE.
Related
how can i find the second smallest value in array in dart?
I have an assignment that needs me to find the smallest and second smallest value in an array but I can't manage to figure it out. !New learner in dart
You can use Quickselect for Θ(n) time complexity:
int findKthSmallest(List<int> nums, int k) {
if (k <= 0 || k > nums.length) {
throw new ArgumentError("k must be positive and less than or equal to the size of nums.");
}
final n = nums.length - k;
int quickSelect(int left, int right) {
var pivot = nums[right];
var p = left;
for (var i = left; i < right; i++) {
if (nums[i] >= pivot) {
var temp = nums[i];
nums[i] = nums[p];
nums[p] = temp;
p++;
}
}
var temp = nums[right];
nums[right] = nums[p];
nums[p] = temp;
if (p > n) {
return quickSelect(left, p - 1);
} else if (p < n) {
return quickSelect(p + 1, right);
} else {
return nums[p];
}
}
return quickSelect(0, nums.length - 1);
}
void main() {
var nums = [5, 3, 1, 3, 2, 7, 9, 10];
var k = 2;
print('findKthSmallest(${nums}, ${k}): ${findKthSmallest([...nums], k)}');
}
Output:
findKthSmallest([5, 3, 1, 3, 2, 7, 9, 10], 2): 2
Find three or more consecutive characters in the same struct in C
I define a struct.
struct test
{
char name[10];
int num;
}temp[20];
I want to only output the data with the same name and three consecutive num or more in temp[20].
for example,the data of the temp[20] are as follows:
temp[0] = { "a",0 };
temp[1] = { "b",1 };
temp[2] = { "a",2 };
temp[3] = { "b",2 };
temp[4] = { "a",3 };
temp[5] = { "b",3 };
temp[6] = { "a",4 };
temp[7] = { "c",1 };
temp[8] = { "c",2 };
temp[9] = { "a",5 };
........
the output is
temp[2] = { "a",2 };
temp[4] = { "a",3 };
temp[6] = { "a",4 };
temp[9] = { "a",5 };
temp[1] = { "b",1 };
temp[3] = { "b",2 };
temp[5] = { "b",3 };
I can only use a triple for loop to output data with the same name and three consecutive num,but four consecutive num can`t output.My code are as follows:
for (int i = 0; i < 20; i++)
{
for (int j = i + 1;j < 20;j++) {
if ((strcmp(temp[i].name, temp[j].name) == 0) && (temp[j].num - temp[i].num == 1))
{
for (int k = 0; k < 20; k++)
{
if ((strcmp(temp[k].name, temp[j].name) == 0) && (temp[k].num - temp[j].num == 1)) {
printf("%s,%d", temp[i].name, temp[i].num);
printf("%s,%d", temp[j].name, temp[j].num);
printf("%s,%d", temp[k].name, temp[k].num);
}
}
}
}
}
what could I do?
Since you haven't specified anything particular about the data's ordering, I assume that it's fine to sort it. Sorting by the name, then number allows you to find consecutive runs of values in linear time. And a typically sorting algorithm can be achieved in N.log(N) time.
Let's assume that N represents the number of elements in your array. You can sort it simply like this:
std::sort(temp, temp + N,
[](const test& a, const test& b) {
int result = strcmp(a.name, b.name);
return result < 0 || result == 0 && a.num < b.num;
});
Using a couple of simple helpers to display the data...
void Show(const test& t)
{
std::cout << t.name << "," << t.num << "\n";
}
void Show(const test* t, int N)
{
for (int i = 0; i < N; ++i)
Show(t[i]);
}
Now, after sorting, a call to Show(temp, N); gives this:
a,0
a,2
a,3
a,4
a,5
b,1
b,2
b,3
c,1
c,2
Now it's quite trivial to walk through the array and find runs of some minimum number. You requested 3, but we can choose any number.
const int minRunLength = 3;
std::cout << "Consecutive runs with matching name (minimum " << minRunLength << "):\n";
for (int i = 0, runLength = 0; i < N; ++i)
{
// Update run length
if (i > 0 &&
strcmp(temp[i-1].name, temp[i].name) == 0 &&
temp[i-1].num + 1 == temp[i].num)
{
++runLength;
}
else
{
runLength = 1;
}
// Show items meeting run length criteria
if (runLength == minRunLength)
{
for (int p = runLength-1; p >= 0; --p)
Show(temp[i - p]);
}
else if (runLength > minRunLength)
{
Show(temp[i]);
}
}
Running this with a minimum length of 3 gives:
Consecutive runs with matching name (minimum 3):
a,2
a,3
a,4
a,5
b,1
b,2
b,3
A minimum of 4 is:
Consecutive runs with matching name (minimum 4):
a,2
a,3
a,4
a,5
Here is a Live Demo that you can play with.
Wave Algorithm (Lee's Algorithm): incorrect final matrix
I'm writing a program calculating the shortest way from point A to point B.
I have a map (matrix) with values:
0 is block (wall, no way to pass);
1 is free way (you can pass);
2 is start point;
In the code below I declare 2 arrays: an array " map"and changed array "visited" while running program demonstrating visited points.
I check the cells in 4 directions (not diagonals) for 1 or 0. If it's 1 (possible to pass), I increase the counter for 1. For do not count the previous cell I'm trying to avoid it by the condition. I realized that in two one-dimensional arrays {1 0 -1 0} and { 0, 1, 0, -1 } to check neighbor points (what mean i check [i+1][j], [i-1][j], [i][j+1] and [i][j-1]).
As a result I wanna see "visited" matrix with a few lines which shows the way to reach to the point B (1, 2, 3, ... 15). I wanna find the way to map[7][7] point.
Right now the error here that I do count++ for the previous position. How to avoid that?
Thank you.
P.S. I wrote a few functions implementing a new array with 0 values, counting free to go cells and printing arrays.
main.c:
#include <stdbool.h>
#include <stdlib.h>
#include <stdio.h>
#define WIDTH 8
#define HEIGHT 8
int mapZero(int map[WIDTH][HEIGHT]);
int mapPrint(int map[WIDTH][HEIGHT]);
int mapInit(int map[WIDTH][WIDTH]);
int findFreeToGoCells(int map[WIDTH][HEIGHT]);
int main(int argc, char * argv[])
{
bool stop;
unsigned int count;
unsigned int max;
int visited[WIDTH][HEIGHT];
int map[WIDTH][HEIGHT] =
{
{ 1, 1, 1, 1, 1, 0, 0, 1 },
{ 0, 1, 1, 1, 1, 1, 0, 1 },
{ 0, 0, 1, 0, 1, 1, 1, 0 },
{ 1, 0, 1, 1, 1, 0, 1, 1 },
{ 0, 0, 0, 1, 0, 0, 0, 1 },
{ 1, 0, 1, 1, 1, 0, 0, 1 },
{ 0, 0, 0, 0, 1, 0, 0, 1 },
{ 0, 1, 1, 1, 1, 1, 1, 1 },
};
mapZero(visited);
printf("Matrix of zeroed-visited cells:\n\n");
mapPrint(visited);
printf("Matrix of the map:\n\n");
mapPrint(map);
printf("Free to go cells: %d\n\n", findFreeToGoCells(map));
max = WIDTH * HEIGHT - 1;
visited[0][0] = map[0][0];
count = 0;
visited[0][0] = 0;
int di[4] = { 1, -1, 0, 0 };
int dj[4] = { 0, 0, 1, -1 };
//do
{
for (int i = 0; i < WIDTH; ++i)
{
for (int j = 0; j < HEIGHT; ++j)
{
if (visited[i][j] == count)
{
for (int k = 0; k < 4; ++k)
{
int i_check = i + di[k];
int j_check = j + dj[k];
if ((i_check >= 0 && i_check < WIDTH) && (j_check >= 0 && j_check < HEIGHT) && (map[i_check][j_check] != 0))
{
visited[i_check][j_check] = count + 1;
}
}
count++;
}
}
}
}// while (visited[7][7] == 0);
if (count > max + 99999)
printf("The way couldn't be found\n");
else
{
printf("Matrix of visited cells:\n\n");
mapPrint(visited);
printf("Free to go cells from [0][0] to [7][7]: %d\n", findFreeToGoCells(visited));
}
/*************************************************************************************/
/*************************************************************************************/
int len;
int x = 7;
int y = 7;
int x_path[WIDTH * HEIGHT];
int y_path[WIDTH * HEIGHT];
len = visited[7][7];
count = len;
while (count > 0)
{
x_path[count] = x;
y_path[count] = y;
count--;
for (int k = 0; k < 4; ++k)
{
int i_check = x + di[k];
int j_check = y + dj[k];
if ((i_check >= 0 && i_check < WIDTH) && (j_check >= 0 && j_check < HEIGHT) && (map[i_check][j_check] == count))
{
x = x + di[k];
y = y + dj[k];
break;
}
}
}
x_path[0] = 0;
y_path[0] = 0;
printf("\nThe shortest way consist of %d cells\nThere are %d the shortest way to reach th the final point\n\n", len, findFreeToGoCells(visited)-len);
system("pause");
return 0;
}
int mapZero(int map[WIDTH][HEIGHT])
{
for (int i = 0; i < WIDTH; ++i)
{
for (int j = 0; j < HEIGHT; ++j)
{
map[i][j] = 0;
}
}
return 0;
}
int mapPrint(int map[WIDTH][HEIGHT])
{
for (int i = 0; i < WIDTH; ++i)
{
for (int j = 0; j < HEIGHT; ++j)
{
printf("%2d ", map[i][j]);
}
printf("\n\n");
}
printf("\n");
return 0;
}
int findFreeToGoCells(int map[WIDTH][HEIGHT])
{
int count = 0;
for (int i = 0; i < WIDTH; ++i)
{
for (int j = 0; j < HEIGHT; ++j)
{
if (map[i][j] != 0) count++;
}
}
return count;
}
Result:
Cell Compete Problems
Here is my assignment:
There is a colony of 8 cells arranged in a straight line where each day every cell competes with its adjacent cells(neighbour). Each day, for each cell, if its neighbours are both active or both inactive, the cell becomes inactive the next day,. otherwise itbecomes active the next day.
Assumptions: The two cells on the ends have single adjacent cell, so
the other adjacent cell can be assumsed to be always inactive. Even
after updating the cell state. consider its pervious state for
updating the state of other cells. Update the cell informationof
allcells simultaneously.
Write a fuction cellCompete which takes takes one 8 element array of
integers cells representing the current state of 8 cells and one
integer days representing te number of days to simulate. An integer
value of 1 represents an active cell and value of 0 represents an
inactive cell.
Program:
int* cellCompete(int* cells,int days)
{
//write your code here
}
//function signature ends
Test Case 1:
INPUT:
[1,0,0,0,0,1,0,0],1
EXPECTED RETURN VALUE:
[0,1,0,0,1,0,1,0]
Test Case 2:
INPUT:
[1,1,1,0,1,1,1,1,],2
EXPECTED RETURN VALUE:
[0,0,0,0,0,1,1,0]
This is the problem statement given above for the problem. The code which I have written for this problem is given below. But the output is coming same as the input.
#include<iostream>
using namespace std;
// signature function to solve the problem
int *cells(int *cells,int days)
{ int previous=0;
for(int i=0;i<days;i++)
{
if(i==0)
{
if(cells[i+1]==0)
{
previous=cells[i];
cells[i]=0;
}
else
{
cells[i]=0;
}
if(i==days-1)
{
if(cells[days-2]==0)
{
previous=cells[days-1];
cells[days-1]=0;
}
else
{
cells[days-1]=1;
}
}
if(previous==cells[i+1])
{
previous=cells[i];
cells[i]=0;
}
else
{
previous=cells[i];
cells[i]=1;
}
}
}
return cells;
}
int main()
{
int array[]={1,0,0,0,0,1,0,0};
int *result=cells(array,8);
for(int i=0;i<8;i++)
cout<<result[i];
}
I am not able to get the error and I think my logic is wrong. Can we apply dynamic programming here If we can then how?
private List<Integer> finalStates = new ArrayList<>();
public static void main(String[] args) {
// int arr[] = { 1, 0, 0, 0, 0, 1, 0, 0 };
// int days = 1;
EightHousePuzzle eightHousePuzzle = new EightHousePuzzle();
int arr[] = { 1, 1, 1, 0, 1, 1, 1, 1 };
int days = 2;
eightHousePuzzle.cellCompete(arr, days);
}
public List<Integer> cellCompete(int[] states, int days) {
List<Integer> currentCellStates = Arrays.stream(states).boxed().collect(Collectors.toList());
return getCellStateAfterNDays(currentCellStates, days);
}
private List<Integer> getCellStateAfterNDays(List<Integer> currentCellStates, int days) {
List<Integer> changedCellStates = new ArrayList<>();
int stateUnoccupied = 0;
if (days != 0) {
for (int i1 = 0; i1 < currentCellStates.size(); i1++) {
if (i1 == 0) {
changedCellStates.add(calculateCellState(stateUnoccupied, currentCellStates.get(i1 + 1)));
} else if (i1 == 7) {
changedCellStates.add(calculateCellState(currentCellStates.get(i1 - 1), stateUnoccupied));
} else {
changedCellStates
.add(calculateCellState(currentCellStates.get(i1 - 1), currentCellStates.get(i1 + 1)));
}
}
if (days == 1) {
System.out.println("days ==1 hit");
finalStates = new ArrayList<>(changedCellStates);
return finalStates;
}
days = days - 1;
System.out.println("Starting recurssion");
getCellStateAfterNDays(changedCellStates, days);
}
return finalStates;
}
private int calculateCellState(int previousLeft, int previousRight) {
if ((previousLeft == 0 && previousRight == 0) || (previousLeft == 1 && previousRight == 1)) {
// the state gets now changed to 0
return 0;
}
// the state gets now changed to 0
return 1;
}
Here is my solution in Java:
public class Colony
{
public static int[] cellCompete(int[] cells, int days)
{
int oldCell[]=new int[cells.length];
for (Integer i = 0; i < cells.length ; i++ ){
oldCell[i] = cells[i];
}
for (Integer k = 0; k < days ; k++ ){
for (Integer j = 1; j < oldCell.length - 1 ; j++ ){
if ((oldCell[j-1] == 1 && oldCell[j+1] == 1) || (oldCell[j-1] == 0 && oldCell[j+1] == 0)){
cells[j] = 0;
} else{
cells[j] = 1;
}
}
if (oldCell[1] == 0){
cells[0] = 0;
} else{
cells[0] = 1;
}
if (oldCell[6] == 0){
cells[7] = 0;
} else{
cells[7] = 1;
}
for (Integer i = 0; i < cells.length ; i++ ){
oldCell[i] = cells[i];
}
}
return cells;
}
}
Your program does not distinguish between the number of days to simulate and the number of cells.
#include <bits/stdc++.h>
using namespace std;
int* cellCompete(int* cells,int days)
{
for(int j=0; j<days; j++)
{
int copy_cells[10];
for(int i=1; i<9; i++)
copy_cells[i]=cells[i-1];
copy_cells[0]=0;copy_cells[9]=0;
for(int i=0; i<8; i++)
cells[i]=copy_cells[i]==copy_cells[i+2]?0:1;
}
return cells;
}
int main()
{
int arr[8]={1,1,1,0,1,1,1,1};
int arr2[8]={1,0,0,0,0,1,0,0};
cellCompete(arr2,1);
for(int i=0; i<8; i++)
{
cout<<arr2[i]<<" ";
}
}
Here's some sweet little python code: def cell(arr, days): new = arr[:] #get a copy of the array n = len(arr) if n == 1: print [0] #when only 1 node, return [0] for _ in range(days): new[0] = arr[1] #determine the edge nodes first new[n - 1] = arr[n - 2] for i in range(1, n-1): new[i] = 1 - (arr[i-1] == arr[i+1]) #logic for the rest nodes arr = new[:] #update the list for the next day return new arr = [1, 1, 1, 0, 1, 1, 1, 1] days = 2 print cell(arr, days)
You can easily do this in Javascript with few lines of code let cells = [1,1,1,0,1,1,1,1]; let numOfDays = 2; let changeState = (cellarr)=> cellarr.map((cur, idx, arr)=> (arr[idx-1] ||0) + (arr[idx+1] || 0)===1?1:0); let newCells =cells; for (let i = 0 ; i <numOfDays; i++) newCells = changeState(newCells); console.log(newCells);
This is a C# version of a possible answer. I really struggled with this for a while for some reason!
I also incorporated some of Janardan's stuff above as it helped spur me in the right direction. (cheers!)
The tricky part of the question was dealing with the fact that you had to persist the state of the cell to figure out the next cell competition which I had originally tried with a second array which was messy.
Note: I chose to use the Array.Copy method as I believe it is slightly more efficient and a lot more readable than copying arrays with a for loop when reading through.
Hopefully this helps someone out in the future!
public int[] cellCompete(int[] cell, int day)
{
//First create an array with an extra 2 cells (these represent the empty cells on either end)
int[] inputArray = new int[cell.Length + 2];
//Copy the cell array into the new input array leaving the value of the first and last indexes as zero (empty cells)
Array.Copy(cell, 0, inputArray, 1, cell.Length);
//This is cool I stole this from the guy above! (cheers mate), this decrements the day count while checking that we are still above zero.
while (day-- > 0)
{
int oldCellValue = 0;
//In this section we loop through the array starting at the first real cell and going to the last real cell
//(we are not including the empty cells at the ends which are always inactive/0)
for (int i = 1; i < inputArray.Length - 1; i++)
{
//if the cells below and above our current index are the same == then the target cell will be inactive/0
//otherwise if they are different then the target cell will be set to active/1
//NOTE: before we change the index value to active/inactive state we are saving the cells oldvalue to a variable so that
//we can use that to do the next "cell competition" comparison (this fulfills the requirement to update the values at the same time)
if (oldCellValue == inputArray[i + 1])
{
oldCellValue = inputArray[i];
inputArray[i] = 0;
}
else
{
oldCellValue = inputArray[i];
inputArray[i] = 1;
}
}
}
//Finally we create a new output array that doesn't include the empty cells on each end
//copy the input array to the output array and Bob's yer uncle ;)...(comments are lies)
int[] outputArray = new int[cell.Length];
Array.Copy(inputArray, 1, outputArray, 0, outputArray.Length);
return outputArray;
}
With C#
public static int[] cellCompete(int[] states, int days)
{
if (days == 0) return states;
int leftValue = 0;
int rigthValue = 0;
for (int i = 0; i < states.Length; i++)
{
if (i == states.Length - 1)
rigthValue = 0;
else
rigthValue = states[i + 1];
if (leftValue == rigthValue){
leftValue = states[i];
states[i] = 0;
}
else{
leftValue = states[i];
states[i] = 1;
}
}
cellCompete(states, days - 1);
return states;
}
I think some of the answers above could be more readable (in addition to being more efficient). Use an additional array and alternate updates between them depending on the number of days. You can return the most recently updated array, which will always be the correct one. Like this:
function cellCompete(states, days) {
const newStates = [];
let originalStates = true;
while (days--) {
changeStates(
originalStates ? states : newStates,
originalStates ? newStates : states,
states.length
);
originalStates = !originalStates;
}
return originalStates ? states : newStates;
}
function changeStates(states, newStates, len) {
newStates[0] = !states[1] ? 0 : 1;
newStates[len-1] = !states[len-2] ? 0 : 1;
for (let i = 1; i < len - 1; i++) {
newStates[i] = states[i-1] === states[i+1] ? 0 : 1;
}
}
Here is my solution in c++ using bitwise operators :
#include <iostream>
using namespace std;
void cellCompete( int *arr, int days )
{
int num = 0;
for( int i = 0; i < 8; i++ )
{
num = ( num << 1 ) | arr[i];
}
for( int i = 0; i < days; i++ )
{
num = num << 1;
num = ( ( ( num << 1 ) ^ ( num >> 1 ) ) >> 1 ) & 0xFF;
}
for( int i = 0; i < 8; i++ )
{
arr[i] = ( num >> 7 - i ) & 0x01;
}
}
int main()
{
int arr[8] = { 1, 0, 0, 0, 0, 1, 0, 0};
cellCompete( arr, 1 );
for(int i = 0; i < 8; i++)
{
cout << arr[i] << " ";
}
}
#include <stdio.h>
int main() {
int days,ind,arr[8],outer;
for(ind=0;ind<8;scanf("%d ",&arr[ind]),ind++); //Reading the array
scanf("%d",&days);
int dupArr[8];
for(outer=0;outer<days;outer++){ //Number of days to simulate
for(ind=0;ind<8;ind++){ //Traverse the whole array
//cells on the ends have single adjacent cell, so the other adjacent cell can be assumsed to be always inactive
if(ind==0){
if(arr[ind+1]==0)
dupArr[ind]=0;
else
dupArr[ind]=1;
}
else if(ind==7){
if(arr[ind-1]==0)
dupArr[ind]=0;
else
dupArr[ind]=1;
}
else{
if((arr[ind-1]==0&&arr[ind+1]==0) || (arr[ind-1]==1&&arr[ind+1]==1)){// if its neighbours are both active or both inactive, the cell becomes inactive the next day
dupArr[ind]=0;
}
else //otherwise it becomes active the next day
dupArr[ind]=1;
}
}
for(ind=0;ind<8;ind++){
arr[ind]=dupArr[ind]; //Copying the altered array to original array, so that we can alter it n number of times.
}
}
for(ind=0;ind<8;ind++)
printf("%d ",arr[ind]);//Displaying output
return 0;
}
Here is my code which i had created some months ago,
You want to create two different arrays, because altering same array element will gives you different results.
func competeCell(cell []uint, days uint) []uint{
n := len(cell)
temp := make([]uint, n)
for i :=0; i < n; i ++ {
temp[i] = cell[i]
}
for days > 0 {
temp[0] = 0 ^ cell[1]
temp[n-1] = 0 ^ cell[n-2]
for i := 1; i < n-2 +1; i++ {
temp[i] = cell[i-1] ^ cell[i +1]
}
for i:=0; i < n; i++ {
cell[i] = temp[i]
}
days -= 1
}
return cell
}
Using c++
#include <list>
#include <iterator>
#include <vector>
using namespace std;
vector<int> cellCompete(int* states, int days)
{
vector<int> result1;
int size=8;
int list[size];
int counter=1;
int i=0;
int temp;
for(int i=0;i<days;i++)//computes upto days
{
vector<int> result;
if(states[counter]==0)
{
temp=0;
list[i]=temp;
//states[i]=0;
result.push_back(temp);
}
else
{
temp=1;
list[i]=temp;
result.push_back(temp);
}
for(int j=1;j<size;j++)
{
if(j==size)
{
if(states[j-1]==0)
{
temp=0;
list[j]=temp;
//states[i]=1;
result.push_back(temp);
}
else
{
temp=1;
list[i]=temp;
//states[i]=1;
result.push_back(temp);
}
}
else if(states[j-1]==states[j+1])
{
temp=0;
list[j]=temp;
//states[i]=1;
result.push_back(temp);
}
else
{
temp=1;
list[j]=temp;
//states[i]=1;
result.push_back(temp);
}
}
result1=result;
for(int i=0;i<size;i++)
{
states[i]=list[i];
}
}
return result1;
}
Java solution
This is solution is Java, which will work any number of Cells and any number of days .
public class Solution
{
public List<Integer> cellCompete(int[] states, int days)
{
List<Integer> inputList = new ArrayList<Integer>();
List<Integer> finalList = new ArrayList<Integer>();
// Covert integer array as list
for (int i :states)
{
inputList.add(i);
}
// for loop for finding status after number of days.
for(int i=1; i<= days; i++)
{
if(i==1)
{
finalList = nextDayStatus(inputList);
}
else
{
finalList = nextDayStatus(finalList);
}
}
return finalList;
}
// find out status of next day, get return as list
public List<Integer> nextDayStatus(List<Integer> input)
{
List<Integer> output = new ArrayList<Integer>();
input.add(0,0);
input.add(0);
for(int i=0; i < input.size()-2; i++)
{
if (input.get(i) == input.get(i+2))
{
output.add(0);
}
else
{
output.add(1);
}
}
return output;
}
}
I know this has been answered, but I gave it a go in Java and am pretty sure it will work for any size states array along with number of days:
public class CellCompete {
public static List<Integer> cellCompete(int[] states, int days) {
List<Integer> resultList = new ArrayList<>();
int active = 1, inactive = 0;
int dayCount = 1;
// Execute for the given number of days
while (days > 0) {
int[] temp = new int[states.length];
System.out.print("Day " + dayCount + ": ");
// Iterate through the states array
for (int i = 0; i < states.length; i++) {
// Logic for first end cell
if (i == 0) {
temp[i] = states[i + 1] == active ? active : inactive;
resultList.add(temp[i]);
System.out.print(temp[i] + ", ");
}
// Logic for last end cell
if (i == states.length - 1) {
temp[i] = states[i - 1] == active ? active : inactive;
resultList.add(temp[i]);
System.out.println(temp[i]);
}
// Logic for the in between cells
if (i > 0 && i < states.length - 1) {
if ((states[i - 1] == active && states[i + 1] == active) || (states[i - 1] == inactive && states[i + 1] == inactive)) {
temp[i] = inactive;
} else {
temp[i] = active;
}
resultList.add(temp[i]);
System.out.print(temp[i] + ", ");
}
}
dayCount++;
days--;
// Reset the states array with the temp array
states = temp;
}
return resultList;
}
public static void main(String[] args) {
int[] states = {1, 1, 0, 1, 0, 1, 0, 0};
int days = 5;
// Total of 40
System.out.println(cellCompete(states, days) );
}
}
Where did the people who wanted optimized solutions go? def Solution(states, days): for i in range(days): for j in range(len(states)): if (j == 0): states[i] = states[1] elif (j == len(states)-1): states[i] = states[-2] else: states[i] = abs(states[i-1] - states[i+1]) return states
By definition, all the cells, including non-existent ones, are in fact booleans:
var cellUpdate = (cells, days) => {
let result = [];
// update states
for(let i = 0; i < cells.length; i++) result.push((!Boolean(cells[i-1]) === !Boolean(cells[i+1])) ? 0 : 1) ;
// repeat for each day
if (days > 1) result = cellUpdate(result, days - 1);
return result;
Here is the best python Solution value=input() n=int(input()) lst=[] for i in value: if "1"in i: lst.append(1) elif "0" in i: lst.append(0) for _ in range(n): store = [] for i in range(8): if i==0: store.append(lst[i+1]) elif i==7: store.append(lst[i-1]) elif lst[i-1]==lst[i+1]: store.append(0) else: store.append(1) lst=store print(store)
Scala solution:
def cellDayCompete(cells: Seq[Int]): Seq[Int] = {
val addEdges = 0 +: cells :+ 0
(addEdges.dropRight(2) zip addEdges.drop(2)).map {
case (left, right) =>
(left - right).abs
}
}
def cellCompete(cells: Seq[Int], days: Int): Seq[Int] = {
if (days == 0) {
cells
} else {
cellCompete(cellDayCompete(cells), days - 1)
}
}
A code run with the example above can be found at Scastie
Just answered this question today and here was my solution in python3 def cellCompete(states, days): for i in range(0, days): #this is where we will hold all the flipped states newStates = [] ''' Algo: if neigbors are the same, append 0 to newStates if they are different append 1 to newStates ''' for currState in range(len(states)): #left and right ptr's left = currState - 1 right = currState + 1 #if at beginning of states, left is automatically inactive if left < 0: if states[right] == 1: newStates.append(1) else: newStates.append(0) #if at end of states, right is automatically inactive elif right > 7: #we know there is always only 8 elems in the states list if states[left] == 1: newStates.append(1) else newStates.append(0) #check to see if neighbors are same or different elif states[left] != states[right]: newStates.append(1) else: newStates.append(0) #Set the states equal to the new flipped states and have it loop N times to get final output. states = newStates return states
def cellCompete(states, days): d = 0 l = len(states) while d < days: new_states = [0] * l for i in range(l): if i == 0 and states[i+1] == 0 or i == l - 1 and states[i-1] == 0: new_states[i] = 0 elif i == 0 and states[i+1] == 1 or i == l - 1 and states[i-1] == 1: new_states[i] = 1 elif states[i+1] == states[i-1]: new_states[i] = 0 else: new_states[i] = 1 states = new_states d = d + 1 return states
static int[] CellCompete(int[] states, int days)
{
int e = states.Length;
int[] newStates = new int[(e+2)];
newStates[0] = 0;
newStates[e+1] = 0;
Array.Copy(states, 0, newStates, 1, e);
for (int d = 0; d < days; d++)
{
states = Enumerable.Range(1, e).Select(x => newStates[x - 1] ^ newStates[x + 1]).ToArray();
newStates[0] = 0;
newStates[e + 1] = 0;
Array.Copy(states, 0, newStates, 1, e);
}
return states;
}
//Here is a working solution for this problem in C#
public class HousesinSeq
{
private string _result;
public string Result
{
get { return _result; }
}
public void HousesActivation(string houses, int noOfDays)
{
string[] housesArr = houses.Split(' ');
string[] resultArr = new string[housesArr.Length];
for (int i = 0; i < noOfDays; i++)
{
for (int j = 0; j < housesArr.Length; j++)
{
if (j == 0)
{
if (housesArr[j + 1] == "0")
{
resultArr[j] = "0";
}
else
{
resultArr[j] = "1";
}
}
else if (j == housesArr.Length - 1)
{
if (housesArr[j - 1] == "0")
{
resultArr[j] = "0";
}
else
{
resultArr[j] = "1";
}
}
else
{
if (housesArr[j + 1] == housesArr[j - 1])
{
resultArr[j] = "0";
}
else
{
resultArr[j] = "1";
}
}
}
resultArr.CopyTo(housesArr, 0);
}
foreach (var item in resultArr)
{
//Console.Write($"{item} ");
_result += item + " ";
}
_result = _result.Trim();
}
}
public class Colony {
public static int[] cellCompete(int[] cell, int day) {
int[] ar = new int[10];
for(int i=1; i<9; i++) {
ar[i] = cell[i-1];
}
while(day-- >0) {
int temp = 0;
for(int i=1; i<9; i++) {
if(Math.abs(temp-ar[i+1])==1) {
temp = ar[i];
ar[i] = 1;
}
else {
temp = ar[i];
ar[i] = 0;
}
}
}
return ar;
}
public static void main(String[] args) {
int[] cell = {1,0,1,1,0,1,0,1};
int day = 1;
cell = cellCompete(cell, day);
for(int i=1; i<9; i++) {
System.out.print(cell[i]+" ");
}
}
}
Error in a "for" loop, I think memory problems
I'm trying to create an algorithm that loops through the loop 'for' ten times, the ultimate goal is to print out a 50x50 matrix with the sum of the values greater than or equal to 49, 0, and 1 represented as '*', but for some strange reason the loop only does 3 iterations instead of 10. ..
#include <stdio.h>
void printsimulaciodeu(int matlefa[50][50]) {
int ii, jj;
for (ii = 0; ii < 50; ii++) {
for (jj = 0; jj < 50; jj++) {
if(matlefa[ii][jj] == 0) {
printf(" 0 ");
}
else if (matlefa[ii][jj] >= 49) {
printf(" %d ", matlefa[ii][jj]-48);
}
else {
printf(" * ");
}
}
printf("\n");
}
}
int main(){
int tabla[50][50]={
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0},
{0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
};
int its;
int i, j;
int disparos_pos[10] = {1, 5, 14, 22, 37, 48, 43, 40, 28, 19};
char disparos_dir[10]= {'i','i','d','i','d','d','i','d','d','i'};
int mataux[50][50];
for (i = 0; i < 50; i++) {
for (j = 0; j < 50; j++) {
mataux[i][j] = tabla[i][j];
}
}
int matsuma[50][50];
for (its = 0; its < 10; its++) {
int pos = disparos_pos[its];
int dir = disparos_dir[its];
i = 0;
j = pos;
int cont = 49;
while (i < 49) {
if (mataux[i+1][j] == 0) {
mataux[i+1][j] = cont;
++i;
}
else {
if (dir == 'i') {
if (mataux[i+1][j] != 0) {
if (mataux[i][j-1] == 0) {
mataux[i][j-1] = cont;
--j;
}
else {
dir = 'd';
}
}
}
if (dir == 'd') {
if (mataux[i+1][j] != 0) {
if (mataux[i][j+1] == 0) {
mataux[i][j+1] = cont;
++j;
}
else {
dir = 'i';
}
}
}
}
}
if (its == 0) {
int m, n;
for (m = 0; m < 50; m++) {
for (n = 0; n < 50; n++) {
matsuma[m][n] = mataux[m][n];
}
}
}
if (its > 0) {
int p, q;
for (p = 0; p < 50; p++) {
for (q = 0; q < 50; q++) {
if (mataux[p][q] == 49) {
if (matsuma[p][q] >= 49) {
matsuma[p][q] = matsuma[p][q]+1;
}
}
}
}
}
}
printsimulaciodeu(matsuma);
}
I thought that one possibility is that the memory for the process is complete and therefore not iterate correctly, is that possible?
please help, I need it to work well for college work.
After a bit of debugging, I figured out the actual problem: Your while loop will run for INFINITE time in 3rd iteration of for loop. Explaination The while loop runs OK until i=33 at i=33 (j = 21, dir = 'i'), the none of the assignment statements (mataux[blah][blah] = blah balh) are executed, as conditons turn out to be false. Ultimate result is that value of 'dir' is toggled between 'i' and 'd', and the loop will run for forever!! Details mataux[i+1][j] = mataux[34][21] = 1 != 0 ---> first IF skipped mataux[i][j-1] = mataux[33][20] = 49 != 0 ---> third IF in first ELSE of WHILE skipped ELSE --> dir = 'd' (100) mataux[i][j+1] = mataux[33][22] = 1 != 0 ---> third IF in last IF block skipped ELSE --> dir = "i" (105) Hope this helps!