#include<stdio.h>
int main()
{
int n ;
printf("Input the number of rows: ");
scanf("%d", &n);
for(int i = 1; i <= n ; i++)
{
for(int j = 1 ; j <= i ; j++)
{
if(j == i + 1)
{
break;
}
printf("%3d", j);
}
for(int j = i - 1 ; j > 0; j--)
{
if(j == 0)
{
break;
}
printf("%3d", j);
}
printf("\n");
}
}
Program session
Number of rows for this output n = 3
My output:
1
1 2 1
1 2 3 2 1
Preferred output:
1
1 2 1
1 2 3 2 1
This here is an exercise where I have to print a pyramid of numbers where the central number of the pyramid is the number of the row. I understand the logic but as you can see I have not been able to fulfill the task successfully. Any tips?
As pointed out by #WeatherWane, you need to add logic to add extra spaces. If you notice carefully, number of spaces on each line(excluding padding you add with %3d is equal to 3 * (n - i)). You can create a simple method like this to add spacing:
void addSpaces(int N, int currentIndex, int padding) {
for (int index = currentIndex; index < N; index++)
for (int spaces = 0; spaces < padding; spaces++)
printf(" ");
}
then you can call it from your first for loop like this:
for(int i = 1; i <= n ; i++)
{
addSpaces(n, i, 3);
for(int j = 1 ; j <= i ; j++)
{
// ...
I tested it and it seems to align it correctly:
c-posts : $ ./a.out
Input the number of rows: 5
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
Related
I am a beginner in programming, I studied all about C, I started to solve problems from hackerrank.com, there I faced a problem to print a pattern like given below
(the output of problem program):
4 4 4 4 4 4 4
4 3 3 3 3 3 4
4 3 2 2 2 3 4
4 3 2 1 2 3 4
4 3 2 2 2 3 4
4 3 3 3 3 3 4
4 4 4 4 4 4 4
the input will be an integer which will provide the data for the length of pattern square, here it is 4 in image,
I tried a lot to type a proper logic and I end up with this useless code bellow:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
int n;
scanf("%d", &n);
int array[n- 1 + n][n - 1 + n];
array[(n - 1 + n) / 2][(n - 1 + n) / 2] = 1;
int f[n];
for(int i = 0; i < n; i++) {
f[i] = i;
}
for(int i = 0; i < n - 1 + n; i++) {
for(int j = 0; j < n - 1 + n; j++) {
array[j][i] = n - f[j]; //top
array[i][j] = n - f[j]; //left
array[(2 * n - 1 - 1) - i][j] = n - f[i]; //bottem
array[j][(2 * n - 1 - 1) - i] = n - f[i]; //rigth
}
}
for(int i = 0; i < n - 1 + n; i++) {
for(int j = 0; j < n - 1 + n; j++) {
printf("%d ", array[i][j]);
}
printf("\n");
}
return 0;
}
my logic was to make all four borders correct in for loop which will end at center, but its not working, I want a new logic or to improve my logic, if you want to help me out then please give me the way to solve problem instead of giving me a direct code.
It is observable that the pattern consists of n stacked squares:
Square #0 is drawn with ns.
Square #1 is drawn with n-1s.
...
Square #n-1 is drawn with 1s.
Implementing the above:
void draw_pattern(const size_t n)
{
const size_t dim = 2*n-1;
int array[dim][dim];
for (size_t i = 0; i < n; ++i) { // Outer square #i
// Row #0 of the outer square #i
for (size_t j = i; j < dim-i; ++j)
array[i][j] = n-i;
// Row #n-1 of the outer square #i
for (size_t j = i; j < dim-i; ++j)
array[dim-i-1][j] = n-i;
// Col #0 of the outer square #i
for (size_t j = i; j < dim-i; ++j)
array[j][i] = n-i;
// Col #n-1 of the outer square #i
for (size_t j = i; j < dim-i; ++j)
array[j][dim-i-1] = n-i;
}
print_array(dim, array);
}
This is print_array():
void print_array(const size_t dim, int array[dim][dim])
{
for (size_t i = 0; i < dim; ++i) {
for(size_t j = 0; j < dim; ++j)
printf("%d ", array[i][j]);
printf("\n");
}
}
Output:
4 4 4 4 4 4 4
4 3 3 3 3 3 4
4 3 2 2 2 3 4
4 3 2 1 2 3 4
4 3 2 2 2 3 4
4 3 3 3 3 3 4
4 4 4 4 4 4 4
The worst case time complexity is O(n2).
When you get a problem like this, try to dumb it down as much a possible. This square can be separated into 8 same, just rotated "slices" that look like:
4 | 4444 | 4444 | 4
43 | 333 | 333 | 34
432 | 22 | 22 | 234
4321 | 1 | 1 | 1234
... and the same for the bottom half, just flipped.
You can see this in the code bellow, to check what line is writing what part of the square, comment it and you will see what section shows zeroes.
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int n;
scanf("%d", &n);
int array[2 * n - 1][2 * n - 1];
for(int i = 0; i < 2 * n - 1; i++){
for(int j = 0; j < 2 * n - 1; j++){
array[i][j] = 0;
}
}
int f[n];
for(int i = 0; i < n; i++)
{
f[i] = i;
}
for(int i = 0; i < n; i++)
{
for(int j = i; j < n; j++)
{
array[i][j] = n - i;
array[j][i] = n - i;//top left
array[j][2*n - i - 2] = n - i;
array[i][2*n - j - 2] = n - i;//bottom left
array[2*n - j - 2][i] = n - i;
array[2*n - i - 2][j] = n - i;//top right
array[2*n - i - 2][2*n - j - 2] = n - i;
array[2*n - j - 2][2*n - i - 2] = n - i;//bottom right
}
}
for(int i = 0; i < n - 1 + n; i++)
{
for(int j = 0; j < n - 1 + n; j++)
{
printf("%d ", array[i][j]);
}
printf("\n");
}
return 0;
}
the code I've written creates a pyramid with given input. I need the first part Incrementing by 2, but the code written does not really Incrementing by 2. Aprrechiate any help.
Code:
int main() {
int i, j, n, a, rows;
printf("Enter a Number: ");
scanf("%d", &rows);
printf("\nPattern for %d:\n\n", rows);
for (i = 0; i <= rows; i++) {
printf("%d | ", i);
for(a = 2; a <= i; a+=2) {
printf("%d ", a);
}
for (j = 0; j <= i; j++) {
if (j == i){
printf("%d ", j);
}
}
for(n = 1; n < i; n++){
printf("%d ", j);
j++;
}
printf("\n");
}
return 0;
}
Needed Output:
0 | 0
1 | 0 1
2 | 0 2 2 3
3 | 0 2 4 3 4 5
4 | 0 2 4 6 4 5 6 7
5 | 0 2 4 6 8 5 6 7 8 9
6 | 0 2 4 6 8 10 6 7 8 9 10 11
Current Output:
0 | 0
1 | 1
2 | 2 2 3
3 | 2 3 4 5
4 | 2 4 4 5 6 7
5 | 2 4 5 6 7 8 9
6 | 2 4 6 6 7 8 9 10 11
I had to add an extra if condition for i = 0 because it was not fitting in the pattern I figured out. I have done it in just 2 loops instead of you using 3 loops.
Here is the brief explanation. Comment on this answer if you don't understand the code still, I will add a detailed explanation.
Case i = 0 is a special case and it is handled separately using an if.
for i > 0, you can notice that there are 2 series in parallel
even numbers always starting from 0 and if you count them then they are equal to value of i in that row.
another series whose first number is the value of i in that row and it also has number of elements equal to value of i in that row.
That's why in both the loops that handle the series they are run from 0 till i-1.
I hope it helps you and please accept as correct answer if it did.
#include <stdio.h>
int main() {
int i, j, n, a, b, c, rows;
printf("Enter a Number: ");
scanf("%d", &rows);
printf("\nPattern for %d:\n\n", rows);
for (i = 0; i <= rows; i++) {
if(i == 0){
printf("0 | 0\n");
continue;
}
printf("%d | ", i);
// For handling the even number series
for(a = 0, b = 0; a < i; a++, b+=2) {
printf("%d ", b);
}
// For handling the incremental series
for (j = 0, c = i; j < i; j++, c++) {
printf("%d ", c);
}
printf("\n");
}
return 0;
}
You try to overcomplicate simple task.
void pyramid(int rows)
{
for(int row = 0; row < rows; row++)
{
for(int even = 0; even < row * 2; even += 2)
{
printf("%d ", even);
}
if(row == 0) printf("%d ", row);
else
{
for(int remain = row; remain < row * 2; remain++)
{
printf("%d ", remain);
}
}
printf("\n");
}
}
https://godbolt.org/z/cdWqsq
I wrote a function to print the below pattern.
For example, if the n value is 4 the pattern is
1
2 7
3 6 8
4 5 9 10
Or if the value of n is 5, then the pattern is
1
2 9
3 8 10
4 7 11 14
5 6 12 13 15
My function gives me the first two block but not the next block. I'm stuck here for long time!
My function is
int printPattern(int n) {
int row, column, fwdCtr = 1, evenCtr = 0, ctr = n;
for(row = 1; row <= n; row++) {
fwdCtr = row;
for(column = 1; column <= row; column++) {
if(column % 2 != 0) {
printf("%d ", fwdCtr++);
} else {
evenCtr = fwdCtr + ctr;
printf("%d ", evenCtr);
ctr = ctr - 2;
}
}
printf("\n");
}
}
What I get is
1
2 7
3 6 4
4 5 5 4
Please give suggestions of changes!
The following code should do it:
#include <stdio.h>
void f(int n)
{
for (int i = 0; i < n; ++i)
{
for (int j=0; j<=i; ++j)
{
// Calculate the numbers used so far by previous columns
int x = 0;
for(int v=0; v<j;++v)
{
x = x + (n-v);
}
if ((j % 2) == 0)
{
// even columns
printf("%d ", x+i-j+1);
}
else
{
// odd columns
printf("%d ", x+n-i);
}
}
printf("\n");
}
}
int main(void)
{
f(5);
return 0;
}
Output:
1
2 9
3 8 10
4 7 11 14
5 6 12 13 15
The easy thing to do is just print the right number based on the row and column and the value of n, like this
int main(void)
{
int n = 20;
for (int row = 0; row < n; row++) {
for (int col = 0; col <= row; col++)
printf("%3d ", 1 + col*n - (col-1)*col/2 + (col%2 ? n-1-row : row-col));
printf("\n");
}
}
I'm trying to create a 2d-array in bubble sort, arranged 25 numbers 5 by 5 in ascending order
my inputs
Enter 25 integers:
Input No.[0][0]: 4
Input No.[0][1]: 5
Input No.[0][2]: 8
Input No.[0][3]: 9
Input No.[0][4]: 4
Input No.[1][0]: 2
Input No.[1][1]: 1
Input No.[1][2]: 0
Input No.[1][3]: 2
Input No.[1][4]: 4
Input No.[2][0]: 6
Input No.[2][1]: 7
Input No.[2][2]: 4
Input No.[2][3]: 5
Input No.[2][4]: 5
Input No.[3][0]: 4
Input No.[3][1]: 8
Input No.[3][2]: 9
Input No.[3][3]: 1
Input No.[3][4]: 2
Input No.[4][0]: 4
Input No.[4][1]: 5
Input No.[4][2]: 2
Input No.[4][3]: 1
Input No.[4][4]: 9
my output shows
Ascending:
4 4 5 8 9
0 1 2 2 4
4 5 5 6 7
1 2 4 8 9
1 2 4 5 9
as you can see its not in proper arranged, it only arranged the 5 numbers each lines not the whole numbers
can anybody help arranged my integers like this
Ascending:
0 1 1 1 2
2 2 2 4 4
4 4 4 4 5
5 5 5 6 7
8 8 9 9 9
this is my code so far
int main(){
int rows = 5, cols = 5;
int arr[rows][cols];
int i,j,k,swap;
printf("Enter 25 integers:\n");
for(i = 0; i < rows; i++){
for(j = 0; j < cols; j++){
printf("Input No.[%d][%d]: ", i+0,j+0);
scanf("%d", &arr[i][j]);
}
}
for(k = 0; k < rows; k++){
for(i = 0 ; i < cols; i++){
for(j = i + 1; j < cols; j++){
if(arr[k][i] > arr[k][j]){
swap = arr[k][i];
arr[k][i] = arr[k][j];
arr[k][j] = swap;
}
}
}
}
printf("Ascending:\n");
for( i = 0 ; i < rows; i++){
for( j = 0 ; j < cols; j++){
printf("%3d", arr[i][j]);
}
printf("\n");
}
getch();
}
Improving on Ahmad's answer, I would like to add the following code (for shorting the table in ascending order):
#include <stdio.h>
#define COL 5
#define ROW 6
int main()
{
int temp, t, i, j;
int arr[ROW][COL]={30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1};
for(t=1; t<(ROW*COL); t++)
{
for(i=0; i<ROW; i++)
{
for(j=0; j<COL-1; j++)
{
if (arr[i][j]>arr[i][j+1])
{
temp=arr[i][j];
arr[i][j]=arr[i][j+1];
arr[i][j+1]=temp;
}
}
}
for(i=0; i<ROW-1; i++)
{
if (arr[i][COL-1]>arr[i+1][0])
{
temp=arr[i][COL-1];
arr[i][COL-1]=arr[i+1][0];
arr[i+1][0]=temp;
}
}
}
for(i=0; i<ROW; i++)
{
printf("\n");
for(j=0; j<COL; j++)
printf("%3d", arr[i][j]);
}
return 0;
}
replace the input with your table and the definitions with the size of your given array and you're done.
output of the above when executed:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
26 27 28 29 30
void twoDimBubbleSort(int** arr, int row, int col) {
for (int i = 0; i < (row * col); ++i) {
for (int j = 0; j < (row * col) - 1; ++j) {
int cr = j / col; // current row
int cc = j % col; // current column
int nr = (j + 1) / col; // next item row
int nc = (j + 1) % col; // next item column
if (arr[cr][cc] > arr[nr][nc])
swap(&arr[cr][cc], &arr[nr][nc]); // any way you want to swap variables
}
}
}
You don't necessarily need to create a 1D array, you can consider your 2D array is a 1D array and transform coordinates when you set/get them.
Consider a structure point with x and y, and ARR_LEN is 5.
int from2Dto1D(point p){ return p.x+ p.y*ARR_LEN;}
Point from1Dto2D(int i){ Point p; p.x = i%ARR_LEN; p.y=i/ARR_LEN; return p;}
Now you can use the normal bubble sorting algorithm with a 1D index on 2D squares array, you just need to convert your index into 2 Point and access/switch data using these Point. (2 because you need a Point with index and a Point with index+1
Put all the array elements from 2-D array to 1-D array then
sort that 1-D array and then put 1-D array in the matrix format
Try this code ....works according to the above given logic
#include<stdio.h>
int main(){
int arr[5][5],l=0;
int result[25],k=0,i,j,temp;
arr[0][0]= 4;
arr[0][1]= 5;
arr[0][2]= 8;
arr[0][3]= 9;
arr[0][4]= 4;
arr[1][0]= 2;
arr[1][1]= 1;
arr[1][2]= 0;
arr[1][3]= 2;
arr[1][4]= 4;
arr[2][0]= 6;
arr[2][1]= 7;
arr[2][2]= 4;
arr[2][3]= 5;
arr[2][4]= 5;
arr[3][0]= 4;
arr[3][1]= 8;
arr[3][2]= 9;
arr[3][3]= 1;
arr[3][4]= 2;
arr[4][0]= 4;
arr[4][1]= 5;
arr[4][2]= 2;
arr[4][3]= 1;
arr[4][4]= 9;
//convert 2 D array in 1 D array
for(i=0;i<5;i++){
printf("\n");
for(j=0;j<5;j++){
printf(" %d",arr[i][j]);
result[k++]=arr[i][j];
}
}
// sort 1 D array
for(i=0;i<25;i++){
for(j=0;j<24;j++){
if(result[j] > result[j+1]){
temp=result[j];
result[j]=result[j+1];
result[j+1]=temp;
}
}
}
/*
for(i=0;i<25;i++){
printf("\n%d",result[i]);
}*/
// convert 1 D array to 2 D array
i=0;
l=0;k=0;
while(i<25){
for(j=0;j<5;j++){
arr[k][j]=result[l];
l++;
}
k++;
i=i+5;
}
//Print matrix i.e 2D array
for(i=0;i<5;i++){
printf("\n");
for(j=0;j<5;j++){
printf(" %d",arr[i][j]);
}
}
}
This works !
#define COL 5
#define ROW 2
int main(){
int temp ;
int arr[2][5]= {2,15,26,14,12,18,1,2,3,4 };
int arr2[10] = {0};
int index = 0 ;
for(int t = 0 ; t<50 ; t++ ){
for (int i =0 ; i < ROW ; i++){
for( int j = 0; j < 5-1 ; j++){
if (arr[i][j] > arr[i][j+1]){
temp = arr[i][j];
arr[i][j] = arr[i][j+1];
arr[i][j+1] = temp;
}
}
//checking for
for( int k = 0 ; k < ROW-1 ; k++){
if (arr[k][COL-1] > arr[k+1][0]){
temp = arr[k][COL-1];
arr[k][COL-1] = arr[k+1][0];
arr[k+1][0] = temp ;
}
}
//---------
}
}
return 0 ;
}
Code
#include <stdio.h>
int main()
{
long long int n, m, p, i, j, total_cost, cost;
scanf("%lld %lld %lld", &n, &m, &p);
long long int ar[n][m];
for(i = 1; i <= n; ++i)
{
for(j = 1; j <= m; ++j)
{
ar[i][j] = j;
}
}
while(p--)
{
scanf("%lld %lld", &i, &j);
ar[i][j] += 1;
}
/*
printf("\n");
for(i = 1; i <= n; ++i)
{
for(j = 1; j <= m; ++j)
{
printf("%d ", ar[i][j]);
}
printf("\n");
}
printf("\n");
*/
if(n == 1 && m == 1)
{
printf("0\n");
return 0;
}
if(m == 1)
{
for(i = 1; i <= n; ++i)
{
printf("0\n");
}
return 0;
}
for(i = 1; i <= n; ++i)
{
total_cost = 0, cost = 0;
for(j = m; j >= 2; --j)
{
cost = ar[i][j] - ar[i][j - 1];
if( cost < 0 )
{
printf("-1\n");
break;
}
total_cost += cost;
}
if(cost >= 0)
{
printf("%lld\n", total_cost);
}
}
return 0;
}
This is giving me WA. Here is the problem description CHEFBM.
Test cases I have checked:
1. 1 5 3
1 2
1 4
1 5
Output
5
2. 4 4 6
2 2
3 2
3 2
4 3
4 4
4 3
Output
3
3
-1
4
3. 1 4 5
1 3
1 2
1 2
1 1
1 1
Output
1
4. 4 4 6
3 2
2 3
3 2
2 4
1 2
3 2
Output
3
4
3
3
5. 4 1 1
1 1
Output
0
0
0
0
6. 1 4 3
1 1
1 1
1 4
Output
-1
7. 1 1 1
1 1
Output
0
8. 1 1 2
1 1
1 1
Output
0
9. 2 2 3
1 1
1 1
1 1
Output
-1
1
And many more. For which test case is this giving wrong answer?
As given in the constraints, n and m both can be 10^5. Allocating a 2-d array a[10^5][10^5], is not feasible here. You are running out of memory and hence you might be getting Run time error or Wrong answer. Your algorithm is O(n^2) so it will not pass in the given time limit. Try to optimize your code.
For further assistance, you can check my code.