Printing array as sub blocks - arrays

I have array and I am trying to print this array as sub blocks, where each block has size = 5.
the out put of this code not as I expected it just print the first 5 values. How to print the array as sub blocks?
int arr[298] = {some int values};
int in = 0;
int siz = 298;
int ii;
int rang = 5;
for (int i = 0; i < siz; i++) {
if (in <= siz) {
for (ii = in; ii < 5; ii++) {
printf("arr=%d \n", arr[ii]);
}
printf("------------\n");
}
ind = ind + rang;
}

Following your request for clarification in the comment section, there are a few problems with your code, for me the biggest one is that it's needlessly complicated, but the one you are looking for is in this line:
ind = ind + rang;
ind is is not declared in your code but I assume you mean in, the first time the inner loop runs in(ind) is 0 so it all goes well, after that in will be 5, you assign it to ii and the condition ii < 5 will never be true again, the body of the loop will never be executed.
I suppose you could fix it by using in as index for the array and scrap rang since it isn't needed, something like this:
int arr[298] = {some int values};
int in = 0;
int siz = 298;
for (int i = 0; i < siz; i++) {
//if (in < siz) { moving this into the for loop
for (int ii = 0; ii < 5 && in < siz; ii++, in++) {
printf("arr=%d \n", arr[in]);
}
printf("------------\n");
//}
}
Live demo: https://godbolt.org/z/YzG9sno1n
But you don't need a nested loop, there are a few ways you can do this, a simple one is to have a variable that controls the block size:
int arr[298] = {some int values};
int siz = 298;
int count = 5;
for (int i = 0; i < siz; i++) {
printf("arr=%d \n", arr[i]);
count--;
if (count == 0) {
printf("------------\n");
count = 5;
}
}
Live demo: https://godbolt.org/z/b4e8vWfhM
In the above code count serves as the control variable, the value in the index is printed 5 times and when it reaches 0 a separator is printed and it resets and starts the new block.
Another possible option is to use the index itself to separate the blocks, you know the remainder of a division is 0 when the numerator is divisible by the denominator, you can use that to your advantage:
int arr[298] = {some int values};
int siz = 298;
for (int i = 0; i < siz; i++) {
if (i % 5 == 0) { // && i != 0 if you want to skip the initial separator
printf("------------\n");
}
printf("arr=%d \n", arr[i]);
}
Live demo : https://godbolt.org/z/nne3z38rY
Finally you can/should use a constant value for size:
#include <stdio.h>
#define SIZE 298
int main() {
int arr[SIZE] = {some int values};
for (int i = 0; i < SIZE; i++) {
if (i % 5 == 0 && i != 0) { // skipping the initial separator
printf("------------\n");
}
printf("arr=%d \n", arr[i]);
}
}
Live demo: https://godbolt.org/z/Mc4Yh4cav

Instead of several for loops, you can use a single while loop.
int arr[298 ]={Some int Values};
int ind =0;
int siz= 298 ;
printf("------------\n");
while(ind<=siz-1){
printf("arr=%d \n",arr[ind]);
ind++;
if(ind%5==0){
printf("------------\n");
}
}
In this, you print the elements through 0 to 297, with a line of dashes printed if the index is divisible by 5, that is after every fifth element.

Related

Count how many 1 in integer's binary

Sample Input
2
2
5
Sample Output
0 1 1
0 1 1 2 1 2
I already knew how to transfer an integer to its binary and count how many 1 in its binary.
But my code only can input one integer each time. I want it to input many numbers, like the sample input and sample output. To make question more easy to understand, so I drew a picture. Thank you!!
Thanks all of you!! But I need some time to understand those code and my English is pretty basic, so I couldn't reply you guys soon. But I will understand and reply you as soon as possible!!thank you :D
#include <stdio.h>
int main()
{
int n,cnt=0,m;
scanf("%d",&n);
while(n>0){
m=n%2;
if(m==1){
cnt++;
}
n/=2;
}
printf("%d",cnt);
return 0;
}
In practice, the process can be simplified and accelerated by noticing that if we know the result for i < 2^p, then, for all values in [2^p, 2^(p+1)-1], we have
count(j) = 1 + count(j-2^p)
This method is useful here as we have to provide the result for all values less or equal to n.
Moreover, in order to avoid performing the same calculation at different times, we first calculate the maximum n value, and perform the calculation for this maximum value.
#include <stdio.h>
int main() {
int t;
int check = scanf("%d", &t);
if (check != 1) return 1;
int nn[t];
for (int i = 0; i < t; ++i) {
int check = scanf("%d", &nn[i]);
if (check != 1) return 1;
}
int nmax = 0;
for (int i = 0; i < t; ++i) {
if (nn[i] > nmax) nmax = nn[i];
}
int count[nmax+1];
count[0] = 0;
count[1] = 1;
int pow2 = 1;
do {
pow2 *= 2;
for (int j = pow2; (j < 2*pow2) && (j <= nmax); j++) {
count[j] = 1 + count[j-pow2];
}
} while (pow2 <= nmax+1);
for (int i = 0; i < t; ++i) {
for (int j = 0; j <= nn[i]; ++j) {
printf ("%d ", count[j]);
}
printf ("\n");
}
return 0;
}
Let the number of measured values be t. Now if you want to take t number of integers as input, you have to run a loop t times and perform necessary tasks (like taking value of n, calculating cnt etc) inside that loop. Check the following code snippet:
int main()
{
int n,cnt=0,m;
int numberOfMeasuredValues;
scanf("%d", &numberOfMeasuredValues);
while(numberOfMeasuredValues > 0){
cnt = 0;
scanf("%d",&n);
// perform necessary calculations
printf("%d\n",cnt);
numberOfMeasuredValues--;
}
return 0;
}
In the code numberOfMeasuredValues is the t I've used in the explanation.
You can pre-allocate number of cases and work on them with outer and inner loops. I've added the code with comments below.
#include <stdio.h>
int main(void)
{
int cases; // Number of cases
scanf("%d", &cases);
int* nums = malloc(cases * sizeof *nums); // Allocate enough space.
/* Take the inputs. */
for (int i = 0; i < cases; i++)
{
scanf("%d", (nums + i));
}
int cnt = 0; // Counter for number of 1s.
// Outer loop for going through cases.
for (int i = 0; i < cases; i++)
{
// Inner loop for couting towards inputted number.
for (int j = 0; j < *(nums + i) + 1; j++)
{
/*
* You can Implement this part by yourself if you wish.
*/
for (int k = 0; k < 32; k++)
{
if ((j >> (31 - k)) & 0x01) cnt++; // Increment counter if digit is 1.
}
printf("%d ", cnt);
cnt = 0;
}
printf("\n");
}
return 0;
}
Here's the following code to your problem
#include<iostream>
using namespace std;
int main()
{
int tcs;
cin >> tcs;
while (tcs--) {
int n;
cin >> n;
for (int i = 0 ; i <= n ; i++) {
cout << __builtin_popcount(i) << " ";
}
cout << "\n";
}
return 0;
}

Why does this program crash, if SPLIT is between 4 and 7

I had a task and the program is working, for the most part, however, it crashes if I put SPLIT value between 4 and 7 (crashes at different values, if I change SIZE, but for sake of simplicity, let's keep it at 10).
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<string.h>
#define SIZE 10
#define SPLIT 4
#define LOW 0
#define HIGH 10
void generateArray(int data[],int size,int low, int high){
srand(time(NULL));
for(int i=0;i<size;++i){
data[i]=rand()%(high-low+1)+low;
}
}
int splitData(int arraySize, int startArray[], int splitPoint, int **firstNewArray, int **secondNewArray){
if(arraySize < 1){
return -1;
}
if(splitPoint < 1 || (splitPoint >= arraySize)){
return -1;
}
if(*firstNewArray != NULL || *secondNewArray != NULL){
return -1;
}
*firstNewArray = malloc(splitPoint * sizeof(int));
*secondNewArray = malloc((arraySize - splitPoint) * sizeof(int));
for(int i = 0; i < arraySize; ++i){
if(i < splitPoint){
(*firstNewArray)[i] = startArray[i];
printf("%d\n",startArray[i]);
}else{
(*secondNewArray)[i] = startArray[i];
printf("%d\n",startArray[i]);
}
}
return 0;
}
int main(){
int arraySize = SIZE ;
int *startArray = malloc(arraySize * sizeof(int));
generateArray(startArray,arraySize,LOW,HIGH);
int splitPoint = SPLIT;
int *firstNewArray = NULL;
int *secondNewArray = NULL;
int result;
result = splitData(arraySize, startArray, splitPoint, &firstNewArray, &secondNewArray);
if(result == 0){
for(int i = 0; i < arraySize; ++i){
if(i < splitPoint){
printf("First array number %d is %d\n",i+1,firstNewArray[i]);
}else{
printf("Second array number %d is %d\n",i,secondNewArray[i]);
}
}
free(firstNewArray);
free(secondNewArray);
}
free(startArray);
return 0;
}
What could be the cause of this behavior and how could I fix it? The task is to split startArray by the value SPLIT into 2 new dynamic arrays, that would be created in a function splitData and both of them could be used outside the function.
You have two issues with your code
first when you display the results:
for(int i = 0; i < arraySize; ++i){
if(i < splitPoint){
printf("First array number %d is %d\n",i+1,firstNewArray[i]);
}else{
printf("Second array number %d is %d\n",i,secondNewArray[i]);
}
}
This will not work specialy if array size is too higth or too low, example splitPoint is 9, this means secondNewArray Size is 1 but in this loop you are accessing secondNewArray[9] where it should be 0, you need to change the loop into something like this
for(int i = 0; i < splitPoint; ++i){
printf("First array number %d is %d\n",i+1,firstNewArray[i]);
}
for(int i = 0; i < SIZE - splitPoint; ++i){
printf("Second array number %d is %d\n",i+splitPoint+1 ,secondNewArray[i]);
}
You have the same isssue in your split function:
for(int i = 0; i < arraySize; ++i){
if(i < splitPoint){
(*firstNewArray)[i] = startArray[i];
printf("%d\n",startArray[i]);
}else{
(*secondNewArray)[i] = startArray[i];
printf("%d\n",startArray[i]);
}
}
In this case also you are accessing regions outside the size of your array, let say split is 9 you will be accessing secondNewArray[9] = startArray[9] where it should be secondNewArray[0] = startArray[9], to fix this you need to do the same thing here where you use different index for each array, the code should look like this:
int j = 0;
int k = 0;
for(int i = 0; i < arraySize; ++i){
if(i < splitPoint) {
(*firstNewArray)[j] = startArray[i];
printf("%d\n",startArray[i]);
j++;
}
else {
(*secondNewArray)[k] = startArray[i];
printf("%d\n",startArray[i]);
k++;
}
}
Take a hard look at the marked line below
for(int i = 0; i < arraySize; ++i){
if(i < splitPoint){
(*firstNewArray)[i] = startArray[i];
printf("%d\n",startArray[i]);
}else{
(*secondNewArray)[i] = startArray[i]; // LOOK HERE
printf("%d\n",startArray[i]);
}
Assuming an array size of 10 and a split point of 4, then *secondNewArray is indexed from 0 to 5; however, you’re trying to assign elements 4 through 9, which is outside the bounds of the array, leading to undefined behavior. You need to adjust the value of i in order to map properly:
(*secondNewArray)[i - splitPoint] = startArray[i];

How to connect puzzles so that right edge has same length as left edge of another puzzle?

I have puzzles that looks like this:
=== ====
=== ===
=== ====
left edge has length from 0 to 10 000 and right also, middle part is from 1 to 10 000.
So the question is if i can build a rectangle? like first puzzle has length of left edge equal to 0 and the last puzzle has right edge of length 0 and in the middle they fit perfectly?
I am given the number of puzzle i have and their params like this:
6
1 9 2
0 3 1
0 4 1
8 9 0
2 9 0
1 5 0
and result can be any of that:
2
0 3 1
1 5 0
or
3
0 3 1
1 9 2
2 9 0
or
2
0 4 1
1 5 0
But if there is no result i have to printf("no result")
I have to do this in C, I thought about doing some tree and searching it with BFS where vertices would have edge lengths and edge would have middle length and when reached 0 i would go all way up and collect numbers but it's hard to code. So i decided to do recursion but im also stuck:
#include<stdio.h>
int main(){
int a;
scanf("%d", &a);//here i get how many puzzles i have
int tab[a][3];//array for puzzles
int result[][3];//result array
int k = 0;//this will help me track how many puzzles has my result array
for(int i = 0; i < a; i++){//upload puzzles to array
for(int j = 0; j < 3; j++){
scanf("%d", &tab[i][j]);
}
}
findx(0, a, tab, result, k);//start of recursion, because start has to be length 0
}
int findx(int x, int a, int *tab[], int *result[], int k){//i am looking for puzzle with length x on start
for(int i = 0; i < a; i++){
if(tab[a][0] == x){//there i look for puzzles with x length at start
if(tab[a][2] == 0){//if i find such puzzle i check if this is puzzle with edge length zero at the end
for(int m = 0; m < 3; m++){//this for loop add to my result array last puzzle
result[k][m] = tab[a][m];
}
return print_result(result, k);//we will return result go to print_result function
}
else{//if i have puzzle with x length on the left and this is not puzzle which ends rectangle i add this puzzle
//to my result array and again look for puzzle with x equal to end length of puzzle i found there
for(int m = 0; m < 3; m++){
result[k][m] = tab[a][m];
k += 1;
}
findx(tab[a][2], a, tab, result, k);
}
}
}
printf("no result");
}
int print_result(int *result[], int k){
printf("%d", &k);//how many puzzles i have
printf("\n");
for(int i = 0; i < k; i++){//printing puzzles...
for(int j = 0; j < 3; j++){
printf("%d ", &result[i][j]);
}
printf("\n");//...in separate lines
}
}
I have an error that result array can't look like this int result[][3] because of of [] but I don't know how many puzzles I'm gonna use so?... and I have implicit declaration for both of my functions. Guys please help, I dont know much about C and its super hard to solve this problem.
I'm not sure I understand the overall logic of the problem, but you definitely are in need of some variable sized containers for result AND tab. Arrays are fixed size and must be defined at compile time. The following should at least compile without warnings:
#include<stdio.h>
#include<stdlib.h>
void print_result(int (*result)[3], int k){
printf("%d", k);//how many puzzles i have
printf("\n");
for(int i = 0; i <= k; i++){//printing puzzles...
for(int j = 0; j < 3; j++){
printf("%d ", result[i][j]);
}
printf("\n");//...in separate lines
}
}
void findx(int x, int a, int (*tab)[3], int (*result)[3], int k){//i am looking for puzzle with length x on start
for(int i = 0; i < a; i++){
if(tab[i][0] == x){//there i look for puzzles with x length at start
if(tab[i][2] == 0){//if i find such puzzle i check if this is puzzle with edge length zero at the end
for(int m = 0; m < 3; m++){//this for loop add to my result array last puzzle
result[k][m] = tab[i][m];
}
print_result(result, k);//we will return result go to print_result function
return;
}
else{//if i have puzzle with x length on the left and this is not puzzle which ends rectangle i add this puzzle
//to my result array and again look for puzzle with x equal to end length of puzzle i found there
for(int m = 0; m < 3; m++){
result[k][m] = tab[i][m];
k += 1;
///** Increase size of result **/
//int (*newptr)[3] = realloc(result, (k+1) * sizeof(int[3]));
//if (newptr)
// result = newptr;
}
findx(tab[i][2], a, tab, result, k);
}
}
}
printf("no result\n");
}
int main(){
int a;
scanf("%d", &a);//here i get how many puzzles i have
int (*tab)[3] = malloc(a * sizeof(int[3]));//array for puzzles
int (*result)[3] = malloc(a * sizeof(int[3]));//array for puzzles
int k = 0;//this will help me track how many puzzles has my result array
for(int i = 0; i < a; i++){//upload puzzles to array
for(int j = 0; j < 3; j++){
scanf("%d", &tab[i][j]);
}
}
findx(0, a, tab, result, k);//start of recursion, because start has to be length 0
}
Note that I changed the tab and result types to (*int)[3]. Due to order of operations, we need parentheses here. Because they are variable size, they require dynamic memory allocations. In the interest of brevity and readability, I did not check the returned values of malloc or realloc. In practice, you should be checking that the returned pointer is not NULL. Because we are using dynamic memory allocations, we should also use free if you plan on doing anything else with this program. Otherwise, it doesn't really matter because exiting the program will free the resources anyway. You actually don't want to free. because we are passing a pointer by value to findx and the realloc can change the address, it may come back with a different address. Also, take note that I needed to include <stdlib.h> for the dynamic memory allocations.
Additional Issues
Your functions print_results and findx are not declared when you call them in main. Your function either need to be above main or have "function prototypes" above main.
In the printfs you do not need the &. You do not want to send the address of the variable to printf. You want to send what will actually be printed.
Now what?
The program still does not provide you with the correct results. It simply outputs 0 as the result every time. This should at least give you a starting point. By changing this line in print_results:
for(int i = 0; i < k; i++){//printing puzzles...
to
for(int i = 0; i <= k; i++){//printing puzzles...
I was at least able to output 0 0 0. This seems more correct because if k is 0, we don't loop at all.
#include<stdio.h>
void findx(int x, int a, int tab[a][3], int result[200000][3], int puzzlesinresult) { //i am looking for puzzle with length x on start
for (int i = 0; i < a; i++) {
if (tab[i][0] == x) { //there i look for puzzles with x length at start
if (tab[i][2] == 0) { //if i find such puzzle i check if this is puzzle with edge length zero at the end
for (int m = 0; m < 3; m++) { //this for loop add to my result array last puzzle
result[puzzlesinresult][m] = tab[i][m];
}
return print_result(result, puzzlesinresult); //we will return result go to print_result function
} else { //if i have puzzle with x length on the left and this is not puzzle which ends rectangle i add this puzzle
//to my result array and again look for puzzle with x equal to end length of puzzle i found there
while (result[puzzlesinresult - 1][2] != tab[i][0] && puzzlesinresult > 0) {
puzzlesinresult -= 1;
}
int isusedpuzzle = 0;
for (int j = 0; j < puzzlesinresult; j++) {
if (result[j][0] == tab[i][0] && result[j][1] == tab[i][1] && result[j][2] == tab[i][2]) {
isusedpuzzle = 1;
} else {
//pass
}
}
if (isusedpuzzle == 0) {
for (int m = 0; m < 3; m++) {
result[puzzlesinresult][m] = tab[i][m];
}
puzzlesinresult += 1;
findx(tab[i][2], a, tab, result, puzzlesinresult);
}
}
}
}
}
void print_result(int result[200000][3], int puzzlesinresult) {
printf("%d\n", puzzlesinresult + 1); //how many puzzles i have
for (int i = 0; i < puzzlesinresult + 1; i++) { //printing puzzles...
for (int j = 0; j < 3; j++) {
printf("%d ", result[i][j]);
}
printf("\n"); //...in separate lines
}
exit(0);
}
int main() {
int a;
scanf("%d", & a); //here i get how many puzzles i have
int tab[a][3]; //array for puzzles
int result[100][3]; //result array
int puzzlesinresult = 0; //this will help me track how many puzzles has my result array
for (int i = 0; i < a; i++) { //upload puzzles to array
for (int j = 0; j < 3; j++) {
scanf("%d", & tab[i][j]);
}
}
for (int i = 0; i < a; i++) { //here i delete puzzles that doesnt contribute anything like 1 x 1,2 x 2,..
if (tab[i][0] == tab[i][2] && tab[i][0] != 0) {
for (int p = i; p < a; p++) {
for (int j = 0; j < 3; j++) {
tab[p][j] = tab[p + 1][j];
}
}
}
}
findx(0, a, tab, result, puzzlesinresult); //start of recursion, because start has to be length 0
printf("NONE");
}
This returns sometimes correct result. If you can find when this program fails I would rly appreciate sharing those cases with me :)

How can I make a new array, by counting the no.of appearances of value and printing it next to that value?

I should make new array out of existing one (ex. 1 0 4 5 4 3 1) so that the new one contains digits already in existing array and the number of their appearances.
So, the new one would look like this: 1 2 0 1 4 2 5 1 3 1 (1 appears 2 times, 0 appears 1 time.... 3 appears 1 time; the order in which they appear in first array should be kept in new one also); I know how to count no. of times a value appears in an array, but how do I insert the no.of appearances? (C language)
#include <stdio.h>
#define max 100
int main() {
int b, n, s, i, a[max], j, k;
printf("Enter the number of array elements:\n");
scanf("%d", &n);
if ((n > max) || (n <= 0)) exit();
printf("Enter the array:\n");
for (i = 0; i < n; i++)
scanf("%d", a[i]);
for (i = 0; i < n; i++) {
for (j = i + 1; j < n;) {
if (a[j] == a[i]) {
for (k = j; k < n; k++) {
a[k] = a[k + 1];
}}}}
//in the last 5 rows i've tried to compare elements, and if they are same, to increment the counter, and I've stopped here since I realised I don't know how to do that for every digit/integer that appears in array//
If you know that the existing array consists of digits between 0 and 9, then you can use the index of the array to indicate the value that you are incrementing.
int in[12] = {1,5,2,5,6,5,3,2,1,5,6,3};
int out[10] = {0,0,0,0,0,0,0,0,0,0};
for (int i = 0; i < 12; ++i)
{
++out[ in[i] ];
}
If you provide any code snippet, its easy for the community to help you.
Try this, even you optimize the no.of loops :)
#include <stdio.h>
void func(int in[], int in_length, int *out[], int *out_length) {
int temp[10] = {0}, i = 0, j = 0, value;
//scan the input
for(i=0; i< in_length; ++i) {
value = in[i];
if(value >= 0 && value <= 9) { //hope all the values are single digits
temp[value]++;
}
}
// Find no.of unique digits
int unique_digits = 0;
for(i = 0; i < 10; ++i) {
if(temp[i] > 0)
unique_digits++;
}
// Allocate memory for output
*out_length = 2 * unique_digits ;
printf("digits: %d out_length: %d \n",unique_digits, *out_length );
*out = malloc(2 * unique_digits * sizeof(int));
//Fill the output
for(i = 0, j = 0; i<in_length && j < *out_length; ++i) {
//printf("\n i:%d, j:%d val:%d cout:%d ", i, j, in[i], temp[in[i]] );
if(temp[in[i]] > 0 ) {
(*out)[j] = in[i];
(*out)[j+1] = temp[in[i]];
temp[in[i]] = 0; //Reset the occurrences of this digit, as we already pushed this digit into output
j += 2;
}
}
}
int main(void) {
int input[100] = {1, 0, 4, 5, 4, 3, 1};
int *output = NULL, output_length = 0, i = 0;
func(input, 7, &output, &output_length );
for(i=0; i < output_length; i+=2) {
printf("\n %d : %d ", output[i], output[i+1]);
}
return 0;
}

C, Showing count of repeating symbols

Need to figure out a code that counts all the repeating symbols in a string. As you can see below, so far so good.
And here starts the tricky part, at the end of the code I want to output symbols in an order they were typed which had for example 2 occurences in a string, and I got problems figuring that out.
int counts[256] = { 0 };
int i;
size = strlen(text);
for (i = 0; i < size; i++) {
counts[(int)(text[i])]++;
}
for (i = 0; i < 256; i++) {
printf("The %d. character has %d occurrences.\n", i, counts[i]);
}
Just iterate through the source string again and for each character look into your counts array.
If you don't want to print the same statistics for every occurence of repeating character, you can reset the corresponding counts value to zero just after you print the statistics, and have an additional check before printing.
for(i = 0; i < size; i++) {
if(counts[(int)(text[i])] == 2)
printf("%d", (int)(text[i]));
The first line loops through your source string for the order of occurences.
The second line checks if it was captured in the counts array as occuring only twice.
If it was we print the char code on the third line.
To only print the character once:
for(i = 0; i < size; i++) {
if(counts[(int)(text[i])] == 2) {
printf("%d", (int)(text[i]));
counts[(int)(text[i])] = 0;
}
}
Here is an implementation of Inspired's answer:
int counts[256] = { 0 };
char text[] = "Hello, world!";
int i, size = strlen(text);
for (i = 0; i < size; i++)
{
counts[(unsigned int)(text[i])]++;
}
for (i = 0; i < size; i++)
{
if (counts[(unsigned int)text[i]] > 1)
{
printf("%c", text[i]);
counts[(unsigned int)text[i]] = 0; // Remove to print repeats.
}
}
Make a key,count pair, like:
#include <string.h>
#include <stdio.h>
int main()
{
char* text = "count this text";
char *keys = new char[strlen(text)];
int* count = new int[strlen(text)];
int last = 0; int j=0;
for(int i=0; i<strlen(text); i++){
for(j=0; j<last; j++){
if(keys[j]==text[i]) break;
}
if(keys[j]==text[i]){
count[j]++;
} else {
keys[last]=text[i];
count[last]=1;
last++;
}
}
for(int i=0; i<last; i++){
printf("%c %d\n", keys[i], count[i]);
}
}
so you keep the order in the text and get the count.
On execution the output is:
c 1
o 1
u 1
n 1
t 4
2
h 1
i 1
s 1
e 1
x 1

Resources