How can I convert a two digits from array to one integer into a new array?
Example: 1245678933658 to [12,45,67,89,33,65,80] add 0 to the last if its an odd.
My try was:
new_array[i]=digits[i]*10+digits[i++]*10/10;
Does it have to be a "one-liner", or would a more readable code also be allowed? Try the following:
const char* digits = "12456789336581";
int digitIndex=0;
int numbers[100];
int currentNumberIndex = 0;
int currentNumber = 0;
while (digits[digitIndex]) {
currentNumber *=10;
currentNumber += digits[digitIndex] - '0';
digitIndex++;
if ( (digitIndex)%2 == 0) // two digits handled?
{
numbers[currentNumberIndex++] = currentNumber;
currentNumber = 0;
}
}
// handle the case that last number contained just one digit:
if (digitIndex%2) {
currentNumber *=10;
numbers[currentNumberIndex++] = currentNumber;
}
for (int i=0; i < currentNumberIndex; i++)
printf("number[%d]: %d\n", i, numbers[i]);
Related
I need to write a program that allows user to enter an array of integers, finds the digit that appears most often in all entered numbers, and removes it from the elements of the array. If several digits appear the same number of times, the smallest of them should be deleted. If all digits of the element of the array are deleted, that element should become zero. In the end, such a modified array is printed.
Example of input and output:
Enter number of elements of the array: 5
Enter the array: 3833 8818 23 33 1288
After deleting, the array is: 8 8818 2 0 1288
Explanation: The numbers 3 and 8 appear the same number of times (6 times each), but 3 is less, so it was removed it from all members of the array. Element 33 consists exclusively of the digits 3, so that it becomes 0.
#include <stdio.h>
int main() {
int i,n,arr[100]; n;
printf("Enter number of elements of the array: ");
scanf("%d", &n);
printf("Enter the array: ");
for(i=0;i<n;i++) {
scanf("%d", &arr[i]);
}
return 0;
}
EDIT: I'm beginner to programming, and this task should be done using only knowledge learned so far in my course which is conditionals, loops, and arrays. This shouldn't be done with strings.
Divide the problem into separate tasks.
Write the code
In the code below I do not treat 0 as having digit 0. It is because it is not possible to remove 0 from 0. You can easily change this behaviour by changing while(){} loop to do{}while()
int removeDigit(int val, int digit)
{
int result = 0;
unsigned mul = 1;
int sign = val < 0 ? -1 : 1;
digit %= 10;
while(val)
{
int dg = abs(val % 10);
if(dg != digit)
{
result += dg * mul;
mul *= 10;
}
val /= 10;
}
return sign * result;
}
void countDigits(int val, size_t *freq)
{
while(val)
{
freq[abs(val % 10)]++;
val /= 10;
}
}
int findMostFrequent(const size_t *freq)
{
size_t max = 0;
for(size_t i = 1; i < 10; i++)
{
if(freq[i] > freq[max]) max = i;
}
return (int)max;
}
int main(void)
{
int table[20];
size_t freq[10] = {0,};
int mostfreq = 0;
srand(time(NULL));
for(size_t i = 0; i < 20; i++)
{
table[i] = rand();
printf("Table[%zu] = %d\n", i, table[i]);
countDigits(table[i], freq);
}
mostfreq = findMostFrequent(freq);
printf("Most frequent digit: %d\n", mostfreq);
for(size_t i = 0; i < 20; i++)
{
table[i] = removeDigit(table[i], mostfreq);
printf("Table[%zu] = %d\n", i, table[i]);
}
}
https://godbolt.org/z/PPj9s341b
I have to solve a problem where one of the important tasks is to reorder the digits of the input in ascending order and we are not allowed to use arrays and lists. I have no problem with that and my code works, but only if we do not consider leading 0, which we should in this problem. The only way I see how to do is to check digit by digit and then add then ordered by multiplying the number by 10 and adding the next digit. (1*10 = 10, 10+3= 13, we got 1 and 3 ordered) However, if we have a 0 in our number this method will not work because if I want to make 0123 with the * 10 method, I won't be able to have the 0 as the first digit never. Does anyone know how to solve this? My code is below:
int ascendingNumbers (int n) { //This function sorts the number on an ascending order
int number = n;
int sortedN = 0;
for (int i = 0; i <= 9; i++) {
int toSortNumber = number;
for (int x = 0; x <= 4; x++) {
int digit = toSortNumber % 10;
if (digit == i) {
if (digit == 0) {
sortedN==10;
}
sortedN *= 10;
sortedN += digit;
}
toSortNumber /= 10;
}
}
return sortedN;
}
Normally I don't do homework problems, but for especially awful ones I'll make an exception.
(Also I'm making an exception to my general rule not to have anything to do with these absurd "desert island" constraints, where you're stranded after a shipwreck and your C compiler's array functionality got damaged in the storm, or something.)
I assume you're allowed to call functions. In that case:
#include <stdio.h>
/* count the number of digits 'd' in 'n'. */
int countdigits(int n, int d)
{
int ret = 0;
/* do/while so consider "0" as "0", not nothing */
do {
if(n % 10 == d) ret++;
n /= 10;
} while(n > 0);
return ret;
}
int main()
{
int i, n;
printf("enter your number:\n");
scanf("%d", &n);
printf("digits: ");
for(i = 0; i < 10; i++) {
int n2 = countdigits(n, i);
int j;
for(j = 0; j < n2; j++) putchar('0' + i);
}
printf("\n");
}
This solution does not involve a function int ascendingNumbers() as you asked about. If you want to handle leading zeroes, as explained in the comments, you can't do it with a function that returns an int.
Your zero problem is solved, check it...
class Main {
public static void main(String[] args) {
int number = 24035217;
int n = number, count = 0;
int sortedN = 0;
while (n != 0) {
n = n / 10;
++count;
}
for (int i = 9; i >= 0; i--) {
int toSortNumber = number;
for (int x = 1; x <= count; x++) {
int digit = toSortNumber % 10;
// printf("\nBefore i = %d, x = %d, toSortNumber = %d, sortedN = %d, digit = %d",i,x,toSortNumber,sortedN,digit);
if (digit == i) {
sortedN *= 10;
sortedN += digit;
}
// printf("\nAfter i = %d, x = %d, toSortNumber = %d, sortedN = %d, digit = %d",i,x,toSortNumber,sortedN,digit);
toSortNumber /= 10;
}
}
System.out.print(sortedN);
}
}
I'm trying to create a function that compares two four digit numbers and
returns the number of similar digits between the two. For example, with a generated number of 4311 and the user entered 1488,
the score should return 2 (4 and 1).
If it was 4311 and the other is 1147,
the score should return three (1, 1 and 4). I don't know why it isn't giving me the right outputs, hope you can help.
int getSameDigitScore(int playerGuess, int generatedNum) {
int score = 0;
int i;
int j;
int k;
int generatedNumArray[4];
int playerGuessArray[4];
// turns playerGuess into an array
while (playerGuess > 0 ) {
i = 0;
playerGuessArray[i] = playerGuess % 10;
i++;
playerGuess /= 10;
}
// turns generatedNum into an array
while (generatedNum > 0) {
i = 0;
generatedNumArray[i] = generatedNum % 10;
i++;
generatedNum /= 10;
}
// compares the two arrays
for (k = 3; k >= 0; k--) {
for (j = 3; j >= 0; j--) {
if (generatedNumArray[k] == playerGuessArray[j]) {
score++;
playerGuessArray[j] = 0;
j = -5;
}
}
}
return score;
}
You are assigning i = 0 inside the while loop while generating the playerGuessArray and generatedNumArray. Due to which the playerGuess and generatedNumArray array will have elements as first digit of your number 0 0 0 .
Move the initialization out of the loop.
int getSameDigitScore(int playerGuess, int generatedNum) {
int score = 0;
int i, j, k, n;
int generatedNumArray[4];
int playerGuessArray[4];
// turns playerGuess into an array
i = 0; // This has been out of while loop
while (playerGuess > 0 ) {
playerGuessArray[i] = playerGuess % 10;
i++;
playerGuess /= 10;
}
// turns generatedNum into an array
int n = 0; // This has been out of the while loop
while (generatedNum > 0) {
generatedNumArray[n] = generatedNum % 10;
n++;
generatedNum /= 10;
}
// compares the two arrays
for (k = 3; k >= 0; k--) {
for (j = 3; j >= 0; j--) {
if (generatedNumArray[k] == playerGuessArray[j]) {
score++;
playerGuessArray[j] = 0;
j = -5;
}
}
}
return score;
}
int main() {
int m;
n = getSameDigitScore(1231, 2342);
printf("Score is: %d\n", m);
}
You're re-initializing increment variable i on every iteration which should be moved out of the while loop. With that moved out the above code works fine.
There are the following issues with the code.
You are initializing the integer i inside the while loop. This needs to be done before the loop for each loop.
You need a separate array to get the output of equal digits. See AnswerArray in code below. Also it is a good design practice to pass this array to the function and clear this array inside the function.
In the last for loop, you should break from the inner loop after getting a match. This is to take care of cases where playerGuess == 1222 and generatedNum = 1111 In the code shown this will result in a score of 1.
See the final code below with some test cases.
int getSameDigitScore(int playerGuess, int generatedNum, int *AnswerArray) {
int score = 0;
int i;
int j;
int k;
int generatedNumArray[4] = {0};
int playerGuessArray[4] = {0};
memset(AnswerArray,0,4*sizeof(int));
// turns playerGuess into an array
i = 0;
while (playerGuess > 0 ) {
playerGuessArray[i] = playerGuess % 10;
i++;
playerGuess /= 10;
}
// turns generatedNum into an array
i = 0;
while (generatedNum > 0) {
generatedNumArray[i] = generatedNum % 10;
i++;
generatedNum /= 10;
}
// compares the two arrays
score=0;
for (k = 3; k >= 0; k--) {
for (j = 3; j >= 0; j--) {
if (generatedNumArray[k] == playerGuessArray[j]) {
AnswerArray[score++] = generatedNumArray[k];
playerGuessArray[j] = -1;
break;
}
}
}
return score;
}
int main(void)
{
int AnswerArray[4],score;
score = getSameDigitScore(4311,1488,AnswerArray);
printf ("\nScore = %d \n Answer Array = ",score);
for (int i=0; i<score; i++)
{
printf ("%d ",AnswerArray[i]);
}
score = getSameDigitScore(4311,1147,AnswerArray);
printf ("\nScore = %d \n Answer Array = ",score);
for (int i=0; i<score; i++)
{
printf ("%d ",AnswerArray[i]);
}
score = getSameDigitScore(1222,1111,AnswerArray);
printf ("\nScore = %d \n Answer Array = ",score);
for (int i=0; i<score; i++)
{
printf ("%d ",AnswerArray[i]);
}
score = getSameDigitScore(1111,1222,AnswerArray);
printf ("\nScore = %d \n Answer Array = ",score);
for (int i=0; i<score; i++)
{
printf ("%d ",AnswerArray[i]);
}
}
The initializing i=0 which you made inside the loop should be outside the loop.
while (playerGuess > 0 ) {
i = 0;
playerGuessArray[i] = playerGuess % 10;
i++;
playerGuess /= 10;
}
If the initialization is inside the looop then,
Everytime playerGuessArray[0] value will be updated.
FYI:
If playerGuess can contain 0 aat the begin of four digit like 0123
For example, playerGuessValue is 0123, Then by using
while (playerGuess > 0 ) {
i = 0;
playerGuessArray[i] = playerGuess % 10;
i++;
playerGuess /= 10;
}
playerGuessArray will contain only [1,2,3] instead of [0,1,2,3].
So, the better solution would be taking two temporary variables and checking last digit one by one.
Like this:
int temp1=playerGuess, temp2=GeneratedNum;
int i=0;
bool flag = true;
while(flag && i < 4){
if(temp1%10 != temp2%10){
flag = false;
}
temp1 /= 10;
temp2 /= 10;
i++;
}
if(flag){
score++;
}
FYI:
Debugging will help you in finding out these little mistakes.So, try to debug your code with multiple inputs and verify your answer.
Here are few reference on how to debug:
https://blog.hartleybrody.com/debugging-code-beginner/
https://www.codementor.io/mattgoldspink/how-to-debug-code-efficiently-and-effectively-du107u9jh%60
Thanks.
I am trying to return the index where the sum of the left hand side is equal to the right hand side. But I get a default of -1. Why doesn't the conditional evaluate to true?
Here is the code:
#include <stdio.h>
int find_even_index(const int *values, int length)
{
int count = 0;
int sumL = 0;
int sumR = 0;
while(count < length)
{
sumR += values[count];
count++;
}
count = 0;
while(count < length)
{
sumL += values[count];
sumR -= values[count];
printf("sumL = %d\n", sumL);
printf("sumR = %d\n", sumR);
if(sumL == sumR)//why doesn't this condition work?
{
return count;
}
count++;
}
return -1;
}
int main (void)
{
int arr[] = { 1,2,3,4,3,2,1 };
printf("%d\n", (find_even_index(arr, 7)));
return 0;
}
Because you are comparing the numbers after adding and subtracting both of them. So in the moment they are about to be the same:
sumL = 6; sumR = 10
You then sum 4 to sumL and subtract 4 to sumR:
sumL = 10; sumR = 6
And then you compare and get different values. You have to check in between the operations.
I've created a solution to problem 4 on Project Euler.
However, what I find is that placing the print statement (that prints the answer) in different locations prints different answers. And for some reason, the highest value of result is 580085. Shouldn't it be 906609? Is there something wrong with my isPalindrome() method?
#include <stdio.h>
#include <stdbool.h>
int isPalindrome(int n);
//Find the largest palindrome made from the product of two 3-digit numbers.
int main(void)
{
int i = 0;
int j = 0;
int result = 0;
int palindrome = 0;
int max = 0;
//Each iteration of i will be multiplied from j:10-99
for(i = 100; i <= 999; i++)
{
for(j = 100; j <= 999; j++)
{
result = i * j;
if(isPalindrome(result) == 0)
{
//printf("Largest Palindrome: %d\n", max); //906609
//printf("Result: %d\n", result); //580085
if(result > max)
{
max = result;
//printf("Largest Palindrome: %d\n", max); //927340
}
printf("Largest Palindrome: %d\n", max); //906609
}
}
}
//printf("Largest Palindrome: %d\n", max); //998001
system("PAUSE");
return 0;
} //End of main
//Determines if number is a palindrome
int isPalindrome(int num)
{
int n = num;
int i = 0;
int j = 0;
int k = 0;
int count = 0;
int yes = 0;
//Determines the size of numArray
while(n/10 != 0)
{
n%10;
count++;
n = n/10;
}
int numArray[count];
//Fill numArray with each digit of num
for(i = 0; i <= count; i++)
{
numArray[i] = num%10;
//printf("%d\n", numArray[i]);
num = num/10;
}
//Determines if num is a Palindrome
while(numArray[k] == numArray[count])
{
k = k + 1;
count = count - 1;
yes++;
}
if(yes >= 3)
{
return 0;
}
}//End of Function
I remember doing that problem a while ago and I simply made a is_palindrome() function and brute-forced it. I started testing from 999*999 downwards.
My approach to detect a palindrome was rather different from yours. I would convert the given number to a string and compare the first char with the nth char, second with n-1 and so on.
It was quite simple (and might be inefficient too) but the answer would come up "instantly".
There is no problem in the code in finding the number.
According to your code fragment:
.
.
}
printf("Largest Palindrome: %d\n", max); //906609
}
}
}
//printf("Largest Palindrome: %d\n", max); //998001
system("PAUSE");
.
.
.
you get the result as soon as a palindrome number is found multiplying the number downwards.
You should store the palindrome in a variable max and let the code run further as there is a possibility of finding a greater palindrome further.
Note that for i=800,j=500, then i*j will be greater when compare with i=999,j=100.
Just understand the logic here.
A few issues in the isPalindrome function :
the first while loop doesn't count the number of digits in the number, but counts one less.
as a result, the numArray array is too small (assuming that your compiler supports creating the array like that to begin with)
the for loop is writing a value past the end of the array, at best overwriting some other (possibly important) memory location.
the second while loop has no properly defined end condition - it can happily compare values past the bounds of the array.
due to that, the value in yes is potentially incorrect, and so the result of the function is too.
you return nothing if the function does not detect a palindrome.
//Determines if number is a palindrome
bool isPalindrome(int num) // Change this to bool
{
int n = num;
int i = 0;
int j = 0;
int k = 0;
int count = 1; // Start counting at 1, to account for 1 digit numbers
int yes = 0;
//Determines the size of numArray
while(n/10 != 0)
{
// n%10; <-- What was that all about!
count++;
n = n/10;
}
int numArray[count];
//Fill numArray with each digit of num
for(i = 0; i < count; i++) // This will crash if you use index=count; Array indices go from 0 to Size-1
{
numArray[i] = num%10;
//printf("%d\n", numArray[i]);
num = num/10;
}
//Determines if num is a Palindrome
/*
while(numArray[k] == numArray[count-1]) // Again count-1 not count; This is really bad though what if you have 111111 or some number longer than 6. It might also go out of bounds
{
k = k + 1;
count = count - 1;
yes++;
}
*/
for(k = 1; k <= count; k++)
{
if(numArray[k-1] != numArray[count-k])
return false;
}
return true;
}//End of Function
That's all I could find.
You also need to change this
if(isPalindrome(result) == 0)
To
if(isPalindrome(result))
The Code's Output after making the modifications: Link
The correct printf is the one after the for,after you iterate through all possible values
You are using int to store the value of the palidrome but your result is bigger then 65536,
you should use unsigned
result = i * j;
this pice of code is wrong :
while(n/10 != 0) {
n%10;
count++;
n = n/10;
}
it should be:
while(n != 0) {
count++;
n = n/10;
}
As well as the changes that P.R sugested.
You could do something like this to find out if the number is palindrome:
int isPalindrom(unsigned nr) {
int i, len;
char str[10];
//convert number to string
sprintf(str, "%d", nr);
len = strlen(str);
//compare first half of the digits with the second half
// stop if you find two digits which are not equal
for(i = 0; i < len / 2 && str[i] == str[len - i - 1]; i++);
return i == len / 2;
}
I converted the number to String so I'd be able to go over the number as a char array:
private static boolean isPalindrom(long num) {
String numAsStr = String.valueOf(num);
char[] charArray = numAsStr.toCharArray();
int length = charArray.length;
for (int i = 0 ; i < length/2 ; ++i) {
if (charArray[i] != charArray[length - 1 - i]) return false;
}
return true;
}
I got correct answer for the problem, but I want to know is my coding style is good or bad from my solution. I need to know how can I improve my coding,if it is bad and more generic way.
#include<stdio.h>
#define MAX 999
#define START 100
int main()
{
int i,j,current,n,prev = 0;
for(i = START;i<=MAX;i++)
{
for(j=START;j<=MAX;j++)
{
current = j * i;
if(current > prev) /*check the current value so that if it is less need not go further*/
{
n = palindrome(current);
if (n == 1)
{
printf("The palindrome number is : %d\n",current);
prev = current; // previous value is updated if this the best possible value.
}
}
}
}
}
int palindrome(int num)
{
int a[6],temp;
temp = num;
/*We need a array to store each element*/
a[5] = temp % 10;
a[4] = (temp/10) %10;
a[3] = (temp/100) %10;
a[2] = (temp/1000) %10;
a[1] = (temp/10000) %10;
if(temp/100000 == 0)
{
a[0] = 0;
if(a[1] == a[5] && a[2] == a[4])
return 1;
}
else
{
a[0] = (temp/100000) %10;
if(a[0] == a[5] && a[1] == a[4] && a[2] == a[3])
return 1;
else
return 0;
}
}
No, the largest should be: 906609.
Here is my program:
def reverse_function(x):
return x[::-1]
def palindrome():
largest_num = max(i * x
for i in range(100, 1000)
for x in range(100, 1000)
if str(i * x) == reverse_function(str(i * x)))
return str(largest_num)
print(palindrome())