Counting and deleting repeated digits of array elements - arrays

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

Related

Problem inserting a 32-digit number into an array

I am trying to write a C program to store a 32-bit number into an Array. For Example, the number: 11000001110010000000000000000000 In the array, arr[0] would be 1 as that is the first digit.
However I am unable to get the desired output. This is my code:
#include<stdio.h>
int main ()
{
int binarynumber;
int arr[32];
printf("Enter A binary Number:\n");
scanf("%d", &binarynumber);
for (int i = 32; i >= 0; i--)
{
arr[i] = binarynumber % 10;
binarynumber /= 10;
}
printf("The first digit is %d", arr[0]);
}
If I were you, I'd read it as a string and iterate through each character and added (and converted) them to the int array. As mentioned, typing the 32 digit long number is too big to store inside an int.
I guess bitwise operations should do the job
#include<stdio.h>
int main ()
{
int binarynumber;
int arr[32];
printf("Enter A binary Number:\n");
scanf("%d", &binarynumber);
int array_size = sizeof(arr)/sizeof(arr[0])
for (int i = 0, j = array_size - 1; i < array_size; i++, j--)
{
int binarynumber_copy = binarynumber;
// shifting number `j` bits right and cutting off one bit by bitwise AND
arr[i] = (binarynumber_copy >> j) & 1;
}
printf("The first digit is %d", arr[0]);
}
First of all if you want user to input 11000001110010000000000000000000, then you can't get that whole number in an integer. So, if the user enters 3251109888 (decimal of 11000001110010000000000000000000), then better to get the number as unsigned int. Also you need to convert it to binary in the loop using %2 and /2. The loop will run from 0 to 31, so that arr[0] contains the LSB. The code will look like this -
int main ()
{
unsigned int number;
int arr[32];
printf("Enter A binary Number:\n");
scanf("%u", &number);
for (int i = 0; i < 32; i++)
{
arr[i] = number % 2;
number /= 2;
}
for (int i = 31; i >= 0; i--) {
printf("%d", arr[i]);
}
printf("\n");
return 0;
}
If you want the user to input as 11000001110010000000000000000000, you need to get that input as string and then convert it into the digit using loop. To do that you can use -
int main ()
{
int arr[32];
char number[33] = {};
scanf("%32s", number); // This will scan max 32 characters
for (int i = 0; i < 32; i++)
{
arr[i] = number[i] - '0';
}
printf("arr[0]: %d\n", arr[0]);
for (int i = 0; i < 32; i++) {
printf("%d", arr[i]);
}
printf("\n");
return 0;
}
You may want to validate whether the input contains only digit or not using isdigit() inside the loop

Program to print sum of primes in C

#include <stdio.h>
#include <math.h>
int main() {
int n, count, sum;
printf("Enter upper bound n \n");
scanf("%d", &n);
for (int a = 1; a <= n; a++) {
count = 0;
sum = 0;
for (int i = 2; i <= sqrt(a); ++i) {
if (a % i == 0) {
count++;
break;
}
}
if (count == 0 && a != 1) {
sum = a + sum;
}
}
printf("%d", sum);
}
The program is my attempt to print summation of primes < n. I am getting sum = 0 every time and I am unable to fix this issue.
The reason you do not get the sum of primes is you reset the value of sum to 0 at the beginning of each iteration. sum will be 0 or the value of the n if n happens to be prime.
Note also that you should not use floating point functions in integer computations: i <= sqrt(a) should be changed to i * i <= a.
The test on a != 1 can be removed if you start the loop at a = 2.
Here is a modified version:
#include <stdio.h>
int main() {
int n = 0, sum = 0;
printf("Enter upper bound n: \n");
scanf("%d", &n);
// special case 2
if (n >= 2) {
sum += 2;
}
// only test odd numbers and divisors
for (int a = 3; a <= n; a += 2) {
sum += a;
for (int i = 3; i * i <= a; i += 2) {
if (a % i == 0) {
sum -= a;
break;
}
}
}
printf("%d\n", sum);
return 0;
}
For large values of n, a much more efficient approach would use an array and perform a Sieve of Eratosthenes, a remarkable greek polymath, chief librarian of the Library of Alexandria who was the first to compute the circumference of the earth, 2300 years ago.
Here is an improved version:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]) {
int n = 0;
long long sum = 0;
if (argc > 1) {
sscanf(argv[1], "%i", &n);
} else {
printf("Enter upper bound n: \n");
scanf("%i", &n);
}
// special case 2
if (n >= 2) {
sum += 2;
}
unsigned char *p = calloc(n, 1);
for (int a = 3; a * a <= n; a += 2) {
for (int b = a * a; b < n; b += a + a) {
p[b] = 1;
}
}
for (int b = 3; b < n; b += 2) {
sum += p[b] * b;
}
free(p);
printf("%lld\n", sum);
return 0;
}
Error about sum getting set to zero inside the loop has been already pointed out in previous answers
In current form also, your code will not return zero always. It will return zero if value of upper bound is given as non prime number. If prime number is given as upper bound, it will return that number itself as sum.
As mentioned in comment you should initialize sum before first loop something like
int n, count, sum=0;
or you can initialize sum in the loop like
for(a=1,sum=0;a <= n; a++)
and remove sum=0; inside the first loop because it changes sum to 0 every time first loop executes. You can check this by inserting this lines to your code
printf("Before sum %d",sum);
sum = 0;
printf("After Sum %d",sum);
make sure sure that if you are initializing sum in the loop, define "a" in outer of the loop if not the sum goes to local variable to for loop and it hides the outer sum.

for loop unexpectedly jumping down in value

Goldbach's conjecture states that every even integer over 4 is the sum of two primes, I am writing a program in C to find these pairs. To do this it first finds all the primes less than a user given number. I have a for loop to iterate from 4 to the user given number and find the pairs within the loop body. When that loop gets to about around 40, suddenly jumps back down by about 30 and then continues to iterate up (with user input 50 it jumped from 38 to 9, with input 60 it jumped from 42 to 7). I can't figure out why this is happening. Here is my code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <sys/types.h>
#include <unistd.h>
struct pair{
int a;
int b;
}pair_t;
int main(){
int N;
int numPrimes = 1;
int *primes = malloc(100*sizeof(int));
int isPrime = 1;
primes[0] = 2;
int timesRealloc = 0;
int availableSlots = 100;
printf("Please enter the largest even number you want to find the Goldbach pair for: \n");
scanf("%d", &N);
struct pair pairs[N/2 + 4];
int j = 0;
int i;
for (i = 3; i <= N; i+=2){
j = 0;
isPrime = 1;
while (primes[j] <= sqrt(i)) {
if (i%primes[j] == 0) {
isPrime = 0;
break;
}
j++;
}
if (isPrime == 1){
primes[numPrimes] = i;
numPrimes++;
}
if (availableSlots == numPrimes){
timesRealloc++;
availableSlots += 100;
primes = realloc(primes, availableSlots*sizeof(int));
}
}
printf("The largest prime I found was %d\n", primes[(numPrimes-1)]);
int k;
for (i=4; i<=N; i+=2){
printf("i is %d, N is %d\n", i, N);
if (i > N){ break; }
for (j=0; j<numPrimes; j++){
for (k=0; k<numPrimes; k++){
int sum = primes[j] + primes[k];
if(sum == i){
pairs[i].a = primes[j];
pairs[i].b = primes[k];
}
}
}
}
for (i=4; i<=N; i+=2){
printf("%d is the sum of %d and %d\n", i, pairs[i].a, pairs[i].b);
}
return 0;
}
You attempt to be space efficient by compressing the pairs array to just hold every other (even) number and start from 4 instead of zero. However, you miscalculate its size and then when you go to use it, you treat it like it hasn't been compressed and that there's a slot for every natural number.
The code suffers from having the prime array calculation in main() along with the other code, this is best separated out. And when it looks for pairs, it doesn't quit when it finds one, nor when it starts getting sums greater than the target. My rework below attempts to address all of these issues:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define INITIAL_SLOTS (100)
struct pair {
int a;
int b;
} pair_t;
int compute_primes(int limit, unsigned **primes, int size) {
int numPrimes = 0;
(*primes)[numPrimes++] = 2;
for (int i = 3; i <= limit; i += 2) {
bool isPrime = true;
for (int j = 0; (*primes)[j] <= i / (*primes)[j]; j++) {
if (i % (*primes)[j] == 0) {
isPrime = false;
break;
}
}
if (isPrime) {
(*primes)[numPrimes++] = i;
}
if (numPrimes == size) {
size *= 2;
*primes = realloc(*primes, size * sizeof(unsigned));
}
}
return numPrimes;
}
int main() {
int N;
printf("Please enter the largest even number you want to find the Goldbach pair for: \n");
scanf("%d", &N);
unsigned *primes = calloc(INITIAL_SLOTS, sizeof(unsigned));
int numPrimes = compute_primes(N, &primes, INITIAL_SLOTS);
printf("The largest prime I found was %d\n", primes[numPrimes - 1]);
struct pair pairs[(N - 4) / 2 + 1]; // compressed data structure
for (int i = 4; i <= N; i += 2) {
int offset = (i - 4) / 2; // compressed index
bool found = false;
for (int j = 0; ! found && j < numPrimes; j++) {
for (int k = 0; ! found && k < numPrimes; k++) {
int sum = primes[j] + primes[k];
if (sum == i) {
pairs[offset].a = primes[j];
pairs[offset].b = primes[k];
found = true;
} else if (sum > i) {
break;
}
}
}
}
for (int i = 4; i <= N; i += 2) {
int offset = (i - 4) / 2; // compressed index
printf("%d is the sum of %d and %d\n", i, pairs[offset].a, pairs[offset].b);
}
free(primes);
return 0;
}
OUTPUT
> ./a.out
Please enter the largest even number you want to find the Goldbach pair for:
10000
The largest prime I found was 9973
4 is the sum of 2 and 2
6 is the sum of 3 and 3
8 is the sum of 3 and 5
10 is the sum of 3 and 7
12 is the sum of 5 and 7
14 is the sum of 3 and 11
...
9990 is the sum of 17 and 9973
9992 is the sum of 19 and 9973
9994 is the sum of 53 and 9941
9996 is the sum of 23 and 9973
9998 is the sum of 31 and 9967
10000 is the sum of 59 and 9941
>

Return which digit occurs the most times

I have to return which digit in a number occurs the most frequently ( though not how many times it occurs )
So far I can only get this, I don't know how to isolate the digit, only to show how many times each digit occurs.
#include <stdio.h>
int frequentDigit(int);
int main()
{
frequentDigit(123032333);
return 0;
}
int frequentDigit(int arg)
{
int tmp; int i; int myArr[9] = { 0 };
tmp = (arg < 0) ? -arg : arg;
do
{
myArr[tmp % 10]++;
tmp /= 10;
} while (tmp != 0);
for (i = 0; i < 9; i++) { printf("\nThere are %d occurances of digit %d", myArr[i], i); }
}
The array where you are storing the frequency of the digits, i.e myArr[]. Its suppose to hold the frequency of all the number from 0...9. And since there are 10 numbers, you would need an array of lenght 10.
int myArr[10];
Later, you need to traverse through the array once, checking for the max element, and saving the index accordingly, to find which number has occured most number of times.
To traverse, the for loop should go till 9
for (i = 0; i <= 9; i++)
Edited
As someone commented, you can find the max value while you are computing the frequencies itself.
int max = -1, max_num = -1;
do
{
myArr[tmp % 10]++;
if( myArr[tmp % 10] > max)
{
max = myArr[tmp % 10];
max_num = tmp % 10;
}
tmp /= 10;
} while (tmp != 0);
printf("%d", max_num);
Its simple. At the end of your code you have an array of frequencies, if you find the max of that you get the most common element
Just use a loop to find the max and print that:
int max = myArr[0]; // start with max = first element
int max_position=0;
for(int i = 1; i<9; i++)
{
if(myArr[i] > max){
max = myArr[i];
max_position=i;
}
}
printf("\The max is %d occuring %d times ", max_position, max_position)

Most frequent element in a sequence using arrays in C

I'm doing an online course on "Programming, Data Structure & Algorithm". I've been given an assignment to "find the most frequent element in a sequence using arrays in C (with some constraints)". They've also provided some test-cases to verify the correctness of the program. But I think I'm wrong somewhere.
Here's the complete question from my online course.
INPUT
Input contains two lines. First line in the input indicates N,
the number of integers in the sequence. Second line contains N
integers, separated by white space.
OUTPUT
Element with the maximum frequency. If two numbers have the
same highest frequency, print the number that appears first in the
sequence.
CONSTRAINTS
1 <= N <= 10000
The integers will be in the range
[-100,100].
And here's the test cases.
Test Case 1
Input:
5
1 2 1 3 1
Output:
1
Input:
6
7 7 -2 3 1 1
Output:
7
And here's the code that I've written.
#include<stdio.h>
int main()
{
int counter[201] = {0}, n, i, input, maximum = 0;
scanf("%d", &n);
for(i = 1; i <= n; i++) {
scanf("%d", &input);
if(input < -100 && input < 100)
++counter[input];
}
maximum = counter[0];
for (i = 1; i < 201; i++) {
if (counter[i] > maximum) {
maximum = counter[i];
}
}
printf("%d", maximum);
return 0;
}
Please tell me where I'm wrong. Thank you.
EDIT:
I've modified the code, as suggested by #zoska. Here's the working code.
#include<stdio.h>
int main()
{
int counter[201] = {0}, n, i, input, maximum = 0;
scanf("%d", &n);
for(i = 1; i <= n; i++) {
scanf("%d", &input);
if(input < 100 && input > 0)
++counter[input + 100];
else
++counter[input];
}
maximum = counter[0];
for (i = 0; i < 201; i++) {
if (counter[i] > maximum) {
maximum = i - 100;
}
}
printf("%d", maximum);
return 0;
}
Additionally to problem pointed out by Paul R is:
You are printing maximum occurrences of number, not the number itself.
You're going to need another variable, which will store the number with maximum occurences. Like :
maximum = count[0];
int number = -100;
for (i = 0; i < 201; i++) {
if (counter[i] > maximum) {
maximum = counter[i];
number = i - 100;
}
}
printf("number %d has maximum occurences: %d", number, maximum);
Also you should iterate through an array from 0 to size-1:
So in all cases of your loops it should be :
for(i = 0; i < 201; i++)
Otherwise you won't be using count[0] and you will only have a range of -99...100.
Try below code
#include<stdio.h>
int main()
{
int counter[201] = {0}, n, i, input, maximum = 0;
scanf("%d", &n);
for(i = 1; i <= n; i++) {
scanf("%d", &input);
if(input >= -100 && input <= 100)
++counter[input + 100];
}
maximum = counter[0];
int index = 0;
for (i = 0; i < 201; i++) {
if (counter[i] >= maximum) {
index = i;
maximum = counter[i];
}
}
printf("number %d occured %d times\n", index-100, maximum);
return 0;
}
I would prefer checking in one loop itself for the maximum value just so that the first number is returned if i have more than one element with maximum number of occurances.
FInd the code as:
#include<stdio.h>
int main()
{
int n,input;
scanf("%d",&n);
int count[201] ={0};
int max=0,found=-1;
for(int i=0;i<n;i++)
{
scanf("%d",&input);
count[input+100]++;
if(max<count[input+100])
{
max= count[input+100];
found=input;
}
}
printf("%d",found);
return 0;
}
But, there is also one condition that if the number of occurance are same for two numbers then the number which appers first in sequence should appear.

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