finding how many times number2 is showing in number1 - c

Question

I am solving an exercise in C and I got stuck. I don't know the logic of the code to get to my solution. For example we enter 2 numbers from input let the numbers be 123451289 and 12 and I want to see how many times number 2 is showing at number 1 (if this is confusing let me know). For the numbers earlier the program outputs 2. I tried solving it here is my code:
#include <stdio.h>
int main() {
int num1, num2, counter = 0;
scanf("%d%d", num1, num2);
if (num1 < num2) {
int temp = num1;
num1 = num2;
num2 = temp;
}
int copy1 = num1;
int copy2 = num2;
while (copy2 > 0) {
counter++; // GETTING THE LENGHT OF THE SECOND NUMBER
copy2 /= 10;
// lastdigits = copy1 % counter //HERE I WANT TO GET THE LAST DIGITS OF THE FIRST NUMBER
// But it does not work
}
}
My question is how can I get the last digits of the first number according to the second one for example if the second number have 3 digits I want to get the last 3 digits of the first number. For the other part I think I can figure it out.
I must solve this problem WITHOUT USING ARRAYS.

Answer 1

The problem: find all the needles (e.g. 12) in a haystack (e.g. 123451289).
This can be done simply without arrays using a modulus of the needle. For 12, this is 100. That is, 12 is two digits wide. Using the modulus, we can
isolate the rightmost N digits of the haystack and compare them against the needle.
We "scan" haystack repeatedly by dividing by 10 until we reach zero.
Here is the code:
#include <stdio.h>
int
main(void)
{
int need, hay, counter = 0;
scanf(" %d %d", &hay, &need);
// ensure that the numbers are _not_ reversed
if (hay < need) {
int temp = need;
need = hay;
hay = temp;
}
// get modulus for needle (similar to number of digits)
int mod = 1;
for (int copy = need; copy != 0; copy /= 10)
mod *= 10;
// search haystack for occurences of needle
// examine the rightmost "mod" digits of haystack and check for match
// reduce haystack digit by digit
for (int copy = hay; copy != 0; copy /= 10) {
if ((copy % mod) == need)
++counter;
}
printf("%d appears in %d exactly %d times\n",need,hay,counter);
return 0;
}
UPDATE:
I'm afraid this does not work for 10 0. –
chqrlie
A one line fix for to the modulus calculation for the 10/0 case. But, I've had to add a special case for the 0/0 input.
Also, I've added a fix for negative numbers and allowed multiple lines of input:
#include <stdio.h>
int
main(void)
{
int need, hay, counter;
while (scanf(" %d %d", &hay, &need) == 2) {
counter = 0;
// we can scan for -12 in -1237812
if (hay < 0)
hay = -hay;
if (need < 0)
need = -need;
// ensure that the numbers are _not_ reversed
if (hay < need) {
int temp = need;
need = hay;
hay = temp;
}
// get modulus for needle (similar to number of digits)
int mod = need ? 1 : 10;
for (int copy = need; copy != 0; copy /= 10)
mod *= 10;
// search haystack for occurences of needle
// examine the rightmost "mod" digits of haystack and check for match
// reduce haystack digit by digit
for (int copy = hay; copy != 0; copy /= 10) {
if ((copy % mod) == need)
++counter;
}
// special case for 0/0 [yecch]
if ((hay == 0) && (need == 0))
counter = 1;
printf("%d appears in %d exactly %d times\n", need, hay, counter);
}
return 0;
}
Here is the program output:
12 appears in 123451289 exactly 2 times
0 appears in 10 exactly 1 times
0 appears in 0 exactly 1 times
UPDATE #2:
Good fixes, including tests for negative numbers... but I'm afraid large numbers still pose a problem, such as 2000000000 2000000000 and -2147483648 8 –
chqrlie
Since OP has already posted an answer, this is bit like beating a dead horse, but I'll take one last attempt.
I've changed from calculating a modulus of needle into calculating the number of digits in needle. This is similar to the approach of some of the other answers.
Then, the comparison is now done digit by digit from the right.
I've also switched to unsigned and allow for the number to be __int128 if desired/supported with a compile option.
I've added functions to decode and print numbers so it works even without libc support for 128 bit numbers.
I may be ignoring [yet] another edge case, but this is an academic problem (e.g. we can't use arrays) and my solution is to just use larger types for the numbers. If we could use arrays, we'd keep things as strings and this would be similar to using strstr.
Anyway, here's the code:
#include <stdio.h>
#ifndef NUM
#define NUM long long
#endif
typedef unsigned NUM num_t;
FILE *xfin;
int
numget(num_t *ret)
{
int chr;
num_t acc = 0;
int found = 0;
while (1) {
chr = fgetc(xfin);
if (chr == EOF)
break;
if ((chr == '\n') || (chr == ' ')) {
if (found)
break;
}
if ((chr >= '0') && (chr <= '9')) {
found = 1;
acc *= 10;
chr -= '0';
acc += chr;
}
}
*ret = acc;
return found;
}
#define STRMAX 16
#define STRLEN 100
const char *
numprt(num_t val)
{
static char strbuf[STRMAX][STRLEN];
static int stridx = 0;
int dig;
char *buf;
buf = strbuf[stridx++];
stridx %= STRMAX;
char *rhs = buf;
do {
if (val == 0) {
*rhs++ = '0';
break;
}
for (; val != 0; val /= 10, ++rhs) {
dig = val % 10;
*rhs = dig + '0';
}
} while (0);
*rhs = 0;
if (rhs > buf)
--rhs;
for (char *lhs = buf; lhs < rhs; ++lhs, --rhs) {
char tmp = *lhs;
*lhs = *rhs;
*rhs = tmp;
}
return buf;
}
int
main(int argc,char **argv)
{
num_t need, hay, counter;
--argc;
++argv;
if (argc > 0)
xfin = fopen(*argv,"r");
else
xfin = stdin;
while (1) {
if (! numget(&hay))
break;
if (! numget(&need))
break;
counter = 0;
// we can scan for -12 in -1237812
if (hay < 0)
hay = -hay;
if (need < 0)
need = -need;
// ensure that the numbers are _not_ reversed
if (hay < need) {
num_t temp = need;
need = hay;
hay = temp;
}
// get number of digits in needle (zero has one digit)
int ndig = 0;
for (num_t copy = need; copy != 0; copy /= 10)
ndig += 1;
if (ndig == 0)
ndig = 1;
// search haystack for occurences of needle
// starting from the right compare digit-by-digit
// "shift" haystack right on each iteration
num_t hay2 = hay;
for (; hay2 != 0; hay2 /= 10) {
num_t hcopy = hay2;
// do the rightmost ndig digits match in both numbers?
int idig = ndig;
int match = 0;
for (num_t need2 = need; idig != 0;
--idig, need2 /= 10, hcopy /= 10) {
// get single current digits from each number
int hdig = hcopy % 10;
int ndig = need2 % 10;
// do they match
match = (hdig == ndig);
if (! match)
break;
}
counter += match;
}
// special case for 0/0 et. al. [yecch]
if (hay == need)
counter = 1;
printf("%s appears in %s exactly %s times\n",
numprt(need), numprt(hay), numprt(counter));
}
return 0;
}
Here's the program output:
12 appears in 123451289 exactly 2 times
123 appears in 123451289 exactly 1 times
1234 appears in 123451289 exactly 1 times
1 appears in 123451289 exactly 2 times
0 appears in 10 exactly 1 times
0 appears in 0 exactly 1 times
1000000000 appears in 1000000000 exactly 1 times
2000000000 appears in 2000000000 exactly 1 times

Answer 2

This looks along the lines of what you're attempting.
You can use the pow() function from math.h to raise 10 to the power of how many digits you need for your modulus operation.
Compile with -lm or make your own function to calculate 10^num_digits
#include <stdio.h>
#include <math.h>
int main() {
int x = 123456789;
double num_digits = 3.0;
int last_digits = x % (int)pow(10.0, num_digits);
printf("x = %d\nLast %d Digits of x = %d\n", x, (int)num_digits, last_digits);
return 0;
}
Outputs:
x = 123456789
Last 3 Digits of x = 789

Answer 3

I think you are trying to ask :- if number1 = 1234567 and number2 = 673, then, length of number2 or number2 has 3 digits, so, you now want the last 3 digits in number1, i.e, '456', if I'm not wrong.
If that is the case, then, what you did to find the number of digits in num2 is correct, i.e,
while (copy2>0) {
counter++; // GETTING THE LENGHT OF THE SECOND NUMBER
copy2/=10;
}
you can do the same for number1 and find out its number of digits, then you can compare whether the number of digits in number2 is less than that in number1. Ex, 3 is less than number of digits in number1, so you can proceed further. Let's say number of digits in number1 is 7 and you want the last 3 digits, so you can do iterate over the digits in number1 till count of digits in number2 and pop out each last digit and store them in an array.
The code:
#include <stdio.h>
int main()
{
int num1,num2;
int count1 = 0, count2 = 0;
scanf("%d",&num1);
scanf("%d",&num2);
if(num1<num2){
int temp = num1;
num1 = num2;
num2 = temp;
}
int copy1 = num1;
int copy2 = num2;
while (copy1>0)
{
count1++;
copy1/=10;
}
while (copy2>0)
{
count2++;
copy2/=10;
}
// printf("num1 has %d digits and num2 has %d digits\n", count1, count2);
if (count1 >= count2)
{
int arr[count2];
int x = count2;
int p = num1;
int i = 0;
while (x > 0)
{
arr[i++] = p%10;
x --;
p/=10;
}
for (int j = 0; j < i; j++)
{
printf("%d ", arr[j]);
}
}
return 0;
}
output : 8 7 6
let's say, num1 = 12345678, num2 = 158, then arr = {8,7,6}.

Answer 4

You must determine the number of digits N of num2 and test if num1 ends with num2 modulo 10N.
Note these tricky issues:
you should not sort num1 and num2: If num2 is greater than num1, the count is obviously 0.
num2 has at least 1 digit even if it is 0.
if num1 and num2 are both 0, the count is 1.
if num2 is greater then INT_MAX / 10, the computation for mod would overflow, but there can only be one match, if num1 == num2.
it is unclear whether the count for 1111 11 should be 2 or 3. We will consider all matches, including overlapping ones.
to handle larger numbers, we shall use unsigned long long instead of int type.
Here is a modified version:
#include <limits.h>
#include <stdio.h>
int main() {
int counter = 0;
unsigned long long num1, num2;
if (scanf("%llu%llu", &num1, &num2) != 2) {
printf("invalid input\n");
return 1;
}
if (num1 == num2) {
/* special case for "0 0" */
counter = 1;
} else
if (num1 > num2 && num2 <= ULLONG_MAX / 10) {
unsigned long long copy1 = num1;
unsigned long long mod = 10;
while (mod < num2) {
mod *= 10;
}
while (copy1 > 0) {
if (copy1 % mod == num2)
counter++;
copy1 /= 10;
}
}
printf("count=%d\n", counter);
return 0;
}
Note that leading zeroes are not supported in either number: 101 01 should produce a count of 1 but after conversion by scanf(), the numbers are 101 and 1 leading to a count of 2. It is non trivial to handle leading zeroes as well as numbers larger than ULLONG_MAX without arrays.

Answer 5

This was the answer that i was looking for but thank you all for helping :)
#include <stdio.h>
#include <math.h>
int main(){
int num1,counter1,counter2,num2,temp,digit,copy1,copy2;
scanf("%d%d",&num1,&num2);
if(num1<num2){
temp = num1;
num1 = num2;
num2 = temp;
}
copy1 = num1;
copy2 = num2;
counter1 = counter2 = 0;
while (copy2>0) {
counter1++;
copy2/=10;
}
counter1 = pow(10,counter1);
if(num1> 1 && num2>1)
while (copy1>0) {
digit = copy1%counter1;
if(digit==num2){
counter2++;
}
copy1/=10;
} else{
if(num2<1){
while (copy1>0) {
digit = copy1%10;
if(digit==copy2){
counter2++;
}
copy1/=10;
}
}
}
printf("%d",counter2);
}