How to iterate over numbers in C - c

Question

I've gotta count how many times a certain digit is repeated in each number in a range. For example, in the numbers between 0 and 20, there is only one occurrence of 1 being repeated twice (11). I originally did this by converting the int to a str and iterating over it, but I would like to be able to solve this in an arithmetic way. Any ideas?

Answer 1

here is a general solution that you can use , your problem didn't contain much information so I assumed that you want to count the number of repeating of each digit in each number.
so what I did is like hashing where the digits in each number will never cross the value 9 , so they are from 0 to 9 so I made that hash table called arr, so what I did is to come to every single digit in number and increment the position of that digit in arr
for example , number 554 will cause arr[5]++; twice and arr[4]++; only once , simple idea of using hash tables.
and then at the end I iterate over the hash array printing the number of occurrence of each digit in each number.
and here is the code :
#include <stdio.h>
#include <math.h>
int main()
{
int arr[6] = {5555, 12112, 14, -3223, 44, 10001};
int tempArr[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
for (int i = 0; i < 6; i++) {
int temp1 = arr[i];
// get the number of occurrences of each digit in each number
do{
tempArr[(abs(temp1) % 10)]++;
temp1 /= 10;
}while(temp1 != 0);
// loop over the array to know how many times each digit occurred
printf("the number of occurrences in number called %d\n", arr[i]);
for (int j = 0; j < 10; j++) {
if(tempArr[j] > 1)
printf("number %d occurred %d times\n", j, tempArr[j]);
// resetting that position of the array
tempArr[j] = 0;
}
}
return 0;
}
and here is the output :
the number of occurrences in number called 5555
number 5 occurred 4 times
the number of occurrences in number called 12112
number 1 occurred 3 times
number 2 occurred 2 times
the number of occurrences in number called 14
the number of occurrences in number called -3223
number 2 occurred 2 times
number 3 occurred 2 times
the number of occurrences in number called 44
number 4 occurred 2 times
the number of occurrences in number called 10001
number 0 occurred 3 times
number 1 occurred 2 times

Answer 2

You can divide your number multiple times by 10:
int number = 72;
int rest;
while(number)
{
rest = number % 10;
printf("%d\n", rest);
number /= 10;
}
Here rest contains '2' and then '7'

Answer 3

how many times a certain digit is repeated in each number in a range.
Pseudo code*1
Get the range: imin, imax (any int pair where imin <= imax)
Get the digit: digit 0 to 9
Iterate m from imin to imax, inclusive
.... Print m
.... Set digit_count = 0
.... Repeat
....... Last digit ld of m is abs(m%10).
....... If ld == digit, increment digit_count.
........ Divide m by 10
.... Until m == 0
.... Print digit_count
Done
*1 As OP did not provide code, seemed best to not provide a code answer.

Answer 4

Riffing on the answer provided by #Abdo Salm presented only to demonstrate a slightly alternative approach. All credit to Abdo.
EDIT: Going beyond single, arbitrary numbers to generating ascending contiguous sequence of numbers to be evaluated.
#include <stdio.h>
#include <string.h>
#include <limits.h>
void process( int n, int cnts[] ) {
// count occurences of each digit (right to left)
while( n )
cnts[ abs(n % 10) ]++, n /= 10;
}
void report( int cnts[], int thresh ) {
char *sep = "";
for( int j = 0; j < 10; j++ )
if( cnts[ j ] > thresh )
printf( "%s%dx%d", sep, cnts[ j ], j ), sep = ", ";
if( !*sep )
printf( "no digit occurred multiple times" );
putchar( '\n' );
}
int main() {
#ifndef PROVE
int tests[] = {
0, 11, 121, 5555, 12112, 14,
-3223, 44, 1223334444,
INT_MIN, INT_MAX,
};
// proof of concept: run through multiple arbitrary test values
for( int i = 0; i < sizeof tests/sizeof tests[0]; i++ ) {
// counter for each digit, all init'd to zero
int cnts[ 10 ];
memset( cnts, 0, sizeof cnts );
process( tests[ i ], cnts );
// report only digits appearing multiple times
printf( "%11d: ", tests[ i ] );
report( cnts, 1 );
}
#else // with "ranges" instead of arbitrary test cases
int ranges[][2] = {
{ 0, 10, },
{ 0, 22, },
{ 110, 133, },
{ 447, 448, },
};
for( int r = 0; r < sizeof ranges/sizeof ranges[0]; r++ ) {
int metacnts[10];
memset( metacnts, 0, sizeof metacnts );
for( int i = ranges[r][0]; i <= ranges[r][1]; i++ ) {
int cnts[ 10 ];
memset( cnts, 0, sizeof cnts );
process( i, cnts );
for( int j = 0; j < 10; j++ )
if( cnts[ j ] > 1 )
metacnts[ j ]++;
}
// report only digits appearing multiple times in numbers between min & max
printf( "Range %3d-%3d (incl) ", ranges[r][0], ranges[r][1] );
report( metacnts, 0 );
}
#endif
return 0;
}
/*
Output from arbitrary sequence:
0: no digit occurred multiple times
11: 2x1
121: 2x1
5555: 4x5
12112: 3x1, 2x2
14: no digit occurred multiple times
-3223: 2x2, 2x3
44: 2x4
1223334444: 2x2, 3x3, 4x4
-2147483648: 3x4, 2x8
2147483647: 3x4, 2x7
Output from arbitrary ranges:
Range 0- 10 (incl) no digit occurred multiple times
Range 0- 22 (incl) 1x1, 1x2
Range 110-133 (incl) 12x1, 1x2, 1x3
Range 447-448 (incl) 2x4
*/