Looping to look for non-repeated digits - c

Question

I have a C program that I am working on which entails to find the first nonrepeated, last nonrepeated digit in an array. I am lost in how to accomplish this. Can anybody help me with the loop control variable to get those two things done ? My loop keeps returning a negative number.
Here is what I have:
for(p = 0; p <= i;p ++ ){
for(u=p+1;u>=1;u--){
if(digits[p] > digits[u]);
firstnon = digits[p];
}
}
if(firstnon){
printf("The first non-repeated digit is: %d",firstnon);
}
else{
printf("There isn't any non-repeated digits");
}
printf("\n");
for(c=0;c<= sizeof(digits)/sizeof(int);c++){
for(k=1;k<sizeof(digits);k++){
if(digits[c]==digits[k]){
lastnon = digits[k];
}
}
}
if(c==k){
printf("The last non-repeated digit is: %d\n",lastnon);
}

Answer 1

Have some problems in your code:
This line: if(digits[p] > digits[u]); don't do what you expect i think, see the last char ;, this kill the propose of the if statement. And the following line don't belong to the if block.
This sizeof(digits) return the size of digits in bytes, if the elements of digits are not bytes, you would be indexing the array with out of bound index (with k).
I don't understand really the search for last non-repeating digit, you are setting lastnon when digits[c] == digits[k], i don't think that it correct if i understand correctly the problem statement.
Here is a naive approach:
#include <stdio.h>
int main() {
int p, u, firstnon = -1, lastnon = -1, digit_repeat;
int digits[10] = { 1, 2, 3, 1, 3, 5, 1, 2, 4, 3};
int digits_size = sizeof(digits) / sizeof(digits[0]);
for (p = 0; p < digits_size; p++) {
digit_repeat = 0;
for (u = 0; u < digits_size; u++) {
if (p != u && digits[p] == digits[u]) {
digit_repeat = 1;
break;
}
}
if (!digit_repeat) {
firstnon = p;
break;
}
}
if (firstnon != -1) {
printf("The first non-repeated digit is: %d", digits[firstnon]);
} else {
printf("There isn't any non-repeated digits");
}
printf("\n");
for (p = digits_size - 1; p >= 0; p--) {
digit_repeat = 0;
for (u = 0; u < digits_size; u++) {
if (p != u && digits[p] == digits[u]) {
digit_repeat = 1;
break;
}
}
if (!digit_repeat) {
lastnon = p;
break;
}
}
if (lastnon != -1) {
printf("The last non-repeated digit is: %d", digits[lastnon]);
} else {
printf("There isn't any non-repeated digits");
}
printf("\n");
return 0;
}
To find the first non repeating digits.
Iterate through the digits.
In every digits, check the other digits (all) for the same digit in another position.
If there is another, continue
If there only one, saved, and break search.
The same apply for last non repeating digits, but the iteration is in reverse.
Some improve version.
#include <stdio.h>
int main() {
int p;
int digits[10] = { 1, 2, 3, 1, 3, 5, 1, 2, 4, 3};
int digits_size = sizeof(digits) / sizeof(digits[0]);
int used_digits[10] = { 0 };
int used_digits_size = sizeof(used_digits) / sizeof(used_digits[0]);
for (p = 0; p < digits_size; p++) {
used_digits[digits[p]]++;
}
for (p = 0; p < used_digits_size; p++) {
if (used_digits[p] == 1) {
printf("The first non-repeated digit is: %d", p);
break;
}
}
if (p == used_digits_size) {
printf("There isn't any non-repeated digits");
}
printf("\n");
for (p = used_digits_size - 1; p >= 0; p--) {
if (used_digits[p] == 1) {
printf("The last non-repeated digit is: %d", p);
break;
}
}
if (p == -1) {
printf("There isn't any non-repeated digits");
}
printf("\n");
return 0;
}
This version use an array to storage the usage of digits. The non-repeating digits are the one with usage == 1. Search this forward and backward to find first and last non repeating digit.