Advent-of-Code/2021/day8/solution-part2.c

#include <stdio.h>
#include <stdbool.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>

unsigned int countCommon( char larger[8], char smaller[8] ){
	unsigned int common=0;
	for ( unsigned int i = 0; i < strlen(smaller); i++ ){
		if ( strchr( larger, smaller[i] ) != NULL ) common++;
	}
	return common;
}

// This will return true if the signals are the same length and contain the same letters
// Order doesn't matter though so ab if ab and ba are given, it will return true
bool sameSignals( char first[8], char second[8] ){
	if ( strlen(first) == strlen(second) && countCommon(first, second) == strlen(first) ){
		return true;
	}
	return false;
}

unsigned int decodeOutput( char signal[10][8], char output[4][8] ){
	// This will hold the mappings.
	// I.e. the string that should output 0 will be heald in mapping[0]
	char mapping[10][8] = {0};

	// 0 - Length 6
	// 1 - Length 2
	// 2 - Length 5
	// 3 - Length 5
	// 4 - Length 4
	// 5 - Length 5
	// 6 - Length 6
	// 7 - Length 3
	// 8 - Length 7
	// 9 - Length 6

	for ( int i = 0; i < 14; i++ ){
		char *source = ( i < 10 ) ? signal[i] : output[i-10];
		switch (strlen(source)) {
			case 2:
				strcpy(mapping[1],source);
				break;
			case 3:
				strcpy(mapping[7],source);
				break;
			case 4:
				strcpy(mapping[4],source);
				break;
			case 7:
				strcpy(mapping[8],source);
				break;
		}
	}
	
	// When we get here, we know what 1, 7, 4 and 8 are


	for ( int i = 0; i < 14; i++ ){
		char *source = ( i < 10 ) ? signal[i] : output[i-10];
		switch (strlen(source)) {
			// We can determine what 2, 3 and 5 are by looking for numbers with a length of 5
			// If it doesn shares all the same digits as 1, it's a 3
			case 5:
				if ( countCommon( source, mapping[1] ) == 2 ){
					strcpy(mapping[3],source);
				} else if (countCommon( source, mapping[4] ) == 3 ) {
					strcpy(mapping[5],source);
				} else {
					strcpy(mapping[2],source);
				}
				break;
			// We can determine what 0, 6 and 9 are by looking for numbers with a length of 6
			// If it doesn't share the same digits as 1, it's a 6
			// Otherwise, if it shares all 4 segments with 4, it's 9
			// Otherwise it's 0
			case 6:
				if ( countCommon( source, mapping[1] ) != 2 ){
					strcpy(mapping[6],source);
				} else if (countCommon( source, mapping[4] ) == 4 ) {
					strcpy(mapping[9],source);
				} else {
					strcpy(mapping[0],source);
				}
				break;
		}
	}

	unsigned int decoded = 0;
	for ( int i = 0; i < 4; i++ ){
		unsigned int multiplyer = pow(10,(3-i));
		for ( int j = 0; j < 10; j++ ){
			if ( sameSignals(output[i], mapping[j] )){
				decoded += (j * multiplyer);
				break;
			}
		}
	}

	return decoded;


}

int main( int argc, char *argv[] ){
	if( argc == 2 ) {


		FILE *fp=fopen(argv[1], "r");

		// This will hold the number of each count

		unsigned int total=0;

		while (!feof (fp)){
			// each didget has a maximum length of 8 (7 + \0)
			char signal[10][8] = {0};
			char output[4][8] = {0};

			char ch;
			char str[8];
			for(unsigned int i = 0; 1 ; i++ ){
				if ( i < 10 ){
					fscanf(fp, "%s%*c", signal[i]);
				} else if (i == 10){
					// This is the pipe character - we can discard it
					char pipe;
					fscanf(fp, "%c%*c", &pipe);
				} else if ( i < 15 ){
					fscanf(fp, "%s%*c", output[i-11]);
				} else {
					break;
				}

			}

			unsigned int decoded = decodeOutput( signal, output );
			printf("%i\n",decoded);
			total += decoded;


		}

		printf("Total: %i\n", total);



		fclose(fp);


		return 0;
	} else {
		printf("You need to provide a file\n");
		return 1;
	}
}