Finally figured out 17 in C

1 files changed, 103 insertions, 0 deletions

diff --git a/17-Number-Letter-Counts/wordnums.c b/17-Number-Letter-Counts/wordnums.c
new file mode 100644
index 0000000..e1eaeba
--- /dev/null
+++ b/17-Number-Letter-Counts/wordnums.c
@@ -0,0 +1,103 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+// Problem 17 - Number Letter Counts
+// So I found the string approach to be a pain in the ass in C (see Python solution)
+// But then I realized, hey, I just need the lengths, so you don't have to pass the string!
+const char *ones[10] = {"","one", "two", "three", "four",
+    "five", "six", "seven", "eight", "nine"};
+const char *teens[10] = {"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", 
+    "sixteen", "seventeen", "eighteen", "nineteen"};
+const char *tens[10] = {"","", "twenty", "thirty", "forty",
+    "fifty", "sixty", "seventy", "eighty", "ninety"};
+const char nd[4] = "AND";
+
+
+int lengthOfName(char *num) {
+    char *word; 
+    // minus 1 for null terminator
+    int digits = strlen(num) - 1;
+    int n = atoi(num);
+    if(digits == 1) {
+        word = (char *) ones[n];
+        printf("%s\n", word);
+        return strlen(word);
+    } else if(digits == 2) {
+        // if the first digit, ie. the tens column is 1
+        if(num[0] == '1'){
+            word = (char *) teens[n % 10];
+            printf("%s\n", word);
+            return strlen(word);
+        }
+        // assuming the first digit is not one, if the second digit is 0
+        else if(num[1] == '0'){
+            word = (char *) tens[n / 10];
+            printf("%s\n", word);
+            return strlen(word);
+        } else {
+            // since the input to the function expects a string, 
+            // an extra '\n' is needed
+            char o[2];
+            int on;
+            word = (char *) tens[n / 10];
+            printf("%s", word);
+            memcpy(o, &num[1], 2);
+            on = lengthOfName(o);
+            return (strlen(word) + on);
+        }
+    } else if(digits == 3) {
+        word = (char *) ones[n / 100];
+        printf("%shundred", word);
+        if(num[1] == '0' && num[2] == '0') {
+            printf("\n");
+            return strlen(word) + strlen("hundred");
+        } else {
+            char t[3];
+            int te;
+            strncpy(t, &num[1], 3);
+            printf("%s", nd);
+            te = lengthOfName(t);
+            // example: the user inputs, num = 121
+            //      one            hundred              AND     twenty one
+            return (strlen(word) + strlen("hundred") + strlen(nd) + te);
+        }
+    }  else if(digits == 4) {
+        word = (char *) ones[n / 1000];
+        printf("%sthousand", word);
+        if(num[1] == '0' && num[2] == '0' && num[3] == '0') {
+            printf("\n");
+            return strlen(word) + strlen("thousand");
+        } /*else {
+            char h[4];
+            int hu;
+            strncpy(h, &num[1], 3);
+            hu = lengthOfName(h);
+            return (strlen(word) + strlen("thousand") + hu);
+        }*/
+    }
+
+    // error condition for debugging, i.e. in case I messed up, return something
+    return -1;
+}
+
+int main() {
+    // 4 is the max for this example, +1 for null terminator or if user tries to cheat
+    char n[5];
+    int out = 0;
+    printf("Number from 1 to 1000: ");
+    fgets(n, 10, stdin);
+    if(atoi(n) > 1000)
+        return printf("only up to 1000 in this example, sorry :(\n");
+    // Test code to make sure the function works
+    //out = lengthOfName(n);
+    //printf("%d Characters... wow you'll hurt your hands typing that!\n", out);
+    
+    for(int i = 1; i <= atoi(n); i++){
+        char curr[5];        
+        sprintf(curr, "%d\n", i);
+        out += lengthOfName(curr);
+    }
+    printf("Sum of charachters: %d\n", out);
+    return 0;
+}