feat: add Letter combinations of phone book problem (#1221)

* feat: add Letter combinations of phone book problem (#17)

* fix: add newline at the end of the file

* fix: add brief description of the algorithm

---------

Co-authored-by: David Leal <halfpacho@gmail.com>

leetcode/DIRECTORY.md

@@ -20,6 +20,7 @@
 |   13 | [Roman to Integer](https://leetcode.com/problems/roman-to-integer)                                                                                     | [C](./src/13.c)   | Easy       |
 |   14 | [Longest Common Prefix](https://leetcode.com/problems/longest-common-prefix)                                                                           | [C](./src/14.c)   | Easy       |
 |   16 | [3Sum Closest](https://leetcode.com/problems/3sum-closest)                                                                                             | [C](./src/16.c)   | Medium     |
+|   17 | [Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number)                                           | [C](./src/17.c)   | Medium     |
 |   20 | [Valid Parentheses](https://leetcode.com/problems/valid-parentheses)                                                                                   | [C](./src/20.c)   | Easy       |
 |   21 | [Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists)                                                                         | [C](./src/21.c)   | Easy       |
 |   24 | [Swap Nodes in Pairs](https://leetcode.com/problems/swap-nodes-in-pairs)                                                                               | [C](./src/24.c)   | Medium     |

leetcode/src/17.c

@@ -0,0 +1,78 @@
+/**
+ * Letter Combinations of a Phone Number problem
+ * The algorithm determines the final size of the return array (combs) and allocates
+ * corresponding letter for each element, assuming that the return array is alphabetically sorted.
+ * It does so by running two loops for each letter:
+ *   - the first loop determines the starting positions of the sequence of subsequent letter positions
+ *   - the second loop determines the length of each subsequent sequence for each letter
+ * The size and space complexity are both O("size of final array"), as even though there are 4 loops,
+ * each element in the final array is accessed only once.
+ */
+
+#include <stdlib.h> // for the malloc() function
+#include <string.h> // for the strlen() function
+
+char *get_letters(char digit) {
+    switch (digit) {
+        case '2':
+            return "abc";
+        case '3':
+            return "def";
+        case '4':
+            return "ghi";
+        case '5':
+            return "jkl";
+        case '6':
+            return "mno";
+        case '7':
+            return "pqrs";
+        case '8':
+            return "tuv";
+        case '9':
+            return "wxyz";
+        default:
+            return "";
+    }
+}
+
+char **letterCombinations(char *digits, int *return_size) {
+    char *cp;
+    int i, j, k, l, ind, k_tot, l_tot, digits_size = 0;
+
+    if (*digits == '\0') {
+        *return_size = 0;
+        return NULL;
+    }
+
+    *return_size = 1;
+    cp = digits;
+    while (*cp != '\0') {
+        *return_size *= strlen(get_letters(*cp));
+        digits_size++;
+        cp++;
+    }
+
+    char **combs = malloc(sizeof(char*) * (*return_size));
+    for (i = 0; i < *return_size; i++) {
+        combs[i] = malloc(sizeof(char) * (digits_size + 1));
+        combs[i][digits_size] = '\0';
+    }
+
+    k_tot = 1;
+    l_tot = (*return_size);
+    for (i = 0; i < digits_size; i++) { // loop accross digits
+        cp = get_letters(digits[i]);
+        l_tot /= strlen(cp);
+        for (j = 0; j < strlen(cp); j++) { // loop accross letters of the digit
+            for (k = 0; k < k_tot; k++) { // loop across the subset starting positions for each letter
+                for (l = 0; l < l_tot; l++) { // loop accross each subset positions for each letter
+                    ind = k * l_tot * strlen(cp) + l + l_tot * j;
+                    combs[ind][i] = cp[j];
+                }
+            }
+        }
+        k_tot *= strlen(cp);
+    }
+
+    return combs;
+}