solution and explaination for 1002.c leetcode

leetcode/src/1002.c

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+/* Given a string array words, return an array of all characters
+that show up in all strings within the words (including duplicates).
+You may return the answer in any order.
+
+ Example 1:
+
+Input: words = ["bella","label","roller"]
+Output: ["e","l","l"]
+Example 2:
+
+Input: words = ["cool","lock","cook"]
+Output: ["c","o"]
+
+
+Constraints:
+
+1 <= words.length <= 100
+1 <= words[i].length <= 100
+words[i] consists of lowercase English letters.
+
+
+Here’s a C solution for the problem. It uses frequency counting to determine the common characters among all words.
+
+Approach:
+Maintain an array min_freq[26] to store the minimum frequency of each letter across all words.
+Initialize min_freq with the frequency of the first word.
+Iterate through the remaining words and update min_freq with the minimum count of each letter.
+Construct the output based on min_freq.
+*/
+
+#include <stdio.h>
+#include <string.h>
+
+#define ALPHABET_SIZE 26
+
+void commonChars(char *words[], int wordsSize) {
+    int min_freq[ALPHABET_SIZE]; // Store minimum frequency of each character
+    int temp_freq[ALPHABET_SIZE]; // Temporary frequency array
+
+    // Initialize min_freq with frequency of characters in the first word
+    memset(min_freq, 0, sizeof(min_freq));
+    for (int i = 0; words[0][i] != '\0'; i++) {
+        min_freq[words[0][i] - 'a']++;
+    }
+
+    // Iterate over remaining words
+    for (int i = 1; i < wordsSize; i++) {
+        memset(temp_freq, 0, sizeof(temp_freq));
+
+        // Count character frequency for the current word
+        for (int j = 0; words[i][j] != '\0'; j++) {
+            temp_freq[words[i][j] - 'a']++;
+        }
+
+        // Update min_freq to store the minimum count
+        for (int k = 0; k < ALPHABET_SIZE; k++) {
+            if (min_freq[k] > temp_freq[k]) {
+                min_freq[k] = temp_freq[k];
+            }
+        }
+    }
+
+    // Print the output based on min_freq
+    for (int i = 0; i < ALPHABET_SIZE; i++) {
+        while (min_freq[i] > 0) {
+            printf("%c ", i + 'a');
+            min_freq[i]--;
+        }
+    }
+    printf("\n");
+}
+
+// Driver Code
+int main() {
+    char *words1[] = {"bella", "label", "roller"};
+    int size1 = sizeof(words1) / sizeof(words1[0]);
+    commonChars(words1, size1); // Output: e l l
+
+    char *words2[] = {"cool", "lock", "cook"};
+    int size2 = sizeof(words2) / sizeof(words2[0]);
+    commonChars(words2, size2); // Output: c o
+
+    return 0;
+}
+/*Explanation:
+The min_freq array keeps track of the minimum occurrence of each letter across all words.
+temp_freq temporarily stores the frequency of letters for the current word.
+The min_freq array is updated by taking the minimum count across all words.
+Finally, the result is printed based on the characters that appear in all words.
+Complexity:
+Time Complexity: 
+𝑂(𝑁×𝑀)
+O(N×M), where 𝑁 N is the number of words and 
+𝑀 M is the average length of each word.
+Space Complexity: 𝑂(1)
+O(1) (since we use fixed-size arrays).*/