From ad8474add6cf9b2348a414a6e842a977bd875a01 Mon Sep 17 00:00:00 2001
From: Nikhil <nikhilsnayak3473@gmail.com>
Date: Tue, 24 Jan 2023 23:47:20 +0530
Subject: [PATCH] Added algorithm to check if a String is Pangram

---
 algorithms/Java/strings/Pangram.java | 61 ++++++++++++++++++++++++++++
 1 file changed, 61 insertions(+)
 create mode 100644 algorithms/Java/strings/Pangram.java

diff --git a/algorithms/Java/strings/Pangram.java b/algorithms/Java/strings/Pangram.java
new file mode 100644
index 00000000..cff8d7b7
--- /dev/null
+++ b/algorithms/Java/strings/Pangram.java
@@ -0,0 +1,61 @@
+/**
+ * The Pangram class is used to check if a given string is a pangram or not. A
+ * pangram is a sentence that contains all the letters of the alphabet.
+ * The class provides a method, isPangram(String str), that takes a string as
+ * input and returns a boolean value indicating whether the string is a pangram
+ * or not.
+ * 
+ * @author Nikhil S
+ */
+
+public class Pangram {
+
+    private static boolean isPangram(String str) {
+        str = str.toLowerCase(); // Convert the input string to lowercase for consistency
+        boolean[] isLetterPresent = new boolean[26]; // A boolean array to indcate if a letter is present or not
+
+        for (int i = 0; i < str.length(); i++) {
+            if (Character.isLetter(str.charAt(i))) { // Check if the character is a english alphabet
+                /*
+                 * The below line performs the logic of updating the isLetterPresent Array based
+                 * on the character
+                 * Each letter in the English alphabets map to unique index of the
+                 * isLetterPresent Array
+                 * isLetterPresent[0] -> 'a';
+                 * isLetterPresent[1] -> 'b';
+                 * .
+                 * .
+                 * .
+                 * .
+                 * isLetterPresent[25] -> 'z';
+                 * 
+                 * Example : 1. Consider the current character as 'a' whose ASCII value is 97.
+                 * Hence 'a' - 'a' => 97 - 97 = 0 => isLetterPresent[0] = true => 'a' is present
+                 * 
+                 * 2. Consider the current character as 'z' whose ASCII value is 122.
+                 * Hence 'z' - 'a' => 122 - 97 = 25 => isLetterPresent[25] = true => 'z' is
+                 * present
+                 */
+                isLetterPresent[str.charAt(i) - 'a'] = true;
+
+            }
+        }
+
+        for (int i = 0; i < 26; i++) { // Check for presence of all the letters i.e., the Array must have all true
+                                       // value
+            if (!isLetterPresent[i]) {
+                return false; // If there is a false value it means the string is not a Pangram hence return
+                              // false
+            }
+        }
+        return true; // If the for loop successfully completes for all values of isLetterPresent[]
+                     // then, the string is Pangram hence return true
+    }
+
+    public static void main(String[] args) {
+
+        System.out.println(isPangram("The quick brown fox jumps over the lazy dog")); // Pangram String; output:true
+        System.out.println(isPangram("The quick brown wolf jumps over the lazy dog"));// Normal String; output:false
+    }
+
+}