Add Abbreviation algorithm (#5790)

src/main/java/com/thealgorithms/dynamicprogramming/Abbreviation.java

@@ -0,0 +1,56 @@
+package com.thealgorithms.dynamicprogramming;
+
+/**
+ * A class that provides a solution to the abbreviation problem.
+ *
+ * Problem: Given two strings, `a` and `b`, determine if string `a` can be
+ * transformed into string `b` by performing the following operations:
+ * 1. Capitalize zero or more of `a`'s lowercase letters (i.e., convert them to uppercase).
+ * 2. Delete any of the remaining lowercase letters from `a`.
+ *
+ * The task is to determine whether it is possible to make string `a` equal to string `b`.
+ *
+ * @author Hardvan
+ */
+public final class Abbreviation {
+    private Abbreviation() {
+    }
+
+    /**
+     * Determines if string `a` can be transformed into string `b` by capitalizing
+     * some of its lowercase letters and deleting the rest.
+     *
+     * @param a The input string which may contain both uppercase and lowercase letters.
+     * @param b The target string containing only uppercase letters.
+     * @return {@code true} if string `a` can be transformed into string `b`,
+     *         {@code false} otherwise.
+     *
+     * Time Complexity: O(n * m) where n = length of string `a` and m = length of string `b`.
+     * Space Complexity: O(n * m) due to the dynamic programming table.
+     */
+    public static boolean abbr(String a, String b) {
+        int n = a.length();
+        int m = b.length();
+
+        boolean[][] dp = new boolean[n + 1][m + 1];
+
+        dp[0][0] = true;
+
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j <= m; j++) {
+                if (dp[i][j]) {
+                    // Case 1: If the current characters match (or can be capitalized to match)
+                    if (j < m && Character.toUpperCase(a.charAt(i)) == b.charAt(j)) {
+                        dp[i + 1][j + 1] = true;
+                    }
+                    // Case 2: If the character in `a` is lowercase, we can skip it
+                    if (Character.isLowerCase(a.charAt(i))) {
+                        dp[i + 1][j] = true;
+                    }
+                }
+            }
+        }
+
+        return dp[n][m];
+    }
+}