diff --git a/DIRECTORY.md b/DIRECTORY.md
index 1b2c40b3c8..2ede552665 100644
--- a/DIRECTORY.md
+++ b/DIRECTORY.md
@@ -160,6 +160,7 @@
   * [NumberOfIslands](Graphs/NumberOfIslands.js)
   * [PrimMST](Graphs/PrimMST.js)
 * **Hashes**
+  * [MD5](Hashes/MD5.js)
   * [SHA1](Hashes/SHA1.js)
   * [SHA256](Hashes/SHA256.js)
 * **Maths**
@@ -300,6 +301,7 @@
   * [KochSnowflake](Recursive/KochSnowflake.js)
   * [LetterCombination](Recursive/LetterCombination.js)
   * [Palindrome](Recursive/Palindrome.js)
+  * [Partition](Recursive/Partition.js)
   * [SubsequenceRecursive](Recursive/SubsequenceRecursive.js)
   * [TowerOfHanoi](Recursive/TowerOfHanoi.js)
 * **Search**
diff --git a/Dynamic-Programming/Abbreviation.js b/Dynamic-Programming/Abbreviation.js
new file mode 100644
index 0000000000..06092216b8
--- /dev/null
+++ b/Dynamic-Programming/Abbreviation.js
@@ -0,0 +1,47 @@
+/**
+ * @description
+ * Given two strings, `source` and `target`, determine if it's possible to make `source` equal
+ * to `target` You can perform the following operations on the string `source`:
+ * 1. Capitalize zero or more of `source`'s lowercase letters.
+ * 2. Delete all the remaining lowercase letters in `source`.
+ *
+ * Time Complexity: (O(|source|*|target|)) where `|source|` => length of string `source`
+ *
+ * @param {String} source - The string to be transformed.
+ * @param {String} target - The string we want to transform `source` into.
+ * @returns {Boolean} - Whether the transformation is possible.
+ * @see https://www.hackerrank.com/challenges/abbr/problem - Related problem on HackerRank.
+ */
+export const isAbbreviation = (source, target) => {
+  const sourceLength = source.length
+  const targetLength = target.length
+
+  // Initialize a table to keep track of possible abbreviations
+  let canAbbreviate = Array.from({ length: sourceLength + 1 }, () =>
+    Array(targetLength + 1).fill(false)
+  )
+  // Empty strings are trivially abbreviatable
+  canAbbreviate[0][0] = true
+
+  for (let sourceIndex = 0; sourceIndex < sourceLength; sourceIndex++) {
+    for (let targetIndex = 0; targetIndex <= targetLength; targetIndex++) {
+      if (canAbbreviate[sourceIndex][targetIndex]) {
+        // If characters at the current position are equal, move to the next position in both strings.
+        if (
+          targetIndex < targetLength &&
+          source[sourceIndex].toUpperCase() === target[targetIndex]
+        ) {
+          canAbbreviate[sourceIndex + 1][targetIndex + 1] = true
+        }
+        // If the current character in `source` is lowercase, explore two possibilities:
+        // a) Capitalize it (which is akin to "using" it in `source` to match `target`), or
+        // b) Skip it (effectively deleting it from `source`).
+        if (source[sourceIndex] === source[sourceIndex].toLowerCase()) {
+          canAbbreviate[sourceIndex + 1][targetIndex] = true
+        }
+      }
+    }
+  }
+
+  return canAbbreviate[sourceLength][targetLength]
+}
diff --git a/Dynamic-Programming/tests/Abbreviation.test.js b/Dynamic-Programming/tests/Abbreviation.test.js
new file mode 100644
index 0000000000..86e0f97336
--- /dev/null
+++ b/Dynamic-Programming/tests/Abbreviation.test.js
@@ -0,0 +1,30 @@
+import { isAbbreviation } from '../Abbreviation.js'
+
+const expectPositive = (word, abbr) =>
+  expect(isAbbreviation(word, abbr)).toBe(true)
+const expectNegative = (word, abbr) =>
+  expect(isAbbreviation(word, abbr)).toBe(false)
+
+describe('Abbreviation - Positive Tests', () => {
+  test('it should correctly abbreviate or transform the source string to match the target string', () => {
+    expectPositive('', '')
+    expectPositive('a', '')
+    expectPositive('a', 'A')
+    expectPositive('abcDE', 'ABCDE')
+    expectPositive('ABcDE', 'ABCDE')
+    expectPositive('abcde', 'ABCDE')
+    expectPositive('abcde', 'ABC')
+    expectPositive('abcXYdefghijKLmnopqrs', 'XYKL')
+    expectPositive('abc123', 'ABC')
+    expectPositive('abc123', 'ABC123')
+    expectPositive('abc!@#def', 'ABC')
+  })
+})
+
+describe('Abbreviation - Negative Tests', () => {
+  test('it should fail to abbreviate or transform the source string when it is not possible to match the target string', () => {
+    expectNegative('', 'A')
+    expectNegative('a', 'ABC')
+    expectNegative('aBcXYdefghijKLmnOpqrs', 'XYKLOP')
+  })
+})
diff --git a/Recursive/Partition.js b/Recursive/Partition.js
new file mode 100644
index 0000000000..51465f8e93
--- /dev/null
+++ b/Recursive/Partition.js
@@ -0,0 +1,39 @@
+/**
+ * @function canPartition
+ * @description Check whether it is possible to partition the given array into two equal sum subsets using recursion.
+ * @param {number[]} nums - The input array of numbers.
+ * @param {number} index - The current index in the array being considered.
+ * @param {number} target - The target sum for each subset.
+ * @return {boolean}.
+ * @see [Partition Problem](https://en.wikipedia.org/wiki/Partition_problem)
+ */
+
+const canPartition = (nums, index = 0, target = 0) => {
+  if (!Array.isArray(nums)) {
+    throw new TypeError('Invalid Input')
+  }
+
+  const sum = nums.reduce((acc, num) => acc + num, 0)
+
+  if (sum % 2 !== 0) {
+    return false
+  }
+
+  if (target === sum / 2) {
+    return true
+  }
+
+  if (index >= nums.length || target > sum / 2) {
+    return false
+  }
+
+  // Include the current number in the first subset and check if a solution is possible.
+  const withCurrent = canPartition(nums, index + 1, target + nums[index])
+
+  // Exclude the current number from the first subset and check if a solution is possible.
+  const withoutCurrent = canPartition(nums, index + 1, target)
+
+  return withCurrent || withoutCurrent
+}
+
+export { canPartition }
diff --git a/Recursive/test/Partition.test.js b/Recursive/test/Partition.test.js
new file mode 100644
index 0000000000..929519999a
--- /dev/null
+++ b/Recursive/test/Partition.test.js
@@ -0,0 +1,24 @@
+import { canPartition } from '../Partition'
+
+describe('Partition (Recursive)', () => {
+  it('expects to return true for an array that can be partitioned', () => {
+    const result = canPartition([1, 5, 11, 5])
+    expect(result).toBe(true)
+  })
+
+  it('expects to return false for an array that cannot be partitioned', () => {
+    const result = canPartition([1, 2, 3, 5])
+    expect(result).toBe(false)
+  })
+
+  it('expects to return true for an empty array (0 elements)', () => {
+    const result = canPartition([])
+    expect(result).toBe(true)
+  })
+
+  it('Throw Error for Invalid Input', () => {
+    expect(() => canPartition(123)).toThrow('Invalid Input')
+    expect(() => canPartition(null)).toThrow('Invalid Input')
+    expect(() => canPartition(undefined)).toThrow('Invalid Input')
+  })
+})