Implemented Partition Problem, Recursive problem (#1582)

* Add Tug of War solution using backtracking

* Updated Documentation in README.md

* Added Tug of war problem link

* Updated Documentation in README.md

* Updated Documentation in README.md

* Refactor tugOfWar: remove unused vars, optimize initialization, and remove redundant checks

* Added Function Export Statment

* Updated Documentation in README.md

* Resolved Code Style --Prettier

* Rename "backtrack" to "recurse"

* Fix test case: The difference needs to be exactly 1.

* Code Modification: subsets should have sizes as close to n/2 as possible

* Updated test-case of TugOfWar

* Changed TugOfWar problem to Partition

* Modified partition problem

* Updated Documentation in README.md

* fixed code style

---------

Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
Co-authored-by: Lars Müller <34514239+appgurueu@users.noreply.github.com>

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 }

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')
+  })
+})