diff --git a/README.md b/README.md
index 5b626f3678..7f052a168e 100644
--- a/README.md
+++ b/README.md
@@ -118,6 +118,7 @@ a set of rules that precisely define a sequence of operations.
   * `B` [Tower of Hanoi](src/algorithms/uncategorized/hanoi-tower)
   * `B` [Square Matrix Rotation](src/algorithms/uncategorized/square-matrix-rotation) - in-place algorithm
   * `B` [Jump Game](src/algorithms/uncategorized/jump-game) - backtracking, dynamic programming (top-down + bottom-up) and greedy examples 
+  * `B` [Unique Paths](src/algorithms/uncategorized/unique-paths) - backtracking and dynamic programming examples 
   * `A` [N-Queens Problem](src/algorithms/uncategorized/n-queens)
   * `A` [Knight's Tour](src/algorithms/uncategorized/knight-tour)
 
@@ -151,6 +152,7 @@ algorithm is an abstraction higher than a computer program.
 * **Dynamic Programming** - build up a solution using previously found sub-solutions
   * `B` [Fibonacci Number](src/algorithms/math/fibonacci)
   * `B` [Jump Game](src/algorithms/uncategorized/jump-game)
+  * `B` [Unique Paths](src/algorithms/uncategorized/unique-paths)
   * `A` [Levenshtein Distance](src/algorithms/string/levenshtein-distance) - minimum edit distance between two sequences
   * `A` [Longest Common Subsequence](src/algorithms/sets/longest-common-subsequence) (LCS)
   * `A` [Longest Common Substring](src/algorithms/string/longest-common-substring)
@@ -166,6 +168,7 @@ algorithm is an abstraction higher than a computer program.
 if it satisfies all conditions, and only then continue generating subsequent solutions. Otherwise, backtrack, and go on a
 different path of finding a solution. Normally the DFS traversal of state-space is being used.
   * `B` [Jump Game](src/algorithms/uncategorized/jump-game)
+  * `B` [Unique Paths](src/algorithms/uncategorized/unique-paths)
   * `A` [Hamiltonian Cycle](src/algorithms/graph/hamiltonian-cycle) - Visit every vertex exactly once
   * `A` [N-Queens Problem](src/algorithms/uncategorized/n-queens)
   * `A` [Knight's Tour](src/algorithms/uncategorized/knight-tour)
diff --git a/src/algorithms/uncategorized/unique-paths/README.md b/src/algorithms/uncategorized/unique-paths/README.md
new file mode 100644
index 0000000000..f40bb81758
--- /dev/null
+++ b/src/algorithms/uncategorized/unique-paths/README.md
@@ -0,0 +1,99 @@
+# Unique Paths Problem
+
+A robot is located at the top-left corner of a `m x n` grid 
+(marked 'Start' in the diagram below).
+
+The robot can only move either down or right at any point in 
+time. The robot is trying to reach the bottom-right corner 
+of the grid (marked 'Finish' in the diagram below).
+
+How many possible unique paths are there?
+
+![Unique Paths](https://leetcode.com/static/images/problemset/robot_maze.png)
+
+## Examples
+
+**Example #1**
+
+```
+Input: m = 3, n = 2
+Output: 3
+Explanation:
+From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
+1. Right -> Right -> Down
+2. Right -> Down -> Right
+3. Down -> Right -> Right
+```
+
+**Example #2**
+
+```
+Input: m = 7, n = 3
+Output: 28
+```
+
+## Algorithms
+
+### Backtracking
+
+First thought that might came to mind is that we need to build a decision tree 
+where `D` means moving down and `R` means moving right. For example in case
+of boars `width = 3` and `height = 2` we will have the following decision tree:
+
+```
+                START
+                /   \
+               D     R
+             /     /   \
+           R      D      R
+         /      /         \
+        R      R            D
+
+       END    END          END
+```
+
+We can see three unique branches here that is the answer to our problem.
+
+**Time Complexity**: `O(2 ^ n)` - roughly in worst case with square board
+of size `n`.
+
+**Auxiliary Space Complexity**: `O(m + n)` - since we need to store current path with
+positions.
+
+### Dynamic Programming
+
+Let's treat `BOARD[i][j]` as our sub-problem.
+
+Since we have restriction of moving only to the right
+and down we might say that number of unique paths to the current
+cell is a sum of numbers of unique paths to the cell above the
+current one and to the cell to the left of current one.
+
+```
+BOARD[i][j] = BOARD[i - 1][j] + BOARD[i][j - 1]; // since we can only move down or right.
+```
+
+Base cases are:
+
+```
+BOARD[0][any] = 1; // only one way to reach any top slot.
+BOARD[any][0] = 1; // only one way to reach any slot in the leftmost column.
+```
+
+For the board `3 x 2` our dynamic programming matrix will look like:
+
+|     | 0   | 1   | 1   |
+|:---:|:---:|:---:|:---:|
+|**0**| 0   | 1   | 1   |
+|**1**| 1   | 2   | 3   |
+
+Each cell contains the number of unique paths to it. We need 
+the bottom right one with number `3`.
+
+**Time Complexity**: `O(m * n)` - since we're going through each cell of the DP matrix.
+
+**Auxiliary Space Complexity**: `O(m * n)` - since we need to have DP matrix.
+
+## References
+
+- [LeetCode](https://leetcode.com/problems/unique-paths/description/)
diff --git a/src/algorithms/uncategorized/unique-paths/__test__/btUniquePaths.test.js b/src/algorithms/uncategorized/unique-paths/__test__/btUniquePaths.test.js
new file mode 100644
index 0000000000..655f8a3215
--- /dev/null
+++ b/src/algorithms/uncategorized/unique-paths/__test__/btUniquePaths.test.js
@@ -0,0 +1,12 @@
+import btUniquePaths from '../btUniquePaths';
+
+describe('btUniquePaths', () => {
+  it('should find the number of unique paths on board', () => {
+    expect(btUniquePaths(3, 2)).toBe(3);
+    expect(btUniquePaths(7, 3)).toBe(28);
+    expect(btUniquePaths(3, 7)).toBe(28);
+    expect(btUniquePaths(10, 10)).toBe(48620);
+    expect(btUniquePaths(100, 1)).toBe(1);
+    expect(btUniquePaths(1, 100)).toBe(1);
+  });
+});
diff --git a/src/algorithms/uncategorized/unique-paths/__test__/dpUniquePaths.test.js b/src/algorithms/uncategorized/unique-paths/__test__/dpUniquePaths.test.js
new file mode 100644
index 0000000000..d5d7bb7d78
--- /dev/null
+++ b/src/algorithms/uncategorized/unique-paths/__test__/dpUniquePaths.test.js
@@ -0,0 +1,12 @@
+import dpUniquePaths from '../dpUniquePaths';
+
+describe('dpUniquePaths', () => {
+  it('should find the number of unique paths on board', () => {
+    expect(dpUniquePaths(3, 2)).toBe(3);
+    expect(dpUniquePaths(7, 3)).toBe(28);
+    expect(dpUniquePaths(3, 7)).toBe(28);
+    expect(dpUniquePaths(10, 10)).toBe(48620);
+    expect(dpUniquePaths(100, 1)).toBe(1);
+    expect(dpUniquePaths(1, 100)).toBe(1);
+  });
+});
diff --git a/src/algorithms/uncategorized/unique-paths/btUniquePaths.js b/src/algorithms/uncategorized/unique-paths/btUniquePaths.js
new file mode 100644
index 0000000000..abe94490e0
--- /dev/null
+++ b/src/algorithms/uncategorized/unique-paths/btUniquePaths.js
@@ -0,0 +1,58 @@
+/**
+ * BACKTRACKING approach of solving Unique Paths problem.
+ *
+ * @param {number} width - Width of the board.
+ * @param {number} height - Height of the board.
+ * @param {number[][]} steps - The steps that have been already made.
+ * @param {number} uniqueSteps - Total number of unique steps.
+ * @return {number} - Number of unique paths.
+ */
+export default function btUniquePaths(width, height, steps = [[0, 0]], uniqueSteps = 0) {
+  // Fetch current position on board.
+  const currentPos = steps[steps.length - 1];
+
+  // Check if we've reached the end.
+  if (currentPos[0] === width - 1 && currentPos[1] === height - 1) {
+    // In case if we've reached the end let's increase total
+    // number of unique steps.
+    return uniqueSteps + 1;
+  }
+
+  // Let's calculate how many unique path we will have
+  // by going right and by going down.
+  let rightUniqueSteps = 0;
+  let downUniqueSteps = 0;
+
+  // Do right step if possible.
+  if (currentPos[0] < width - 1) {
+    steps.push([
+      currentPos[0] + 1,
+      currentPos[1],
+    ]);
+
+    // Calculate how many unique paths we'll get by moving right.
+    rightUniqueSteps = btUniquePaths(width, height, steps, uniqueSteps);
+
+    // BACKTRACK and try another move.
+    steps.pop();
+  }
+
+  // Do down step if possible.
+  if (currentPos[1] < height - 1) {
+    steps.push([
+      currentPos[0],
+      currentPos[1] + 1,
+    ]);
+
+    // Calculate how many unique paths we'll get by moving down.
+    downUniqueSteps = btUniquePaths(width, height, steps, uniqueSteps);
+
+    // BACKTRACK and try another move.
+    steps.pop();
+  }
+
+  // Total amount of unique steps will be equal to total amount of
+  // unique steps by going right plus total amount of unique steps
+  // by going down.
+  return rightUniqueSteps + downUniqueSteps;
+}
diff --git a/src/algorithms/uncategorized/unique-paths/dpUniquePaths.js b/src/algorithms/uncategorized/unique-paths/dpUniquePaths.js
new file mode 100644
index 0000000000..75d33edce7
--- /dev/null
+++ b/src/algorithms/uncategorized/unique-paths/dpUniquePaths.js
@@ -0,0 +1,40 @@
+/**
+ * DYNAMIC PROGRAMMING approach of solving Unique Paths problem.
+ *
+ * @param {number} width - Width of the board.
+ * @param {number} height - Height of the board.
+ * @return {number} - Number of unique paths.
+ */
+export default function dpUniquePaths(width, height) {
+  // Init board.
+  const board = Array(height).fill(null).map(() => {
+    return Array(width).fill(0);
+  });
+
+  // Base case.
+  // There is only one way of getting to board[0][any] and
+  // there is also only one way of getting to board[any][0].
+  // This is because we have a restriction of moving right
+  // and down only.
+  for (let rowIndex = 0; rowIndex < height; rowIndex += 1) {
+    for (let columnIndex = 0; columnIndex < width; columnIndex += 1) {
+      if (rowIndex === 0 || columnIndex === 0) {
+        board[rowIndex][columnIndex] = 1;
+      }
+    }
+  }
+
+  // Now, since we have this restriction of moving only to the right
+  // and down we might say that number of unique paths to the current
+  // cell is a sum of numbers of unique paths to the cell above the
+  // current one and to the cell to the left of current one.
+  for (let rowIndex = 1; rowIndex < height; rowIndex += 1) {
+    for (let columnIndex = 1; columnIndex < width; columnIndex += 1) {
+      const uniquesFromTop = board[rowIndex - 1][columnIndex];
+      const uniquesFromLeft = board[rowIndex][columnIndex - 1];
+      board[rowIndex][columnIndex] = uniquesFromTop + uniquesFromLeft;
+    }
+  }
+
+  return board[height - 1][width - 1];
+}