diff --git a/Main.js b/Main.js
index f0111d6..7ee020c 100644
--- a/Main.js
+++ b/Main.js
@@ -2,6 +2,7 @@
 var permutationWithoutDuplicates = require("./PermutationsWithoutDuplicates.js");
 var permutationWithDuplicates = require("./PermutationsWithDuplicates.js");
 var subsets = require("./Subsets.js");
+var uniquePaths = require("./UniquePaths.js");
 var floodFill = require("./FloodFill.js")
 
 // Invocation
@@ -9,4 +10,5 @@ var floodFill = require("./FloodFill.js")
 // permutationWithoutDuplicates.main();
 // permutationWithDuplicates.main();
 // subsets.main();
-// floodFill.main();
\ No newline at end of file
+// uniquePaths.main();
+// floodFill.main();
diff --git a/UniquePaths.js b/UniquePaths.js
new file mode 100644
index 0000000..a80a38e
--- /dev/null
+++ b/UniquePaths.js
@@ -0,0 +1,105 @@
+/*
+
+62. Unique Paths 
+https://leetcode.com/problems/unique-paths/description/
+
+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?
+
+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
+
+*/
+
+// Solution 1
+// This solution is a naive solution implementing a binary tree and visiting each node.
+
+var uniquePaths1 = function(m, n) {
+  return uniquePathsAux(0, 0, m, n)
+};
+
+var uniquePathsAux = function(row, col, rowLength, colLength) {
+  if(row >= rowLength || col >= colLength) {
+    return 0;
+  }
+  if(row == rowLength - 1 && col == colLength - 1) {
+    return 1;
+  }
+  
+  return uniquePathsAux(row + 1, col, rowLength, colLength) + 
+    uniquePathsAux(row, col + 1, rowLength, colLength)
+};
+
+// Solution 2
+// This solution is solution 1 but memoized.
+var uniquePaths2 = function(m, n) {
+  var memo = {};
+  return uniquePathsAux2(0, 0, m, n, memo)
+};
+
+var uniquePathsAux2 = function(row, col, rowLength, colLength, memo) {
+  if(memo[memoKey(row, col)]) {
+    return memo[row + "-" + col];
+  }
+  if(row >= rowLength || col >= colLength) {
+    return 0;
+  }
+  if(row == rowLength - 1 && col == colLength - 1) {
+    return 1;
+  }
+  
+  var result = uniquePathsAux(row + 1, col, rowLength, colLength, memo) + 
+    uniquePathsAux(row, col + 1, rowLength, colLength, memo);
+  memo[memoKey(row, col)] = result;
+  return result;
+};
+
+var memoKey = function(row, col) {
+  return row + "-" + col;
+}
+
+// Solution 3
+// This solution uses Dinamic Programming
+var uniquePaths3 = function(m, n) {
+  var matrix = [];
+  for(var i = 0; i < m; i++) {
+    matrix[i] = [];
+    for(var j = 0; j < n; j++) {
+      if(i == 0 || j == 0) {
+        matrix[i][j] = 1;
+      } else{
+        matrix[i][j] = 0;   
+      }
+    }
+  }
+  
+  for(var row = 1; row < m; row++) {
+    for(var col = 1; col < n; col++) {
+      matrix[row][col] = matrix[row - 1][col] + matrix[row][col - 1]
+    }
+  }
+
+  return matrix[m - 1][n - 1];
+};
+
+var main = function() {
+  console.log(uniquePaths1(10,4));
+  console.log(uniquePaths2(10,4));
+  console.log(uniquePaths3(10,4));
+}
+
+module.exports.main = main;
\ No newline at end of file