Add unique-paths solution

Author: Jeehay28

unique-paths/Jeehay28.js

@@ -0,0 +1,86 @@
+/**
+ * @param {number} m
+ * @param {number} n
+ * @return {number}
+ */
+
+// 🍎 DP
+// Time Complexity: O(m*n)
+// Space Complexity: O(n)
+
+// Row 0: [1, 1, 1, 1]
+// Row 1: [1, 2, 3, 4]
+// Row 2: [1, 3, 6, 10]
+
+// Initial dp: [1, 1, 1, 1]
+// left = 1 (always starts with 1 for the first column)
+
+// col = 1: left += dp[1] → left = 1 + 1 = 2 → dp[1] = left
+// col = 2: left += dp[2] → left = 2 + 1 = 3 → dp[2] = left
+// col = 3: left += dp[3] → left = 3 + 1 = 4 → dp[3] = left
+
+// dp after row 1: [1, 2, 3, 4]
+
+// Initial dp: [1, 2, 3, 4]
+// left = 1
+
+// col = 1: left += dp[1] → left = 1 + 2 = 3 → dp[1] = left
+// col = 2: left += dp[2] → left = 3 + 3 = 6 → dp[2] = left
+// col = 3: left += dp[3] → left = 6 + 4 = 10 → dp[3] = left
+
+// dp after row 2: [1, 3, 6, 10]
+
+var uniquePaths = function (m, n) {
+  let dp = new Array(n).fill(1);
+
+  for (let row = 1; row < m; row++) {
+    let left = 1;
+    for (let col = 1; col < n; col++) {
+      left += dp[col];
+      dp[col] = left;
+    }
+  }
+
+  return dp[n - 1];
+};
+
+/**
+ * @param {number} m
+ * @param {number} n
+ * @return {number}
+ */
+
+// 🍎 DFS
+// Time Complexity: O(m*n)
+// Space Complexity: O(m*n)
+
+// var uniquePaths = function (m, n) {
+//   const memo = {};
+
+//   const dfs = (row, col) => {
+//     if (row === m - 1 && col === n - 1) {
+//       return 1;
+//     }
+
+//     let cnt = 0;
+
+//     const key = `${row}, ${col}`;
+//     if (key in memo) {
+//       return memo[key];
+//     }
+
+//     if (row + 1 < m) {
+//       cnt += dfs(row + 1, col);
+//     }
+
+//     if (col + 1 < n) {
+//       cnt += dfs(row, col + 1);
+//     }
+
+//     memo[key] = cnt;
+
+//     return cnt;
+//   };
+
+//   return dfs(0, 0);
+// };