Update the solution of 718. Maximum Length of Repeated Subarray

Author: enginebai

leetcode/718.maximum-length-of-repeated-subarray.md

@@ -56,6 +56,50 @@ fun findLength(nums1: IntArray, nums2: IntArray): Int {
 * **Space Complexity**: `O(m * n)` for 2D dp array.
 
 ### Dynamic Programming (Space Optimization)
-> TODO: to implement when reviewing.
+Since every `dp[i][j]` depends only on `dp[i - 1][j - 1]`, we can store as `dp[2][j]` for space optimization.
+
+
+```js
+     j, j+1
+i    X
+       \
+i+1      Y
+
+n1 = "23"
+n2 = "123"
+      
+   0, 1, 2, 3  // index
+0  0  X
+1        Y     // Y depends on X
+2           Z  // Z depends on Y
+
+// So we store for X, Y row, then shift to Y, Z row.
+```
+
+```kotlin
+fun findLength(nums1: IntArray, nums2: IntArray): Int {
+    // TODO: Here we should use the shorter array as the first array to reduce the space complexity.
+    val dp = Array(2) { IntArray(nums2.size + 1) { 0 }
+    val max = 0
+    for (i in 1..nums1.size) {
+        for (j in 1..nums2.size) {
+            if (nums1[i - 1] == nums2[j - 1]) {
+                dp[0][j] = 1 + dp[1][j - 1]
+            }
+            max = maxOf(max, dp[0][j])
+        }
+
+        // shift dp[1] to dp[0]
+        for (j in 0..nums2.size) {
+            dp[1][j] = dp[0][j]
+            // This is nessary to clean up the value of dp[0][j] for next iteration.
+            dp[0][j] = 0
+        }
+    }
+    return max
+}
+```
+
+* **Time Complexity**: `O(m * n)` where `m` and `n` are the length of two strings, respectively.
+* **Space Complexity**: `O(min(n, n)`.
 
-> TODO: There are some different optimal solution, try to practice.

topics/problems-solutions.md

@@ -530,6 +530,7 @@
 > * https://leetcode.com/problems/russian-doll-envelopes/ 4k h
 > * https://leetcode.com/problems/capacity-to-ship-packages-within-d-days 7k m
 > * https://leetcode.com/problems/sort-an-array/ 4k m
+> * https://leetcode.com/problems/minimize-the-maximum-difference-of-pairs m
 
 ## Design
 | Problem          | Difficulty |