Implement longest-arithmetic-subsequence

name: longest-arithmetic-subsequence
url: https://leetcode.com/problems/longest-arithmetic-subsequence
difficulty: 2

time: 448 ms
time-rank: 20.7 %
time-complexity: O(N^2)

space: 141.3 MB
space-rank: 66.83 %
space-complexity: O(N*M)

Fixes #325

Signed-off-by: Christopher Friedt <[email protected]>

Author: cfriedt

longest-arithmetic-subsequence-test.cpp

@@ -0,0 +1,31 @@
+/*
+ * Copyright (c) 2021, Christopher Friedt
+ *
+ * SPDX-License-Identifier: MIT
+ */
+
+#include <gtest/gtest.h>
+
+#include "longest-arithmetic-subsequence.cpp"
+
+using namespace std;
+
+TEST(LongestArithmeticSubsequence, 3_6_9_12) {
+  vector<int> A = {3, 6, 9, 12};
+  EXPECT_EQ(4, Solution().longestArithSeqLength(A));
+}
+
+TEST(LongestArithmeticSubsequence, 9_4_7_2_10) {
+  vector<int> A = {9, 4, 7, 2, 10};
+  EXPECT_EQ(3, Solution().longestArithSeqLength(A));
+}
+
+TEST(LongestArithmeticSubsequence, 20_1_15_3_10_5_8) {
+  vector<int> A = {20, 1, 15, 3, 10, 5, 8};
+  EXPECT_EQ(4, Solution().longestArithSeqLength(A));
+}
+
+TEST(LongestArithmeticSubsequence, 83_20_17_43_52_78_68_45) {
+  vector<int> A = {83, 20, 17, 43, 52, 78, 68, 45};
+  EXPECT_EQ(2, Solution().longestArithSeqLength(A));
+}

longest-arithmetic-subsequence.cpp

@@ -0,0 +1,48 @@
+/*
+ * Copyright (c) 2021, Christopher Friedt
+ *
+ * SPDX-License-Identifier: MIT
+ */
+
+// clang-format off
+// name: longest-arithmetic-subsequence
+// url: https://leetcode.com/problems/longest-arithmetic-subsequence
+// difficulty: 2
+// clang-format on
+
+#include <vector>
+
+using namespace std;
+
+class Solution {
+public:
+  int longestArithSeqLength(vector<int> &A) {
+    /*
+    Let N be the length of A.
+
+    Since A[i] is in [0, 500], then we can consider all differences, d,
+    to be constrained to the range [-500,500].
+
+    let LASL(i,d) == be the longest arithmetic subsequence in the first
+    i ordered elements where the difference between each element is d.
+
+    LASL(0, d) = 0, is a base case for all d
+    */
+    const int N = A.size();
+    constexpr int M = 500 - (-500) + 1;
+    constexpr int M_2 = 500;
+
+    vector<vector<int>> dp(N, vector<int>(M, 0));
+
+    int max_len = 1;
+    for (int i = 1; i < N; ++i) {
+      for (int j = 0; j < i; ++j) {
+        int d = A[i] - A[j];
+        dp[i][d + M_2] = dp[j][d + M_2] + 1;
+        max_len = max(max_len, dp[i][d + M_2]);
+      }
+    }
+
+    return max_len + 1;
+  }
+};