Implement min-cost-climbing-stairs

name: min-cost-climbing-stairs
url: https://leetcode.com/problems/min-cost-climbing-stairs
difficulty: 1

time: 98.58 ms
time-rank: -1 %
time-complexity: O(N)

space: 13.5 MB
space-rank: 100.0 %
space-complexity: O(1)

Fixes #301

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

Author: cfriedt

min-cost-climbing-stairs-test.cpp

@@ -0,0 +1,26 @@
+/*
+ * Copyright (c) 2021, Christopher Friedt
+ *
+ * SPDX-License-Identifier: MIT
+ */
+
+#include <gtest/gtest.h>
+
+#include "min-cost-climbing-stairs.cpp"
+
+using namespace std;
+
+TEST(MinCostClimbingStairs, 10_15_20) {
+  vector<int> cost = {10, 15, 20};
+  EXPECT_EQ(15, Solution().minCostClimbingStairs(cost));
+}
+
+TEST(MinCostClimbingStairs, 1_100_1_1_1_100_1_1_100_1) {
+  vector<int> cost = {1, 100, 1, 1, 1, 100, 1, 1, 100, 1};
+  EXPECT_EQ(6, Solution().minCostClimbingStairs(cost));
+}
+
+TEST(MinCostClimbingStairs, 0_1_1_1_1) {
+  vector<int> cost = {0, 1, 1, 1};
+  EXPECT_EQ(1, Solution().minCostClimbingStairs(cost));
+}

min-cost-climbing-stairs.cpp

@@ -0,0 +1,72 @@
+/*
+ * Copyright (c) 2021, Christopher Friedt
+ *
+ * SPDX-License-Identifier: MIT
+ */
+
+// clang-format off
+// name: min-cost-climbing-stairs
+// url: https://leetcode.com/problems/min-cost-climbing-stairs
+// difficulty: 1
+// clang-format on
+
+#include <vector>
+
+using namespace std;
+
+class Solution {
+public:
+  int minCostClimbingStairs(vector<int> &cost) {
+    // clang-format off
+    // C(i) = min(C(i - 1), C(i - 2)) + cost[i]
+    //      = cost[i], i < 2
+    //
+    // The exit criteria are to successfully arrive at either N-1 or N-2
+    // since from N-2 it's possible to climb 2 stairs at once to get to N.
+    //
+    // Eg.               v--- goal is 1 past the last array element
+    // cost = [10,15,20]
+    // i  cost[i]  C(i)   C(i - 1)  C(i - 2)  comment
+    // ----------------------------------------------
+    // 0  10       10     0         0
+    // 1  15       15    10         0         stop here, because i >= N - 2 && C(1) < cost[2] OR set C(2) = C(1)
+    // 2  20       15*   15         10
+    //
+    // Eg.
+    // cost = [1,100,1,1,1,100,1,1,100,1]
+    // i  cost[i]  C(i)   C(i - 1)  C(i - 2)  comment
+    // ----------------------------------------------
+    // 0  1        1      0         0
+    // 1  100      100    1         0
+    // 2  1        2      100       1
+    // 3  1        3      2         100
+    // 4  1        3      3         2
+    // 5  100      103    3         3
+    // 6  1        4      103       3
+    // 7  1        5      4         103
+    // 8  100      104    5         4
+    // 9  1        6      104       5
+	// Eg.
+	// cost = [0,1,1,1]
+	// i  cost[i]  C(i)   C(i - 1)  C(i - 2)  comment
+	// ----------------------------------------------
+	// 0  0        0      0         0
+	// 1  1        1      0         0
+	// 2  1        1      1         0
+	// 3  1        1*     1         1
+    // clang-format on
+    const int N = cost.size();
+    for (int i = 2; i < N; ++i) {
+      int t = cost[i] + min(cost[i - 1], cost[i - 2]);
+      if (i == N - 1) {
+        if (cost[i - 1] < t) {
+          cost[i] = cost[i - 1];
+          break;
+        }
+      }
+      cost[i] = t;
+    }
+
+    return cost[N - 1];
+  }
+};