Implement minimum-deletion-cost-to-avoid-repeating-letters

name: minimum-deletion-cost-to-avoid-repeating-letters
url: https://leetcode.com/problems/minimum-deletion-cost-to-avoid-repeating-letters
difficulty: medium

time: 264ms
time-rank: 95.54%
time-complexity: O(N)

space: 95.9MB
space-rank: 24.79%
space-complexity: O(1)

Fixes #280

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

Author: cfriedt

minimum-deletion-cost-to-avoid-repeating-letters-test.cpp

@@ -0,0 +1,36 @@
+/*
+ * Copyright (c) 2021, Christopher Friedt
+ *
+ * SPDX-License-Identifier: MIT
+ */
+
+#include <gtest/gtest.h>
+
+#include "minimum-deletion-cost-to-avoid-repeating-letters.cpp"
+
+using namespace std;
+
+TEST(MinimumDeletionCostToAvoidRepeatingLetters, abaac__1_2_3_4_5) {
+	string s = "abaac";
+	vector<int> cost = {
+		1, 2, 3, 4, 5
+	};
+	EXPECT_EQ(3, Solution().minCost(s, cost));
+}
+
+TEST(MinimumDeletionCostToAvoidRepeatingLetters, abc__1_2_3) {
+	string s = "abc";
+	vector<int> cost = {
+		1, 2, 3,
+	};
+	EXPECT_EQ(0, Solution().minCost(s, cost));
+}
+
+TEST(MinimumDeletionCostToAvoidRepeatingLetters, aabaa__1_2_3_4_1) {
+	string s = "aabaa";
+	vector<int> cost = {
+		1, 2, 3, 4, 1
+	};
+	EXPECT_EQ(2, Solution().minCost(s, cost));
+}
+

minimum-deletion-cost-to-avoid-repeating-letters.cpp

@@ -0,0 +1,50 @@
+/*
+ * Copyright (c) 2021, Christopher Friedt
+ *
+ * SPDX-License-Identifier: MIT
+ */
+
+#include <algorithm>
+#include <vector>
+
+using namespace std;
+
+// name: minimum-deletion-cost-to-avoid-repeating-letters
+// url: https://leetcode.com/problems/minimum-deletion-cost-to-avoid-repeating-letters
+// difficulty: 2
+
+class Solution {
+public:
+   
+    
+    int minCost(string s, vector<int>& cost) {
+/*
+i  p s[i] c[i] mn  mx   
+=====================
+0  -  a   1    1   1
+1  a  a   2    3   2
+2  a  b   3    1   1
+3  b  a   4    5   4
+4  a  a   1    6   4
+*/
+        int max_cost;
+        int min_cost = 0;
+        char prev = '\0';
+        for(size_t i = 0, N = s.size(); i < N; prev = s[i], ++i) {
+            if (s[i] == prev) {
+                max_cost = max(max_cost, cost[i]);
+                min_cost += cost[i];
+                if (i == N - 1 || s[i + 1] != prev) {
+                    min_cost -= max_cost;
+                }
+            } else {
+                max_cost = cost[i];
+                if (i < N - 1 && s[i+1] == s[i]) {
+                    min_cost += cost[i];
+                }
+            }
+        }
+
+        return min_cost;
+    }
+};