Implement decode-string

name: decode-string
url: https://leetcode.com/problems/decode-string
difficulty: 2

time: 0 ms
time-rank: 100 %
time-complexity: O(N)

space: 6.5 MB
space-rank: 100 %
space-complexity: O(N)

Fixes #292

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

Author: cfriedt

decode-string-test.cpp

@@ -0,0 +1,27 @@
+/*
+ * Copyright (c) 2021, Christopher Friedt
+ *
+ * SPDX-License-Identifier: MIT
+ */
+
+#include <gtest/gtest.h>
+
+#include "decode-string.cpp"
+
+using namespace std;
+
+TEST(DecodeString, 3_a_2_bc_) {
+  EXPECT_EQ("aaabcbc", Solution().decodeString("3[a]2[bc]"));
+}
+
+TEST(DecodeString, 3_a2_c__) {
+  EXPECT_EQ("accaccacc", Solution().decodeString("3[a2[c]]"));
+}
+
+TEST(DecodeString, 2_abc_3_cd_ef) {
+  EXPECT_EQ("abcabccdcdcdef", Solution().decodeString("2[abc]3[cd]ef"));
+}
+
+TEST(DecodeString, abc3_cd_xyz) {
+  EXPECT_EQ("abccdcdcdxyz", Solution().decodeString("abc3[cd]xyz"));
+}

decode-string.cpp

@@ -0,0 +1,79 @@
+/*
+ * Copyright (c) 2021, Christopher Friedt
+ *
+ * SPDX-License-Identifier: MIT
+ */
+
+// clang-format off
+// name: decode-string
+// url: https://leetcode.com/problems/decode-string
+// difficulty: 2
+// clang-format on
+
+#include <vector>
+
+using namespace std;
+
+class Solution {
+public:
+  string decodeString(string s) {
+    // maintain stacks (lifo queues) for both lefts and rights
+    // it takes O(N) time to go through s and populate them
+
+    // possible optimization:
+    // technically speaking, there can be maximally N/4 lefts and N/4 rights
+    // so maybe it just makes sense to preallocate that amount and keep
+    // an L and R handy.
+    // the question says that s will be maximally 30 characters.
+    // so that would allow us to be O(1) space (except for the output
+    // which is proportional in length to the input, so O(N)).
+
+    stringstream ss;
+    vector<size_t> nums;
+    uint8_t lefts = 0;
+    vector<string> strs;
+    string num;
+    bool have_number = false;
+
+    for (size_t i = 0, N = s.size(); i < N; ++i) {
+      if ('[' == s[i]) {
+        // we can now parse the number string that precedes the '[' character
+        nums.push_back(stoi(num));
+        num.clear();
+        ++lefts;
+        strs.push_back("");
+
+      } else if (']' == s[i]) {
+
+        auto j = nums.back();
+        string t = strs.back();
+
+        nums.pop_back();
+        lefts--;
+        strs.pop_back();
+
+        if (nums.size() > 0) {
+          for (; j > 0; --j) {
+            strs.back() += t;
+          }
+        } else {
+          for (; j > 0; --j) {
+            ss << t;
+          }
+        }
+
+      } else if ('0' <= s[i] && s[i] <= '9') {
+        have_number = true;
+        num.push_back(s[i]);
+      } else if ('a' <= s[i] && s[i] <= 'z') {
+        if (!have_number || lefts == 0) {
+          ss << s[i];
+        } else {
+          strs.back().push_back(s[i]);
+        }
+      }
+    }
+
+    return ss.str();
+  }
+};