Baozi Training Google Code Jam 2020 Qualification Round: Nesting Depth Solution

Problem

tl;dr: Given a string of digits S, insert a minimum number of opening and closing parentheses into it such that the resulting string is balanced and each digit d is inside exactly d pairs of matching parentheses.

Let the nesting of two parentheses within a string be the substring that occurs strictly between them. An opening parenthesis and a closing parenthesis that is further to its right are said to match if their nesting is empty, or if every parenthesis in their nesting matches with another parenthesis in their nesting. The nesting depth of a position p is the number of pairs of matching parentheses m such that p is included in the nesting of m.

For example, in the following strings, all digits match their nesting depth: 0((2)1)(((3))1(2))((((4))))((2))((2))(1). The first three strings have minimum length among those that have the same digits in the same order, but the last one does not since ((22)1) also has the digits 221 and is shorter.

Given a string of digits S, find another string S', comprised of parentheses and digits, such that:

• all parentheses in S' match some other parenthesis,
• removing any and all parentheses from S' results in S,
• each digit in S' is equal to its nesting depth, and
• S' is of minimum length.

Input

The first line of the input gives the number of test cases, TT lines follow. Each line represents a test case and contains only the string S.

Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the string S' defined above.

Limits

Time limit: 20 seconds per test set.
Memory limit: 1GB.
1 ≤ T ≤ 100.
1 ≤ length of S ≤ 100.

Test set 1 (Visible Verdict)

Each character in S is either 0 or 1.

Test set 2 (Visible Verdict)

Each character in S is a decimal digit between 0 and 9, inclusive.

Sample

 Input Output 4 0000 101 111000 1  Case #1: 0000 Case #2: (1)0(1) Case #3: (111)000 Case #4: (1) 

The strings ()0000()(1)0(((()))1) and (1)(11)000 are not valid solutions to Sample Cases #1, #2 and #3, respectively, only because they are not of minimum length. In addition, 1)( and )(1 are not valid solutions to Sample Case #4 because they contain unmatched parentheses and the nesting depth is 0 at the position where there is a 1.

You can create sample inputs that are valid only for Test Set 2 by removing the parentheses from the example strings mentioned in the problem statement.

Video Tutorial

You can find the detailed video tutorial here

Thought Process

Simple simulation. Key points
• An integer could be used for depth to record how many parentheses we need to append. (I used a stack in the code during the competition)
• We could use a dummy node as the depth is 0 instead of starting from index 1.

Solutions

Simulation solution

 1 import java.io.BufferedReader;
3 import java.util.Scanner;
4 import java.util.Stack;
5
6 public class Solution {
7     public static void main(String[] args) {
9
10         int testCases = s.nextInt();
12         int caseNum = 1;
13         Solution nestingDepth = new Solution();
14
15         while (caseNum <= testCases) {
16             String n = s.nextLine();
17             System.out.println(String.format("Case #%d: %s", caseNum, nestingDepth.minBrackets(n)));
18             caseNum++;
19         }
20     }
21
22     private String minBrackets(String s) {
23         int[] digits = new int[s.length()];
24         for (int i = 0; i < s.length(); i++) {
25             digits[i] = s.charAt(i) - '0';
26         }
27         if (digits == null || digits.length == 0) {
28             return null;
29         }
30
31         Stack<Character> stack = new Stack<>();
32         StringBuilder sb = new StringBuilder();
33         for (int i = 0; i < digits[0]; i++) {
34             stack.push('(');
35             sb.append("(");
36         }
37
38         sb.append(digits[0]);
39
40         for (int i = 1; i < digits.length; i++) {
41             if (digits[i] > digits[i - 1]) {
42                 for (int k = 0; k < digits[i] - digits[i - 1]; k++) {
43                     stack.push('(');
44                     sb.append("(");
45                 }
46             } else if (digits[i] < digits[i - 1]) {
47                 for (int k = 0; k < digits[i - 1] - digits[i]; k++) {
48                     stack.pop();
49                     sb.append(")");
50                 }
51             }
52
53             sb.append(digits[i]);
54         }
55
56         while (!stack.isEmpty()) {
57             stack.pop();
58             sb.append(")");
59         }
60
61         return sb.toString();
62     }
63 }

Time Complexity: O(N)
Space Complexity: O(N) since we used extra extra space to store the output

References

posted @ 2020-05-11 06:50  包子模拟面试  阅读(214)  评论(0编辑  收藏  举报