Lazy Run-Length Decoding - Google Top Interview Questions
Problem Statement :
Implement a data structure RunLengthDecoder which implements the following methods:
RunLengthDecoder(string s) which takes a string s that is run-length encoded.
string value(int i) which returns the character at index i of the run-length decoded version of s.
string valueInRange(string c, int i, int j) which returns the first lowercase alphabet character that is greater than or equal to c from the range [i, j) of the decoded string.
You can assume that the range contains characters that are in ascending order. If no such character exists, return "?"
Constraints
n ≤ 100,000 where n is the length of s
k ≤ 100,000 where k is the number of calls to value and valueInRange
Example 1
Input
methods = ["constructor", "value", "value", "valueInRange", "valueInRange", "valueInRange"]
arguments = [["3a3b2c1d1a"], [0], [4], ["a", 0, 9], ["b", 3, 9], ["e", 3, 9]]`
Output
[None, "a", "b", "a", "b", "?"]
Explanation
r = RunLengthDecoder("3a3b2c1d1a") # In decoded version it's "aaabbbccda"
r.value(0) == "a"
r.value(4) == "b"
r.valueInRange("a", 0, 9) == "a"
r.valueInRange("b", 3, 9) == "b"
r.valueInRange("e", 3, 9) == "?"
Solution :
Solution in C++ :
class RunLengthDecoder {
public:
map<char, vector<pair<int, int>>> charToRanges;
vector<pair<int, int>> ranges;
vector<char> chars;
RunLengthDecoder(string s) {
int curr = 0;
int amt = 0;
for (int i = 0; i < s.size(); i++) {
if (s[i] >= 'a' && s[i] <= 'z') {
charToRanges[s[i]].emplace_back(amt, amt + curr);
ranges.emplace_back(amt, amt + curr);
chars.push_back(s[i]);
amt += curr;
curr = 0;
} else {
curr *= 10;
curr += s[i] - '0';
}
}
assert(curr == 0);
}
string value(int i) {
auto it = upper_bound(ranges.begin(), ranges.end(), make_pair(i + 1, (int)-1e9));
it--;
int idx = it - ranges.begin();
return string(1, chars[idx]);
}
string valueInRange(string c, int i, int j) {
for (char ch = c[0]; ch <= 'z'; ch++) {
auto it = upper_bound(charToRanges[ch].begin(), charToRanges[ch].end(),
make_pair(j, (int)-1e9));
if (it != charToRanges[ch].begin()) it--;
if (it == charToRanges[ch].end()) continue;
pair<int, int> key = *it;
if (key.second <= i || key.first >= j) continue;
return string(1, ch);
}
return "?";
}
};
Solution in Java :
import java.util.*;
class RunLengthDecoder {
private StringBuilder sb = new StringBuilder();
private List<Integer> begins = new ArrayList<>();
public RunLengthDecoder(String s) {
for (int i = 0, end = 0; i != s.length(); i++) {
int j = i;
while (Character.isDigit(s.charAt(j + 1))) j++;
begins.add(end);
int len = Integer.parseInt(s.substring(i, j + 1));
if (len != 0) {
end += Integer.parseInt(s.substring(i, j + 1));
sb.append(s.charAt(j + 1));
}
i = j + 1;
}
}
public String value(int i) {
int idx = Collections.binarySearch(begins, i);
if (idx < 0)
idx = -2 - idx;
return sb.substring(idx, idx + 1);
}
public String valueInRange(String c, int i, int j) {
int idx = Collections.binarySearch(begins, i);
if (idx < 0)
idx = -2 - idx;
final char ch = c.charAt(0);
for (int round = 0; idx != sb.length() && round != 26; round++, idx++) {
if (begins.get(idx) >= j)
break;
if (sb.charAt(idx) >= ch)
return Character.toString(sb.charAt(idx));
}
return "?";
}
}
Solution in Python :
class RunLengthDecoder:
def __init__(self, s):
self.chars = []
self.indexes = [0]
char_count = 0
for char in s:
if not char.isdigit():
self.chars.append(char)
self.indexes.append(self.indexes[-1] + char_count)
char_count = 0
else:
char_count = char_count * 10 + int(char)
self.next_indexes = [[self.indexes[-1]] * 26 for _ in range(len(s) + 1)]
for i in range(len(self.chars) - 1, -1, -1):
self.next_indexes[i] = self.next_indexes[i + 1][:]
char_num = ord(self.chars[i]) - ord("a")
self.next_indexes[i][char_num] = self.indexes[i]
def value(self, i):
char_index = max(0, bisect_right(self.indexes, i) - 1)
return self.chars[char_index]
def valueInRange(self, c, i, j):
char_index = max(0, bisect_right(self.indexes, i) - 1)
next_indexes = self.next_indexes[char_index]
char_num = ord(c) - ord("a")
for num in range(char_num, 26):
if next_indexes[num] < j:
return chr(num + ord("a"))
return "?"
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