Leetcode 880: Decoded String at Index
You are given an encoded string consisting of letters and digits. The string is decoded such that for each digit, the preceding segment of the decoded string is repeated (digit - 1) additional times. Your task is to return the kth character (1-indexed) in the decoded string, where k is a positive integer. The decoded string could be extremely large, so you are not expected to fully decode the string.
Problem
Approach
Steps
Complexity
Input: The input consists of an encoded string containing lowercase English letters and digits between 2 and 9. The string always starts with a letter.
Example: Input: s = 'abc3de2', k = 5
Constraints:
• 2 <= s.length <= 100
• s consists of lowercase English letters and digits 2 through 9.
• s starts with a letter.
• 1 <= k <= 10^9
• It is guaranteed that k is less than or equal to the length of the decoded string.
• The decoded string will have fewer than 2^63 letters.
Output: You should return a string representing the kth character in the decoded string.
Example: Output: 'd'
Constraints:
• The output is a single lowercase letter.
Goal: To efficiently determine the kth character without fully decoding the string.
Steps:
• First, calculate the total length of the decoded string by iterating over the encoded string.
• Identify the range in which the kth character lies by adjusting the index in reverse.
• When a digit is encountered, divide the total length by that digit to adjust the range, and continue until the correct position is found.
Goal: The problem has constraints that ensure that the decoded string is large enough that fully decoding it is impractical. The goal is to calculate the position of the kth character without decoding the entire string.
Steps:
• The number of characters to decode can grow very large, so a brute-force solution that fully decodes the string would be inefficient.
Assumptions:
• It is guaranteed that k is less than or equal to the length of the decoded string.
• The input string contains only lowercase letters and digits between 2 and 9.
• Input: Input: s = 'abc3de2', k = 5
• Explanation: The string 'abc3de2' decodes to 'abcdeabcde'. The 5th character is 'd'.
• Input: Input: s = 'xy2z3', k = 6
• Explanation: The string 'xy2z3' decodes to 'xyzxyzxyzxyz'. The 6th character is 'z'.
Approach: The key observation is that you do not need to fully decode the string. Instead, track the length of the string dynamically as you process the encoded string. By identifying when the kth character falls within a segment, you can reverse the encoding process to find the character.
Observations:
• Instead of decoding the string fully, keep track of the lengths of the decoded segments as you parse the string.
• Reverse the decoding process to determine the position of the kth character efficiently.
Steps:
• Iterate through the encoded string and calculate the total length of the decoded string as you go.
• When encountering a digit, multiply the length of the current segment by the digit to account for repetition.
• When encountering a letter, decrement the length and check if it corresponds to the kth character. If so, return it.
Empty Inputs:
• There will always be at least one letter, so empty input will not occur.
Large Inputs:
• The algorithm must efficiently handle cases where the length of the decoded string exceeds the practical limit for direct storage.
Special Values:
• Handle cases where the kth character is located within a repetitive segment.
Constraints:
• Make sure the solution works within the problem's constraints, particularly the large value of k.
class Solution {
public:
string decodeAtIndex(string s, int k) {
long N = 0, i;
for(i = 0; N < k; i++)
N = isdigit(s[i])? (s[i] - '0') * N: N +1;
while(i--) {
if(isdigit(s[i])) {
N /= (s[i] - '0');
k %= N;
}
else if (k % N-- == 0) {
return string(1, s[i]);
}
}
return "hello yasir";
}
1 : Access Modifier
public:
Access modifier `public` allows the method `decodeAtIndex` to be accessed from outside the class.
2 : Function
string decodeAtIndex(string s, int k) {
This function `decodeAtIndex` takes a string `s` and an integer `k`, and decodes the string to find the k-th character.
3 : Variable Declaration
long N = 0, i;
Declares two variables `N` (used to track the length of the decoded string) and `i` (used for iteration).
4 : Loop
for(i = 0; N < k; i++)
A loop that iterates through the string `s`, updating `N` to track the decoded length until it reaches `k`.
5 : Conditional Expression
N = isdigit(s[i])? (s[i] - '0') * N: N +1;
Updates `N` based on whether the character at `s[i]` is a digit or not. If it's a digit, `N` is multiplied; otherwise, it's incremented.
6 : Loop
while(i--) {
A while loop that iterates backwards through the string to find the character corresponding to the k-th position.
7 : Conditional Expression
if(isdigit(s[i])) {
Checks if the current character is a digit.
8 : Arithmetic Operation
N /= (s[i] - '0');
Divides `N` by the value of the digit at `s[i]`.
9 : Arithmetic Operation
k %= N;
Calculates the remainder of `k` after dividing it by `N` to find the new position of `k`.
10 : Conditional Expression
else if (k % N-- == 0) {
Checks if `k` is divisible by `N` and decrements `N` after the check.
11 : Return Statement
return string(1, s[i]);
Returns the character at `s[i]` as a string when the k-th position is found.
12 : Return Statement
return "hello yasir";
Returns a default string "hello yasir" if the character is not found.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear with respect to the length of the input string, as we only need to process the string once.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant since we are only keeping track of a few variables and do not need to store the decoded string.
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