Leetcode 1849: Splitting a String Into Descending Consecutive Values
Given a string s consisting of digits, check if it is possible to split it into two or more non-empty substrings such that the numerical values of the substrings are in strictly descending order and the difference between the values of adjacent substrings is exactly 1.
Problem
Approach
Steps
Complexity
Input: The input consists of a string s made up of digits.
Example: "5432"
Constraints:
• 1 <= s.length <= 20
• s consists only of digits
Output: Return true if it is possible to split the string into substrings satisfying the conditions; otherwise, return false.
Example: true
Constraints:
Goal: To find a valid way to split the string into substrings that meet the conditions.
Steps:
• Start by iterating through the string and attempting to split it into substrings.
• For each split, check if the substrings' numerical values are in strictly descending order with a difference of exactly 1 between adjacent values.
• Return true if a valid split is found, otherwise return false.
Goal: The input string s is guaranteed to have a length between 1 and 20, and it only contains digits.
Steps:
• 1 <= s.length <= 20
• s consists only of digits.
Assumptions:
• The input string will always contain at least one character.
• Input: "1000001"
• Explanation: The string can be split into the substrings ['1', '00000', '1'], which correspond to the values [1, 0, 1]. These values are in descending order with a difference of 1, so the split is valid.
Approach: We will attempt to split the string in different ways and check if any of those splits satisfy the conditions of strictly descending values and adjacent differences of exactly 1.
Observations:
• We need to check all possible ways of splitting the string and validate if the split satisfies the numerical conditions.
• We can approach this problem using a recursive backtracking method, where we try different splits and check if they meet the criteria.
Steps:
• Start with an empty result array.
• Iterate over possible substrings of s.
• For each substring, convert it to an integer and check if it fits the descending order and difference condition with the previous substring.
• If a valid split is found, return true. If no valid split is found after checking all possibilities, return false.
Empty Inputs:
• The input string will never be empty, as the length is guaranteed to be at least 1.
Large Inputs:
• The string length will not exceed 20, so efficiency for large inputs is not a concern.
Special Values:
• Consider cases where the string consists of repeating digits or where it starts with zeros.
Constraints:
• The solution should handle strings with lengths from 1 to 20 efficiently.
string s;
bool splitString(string s) {
this->s = s;
double tmp = 0;
for(int i = 0; i < s.size() - 1; i++) {
tmp = tmp * 10 + (s[i] - '0');
if(bt(i + 1, tmp)) return true;
}
return false;
}
bool bt(int idx, double prv) {
if(idx == s.size()) return true;
double tmp = 0;
for(int i = idx; i < s.size(); i++) {
tmp = tmp * 10 + (s[i] - '0');
if(tmp - prv > -1) break;
if(tmp - prv == -1) {
if(bt(i + 1, tmp))
return true;
}
}
return false;
}
1 : Variable Declaration
string s;
Declares a string variable `s` which will hold the input string to be split.
2 : Function Definition
bool splitString(string s) {
Defines the `splitString` function, which takes a string `s` as input and returns a boolean indicating whether it can be split into a sequence of numbers with a difference of 1 between consecutive numbers.
3 : Assignment
this->s = s;
Assigns the input string `s` to the member variable `this->s`.
4 : Variable Initialization
double tmp = 0;
Initializes a temporary variable `tmp` of type double to store the value of the current number being formed during the iteration.
5 : Loop Initialization
for(int i = 0; i < s.size() - 1; i++) {
Starts a loop that iterates through the string `s`, except for the last character. The loop is used to form numbers from the string.
6 : String to Number Conversion
tmp = tmp * 10 + (s[i] - '0');
Forms a number by converting the current character `s[i]` to its integer value and updating `tmp`.
7 : Backtracking Function Call
if(bt(i + 1, tmp)) return true;
Calls the backtracking function `bt` with the next index `i + 1` and the current value of `tmp` to check if the string can be split starting from that index.
8 : Return Statement
return false;
If no valid split is found, the function returns false, indicating that the string cannot be split into the desired sequence.
9 : Backtracking Function Definition
bool bt(int idx, double prv) {
Defines the helper function `bt` (backtracking), which attempts to find a valid split starting at index `idx` and with the previous number `prv`.
10 : Base Case
if(idx == s.size()) return true;
Checks if the current index has reached the end of the string. If so, it returns true, indicating a successful split.
11 : Variable Initialization
double tmp = 0;
Initializes a temporary variable `tmp` to store the current number formed from the string.
12 : Loop Initialization
for(int i = idx; i < s.size(); i++) {
Starts a loop that iterates from the current index `idx` to the end of the string `s`. It attempts to form the next possible number.
13 : String to Number Conversion
tmp = tmp * 10 + (s[i] - '0');
Forms a number by converting the current character `s[i]` to its integer value and updating `tmp`.
14 : Backtracking Pruning
if(tmp - prv > -1) break;
Prunes the search space. If the difference between the current number `tmp` and the previous number `prv` is greater than -1, it breaks the loop because the sequence cannot continue.
15 : Backtracking Condition
if(tmp - prv == -1) {
Checks if the difference between the current number `tmp` and the previous number `prv` is exactly -1, which is the condition for a valid split.
16 : Recursive Call
if(bt(i + 1, tmp))
Makes a recursive call to `bt` with the next index `i + 1` and the current value of `tmp` to check if the rest of the string can be split.
17 : Return True
return true;
If the recursive call returns true, indicating a valid split, the function returns true.
18 : Return False
return false;
If no valid split is found, the function returns false.
Best Case: O(n^2)
Average Case: O(n^2)
Worst Case: O(n^2)
Description: The time complexity is quadratic in terms of the length of the string because we check all possible splits and validate each one.
Best Case: O(1)
Worst Case: O(n)
Description: The space complexity is O(n) due to the storage of the intermediate substrings during the split process.
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