Leetcode 2746: Decremental String Concatenation
You are given an array of distinct strings. Your task is to perform n-1 join operations optimally to minimize the length of the final string.
Problem
Approach
Steps
Complexity
Input: The input consists of an array `words` of size `n` where each element is a distinct string.
Example: words = ["xx", "xy", "yz"]
Constraints:
• 1 <= words.length <= 1000
• 1 <= words[i].length <= 50
• Each character in words[i] is a lowercase English letter
Output: Return the minimum possible length of the final string after performing the join operations.
Example: 4
Constraints:
• The output will be an integer representing the minimum possible length.
Goal: Minimize the final length of the string by choosing optimal join operations at each step.
Steps:
• Start with the first string in the array.
• For each subsequent string, try both possible join operations (str(i-1) + words[i] or words[i] + str(i-1)) and choose the one that minimizes the length.
• Repeat this process for all strings.
Goal: The problem must be solved under the provided constraints.
Steps:
• 1 <= words.length <= 1000
• 1 <= words[i].length <= 50
• Each character in words[i] is a lowercase English letter
Assumptions:
• Each string in the array is distinct.
• The array may contain up to 1000 strings.
• Input: words = ["xx", "xy", "yz"]
• Explanation: By performing the join operations in the optimal order, we minimize the final length to 4.
Approach: The approach to solving this problem involves dynamic programming with memoization to store the results of subproblems.
Observations:
• The problem asks us to minimize the final string length by performing optimal join operations.
• We need to consider two possible join operations for each string and choose the one that minimizes the length.
• A dynamic programming approach with memoization can help solve this problem efficiently.
Steps:
• Use dynamic programming to store the minimum length at each stage of the concatenation process.
• Consider the two possible join operations at each step and update the result accordingly.
Empty Inputs:
• An empty input array should return 0.
Large Inputs:
• Handle arrays with up to 1000 strings efficiently.
Special Values:
• Consider cases where all strings have the same characters.
Constraints:
• Ensure that the solution works within the time limits for large inputs.
int n ;
int dp[1010][26][26] ;
int rec(int level , int first , int last , vector <string>& arr){
if(level == n){
return 0 ;
}
if(dp[level][first][last] != -1){
return dp[level][first][last] ;
}
int ans = 1e9 ;
int t = (int)arr[level].size() ;
if(arr[level][t-1] - 'a' == first){
ans = min(ans , t-1 + rec(level+1 , arr[level][0] - 'a' , last , arr) ) ;
}
ans = min(ans , t + rec(level+1 , arr[level][0] - 'a' , last , arr)) ;
if(arr[level][0] - 'a' == last){
ans = min(ans , t-1 + rec(level+1 , first , arr[level][t-1] - 'a' , arr)) ;
}
ans = min(ans , t + rec(level+1 , first , arr[level][t-1] - 'a' , arr)) ;
return dp[level][first][last] = ans ;
}
int minimizeConcatenatedLength(vector<string>& words) {
n = (int)words.size() ;
memset(dp , -1 , sizeof(dp)) ;
int t = (int)words[0].size() ;
int ans = t + rec(1 , words[0][0] -'a' , words[0][t-1]-'a' , words) ;
return ans ;
}
1 : Variable Initialization
int n ;
Declares the integer variable `n` to store the number of words in the input list.
2 : Array Initialization
int dp[1010][26][26] ;
Declares a 3D array `dp` to store intermediate results for dynamic programming (with dimensions for words, first and last characters).
3 : Function Definition
int rec(int level , int first , int last , vector <string>& arr){
Defines the recursive function `rec` that calculates the minimum concatenated length from a specific `level` of the word list, considering the `first` and `last` characters of the current word.
4 : Base Case
if(level == n){
Checks if the function has processed all words in the list.
5 : Base Case Return
return 0 ;
Returns 0 if all words have been processed, indicating no more concatenation is needed.
6 : Memoization Check
if(dp[level][first][last] != -1){
Checks if the result for this specific subproblem has already been computed and stored.
7 : Memoization Return
return dp[level][first][last] ;
Returns the stored result from the `dp` array if it has been previously computed.
8 : Initialization of Answer
int ans = 1e9 ;
Initializes the variable `ans` to a large number, representing the best possible answer for the current subproblem.
9 : String Length Calculation
int t = (int)arr[level].size() ;
Calculates the length `t` of the current word at `level`.
10 : First Character Check
if(arr[level][t-1] - 'a' == first){
Checks if the last character of the current word matches the `first` character from the previous word.
11 : Recursive Call with Matching Last Character
ans = min(ans , t-1 + rec(level+1 , arr[level][0] - 'a' , last , arr) ) ;
Recursively calculates the minimum concatenated length if the last character of the current word matches the first character of the previous word.
12 : Recursive Call without Matching Last Character
ans = min(ans , t + rec(level+1 , arr[level][0] - 'a' , last , arr)) ;
Recursively calculates the minimum concatenated length if the last character does not match.
13 : Last Character Check
if(arr[level][0] - 'a' == last){
Checks if the first character of the current word matches the `last` character of the previous word.
14 : Recursive Call with Matching First Character
ans = min(ans , t-1 + rec(level+1 , first , arr[level][t-1] - 'a' , arr)) ;
Recursively calculates the minimum concatenated length if the first character of the current word matches the last character of the previous word.
15 : Recursive Call without Matching First Character
ans = min(ans , t + rec(level+1 , first , arr[level][t-1] - 'a' , arr)) ;
Recursively calculates the minimum concatenated length if the first character does not match.
16 : Memoization Store
return dp[level][first][last] = ans ;
Stores the computed result for the current subproblem in the `dp` array.
17 : Function Definition
int minimizeConcatenatedLength(vector<string>& words) {
Defines the main function to minimize the concatenated length of the words.
18 : Words Size Calculation
n = (int)words.size() ;
Calculates the number of words in the input list.
19 : Memory Initialization
memset(dp , -1 , sizeof(dp)) ;
Initializes the `dp` array with -1 to indicate that no subproblem has been solved yet.
20 : Word Length Calculation
int t = (int)words[0].size() ;
Calculates the length of the first word in the list.
21 : Recursive Call to Minimize Length
int ans = t + rec(1 , words[0][0] -'a' , words[0][t-1]-'a' , words) ;
Starts the recursion with the first word and calculates the minimized concatenated length.
22 : Return Result
return ans ;
Returns the minimized concatenated length as the result.
Best Case: O(n)
Average Case: O(n * m)
Worst Case: O(n * m * 26 * 26)
Description: Where n is the number of strings, and m is the average length of each string.
Best Case: O(n)
Worst Case: O(n * 26 * 26)
Description: The space complexity is dominated by the DP table storing results for each string and character pair.
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