## Minimum number of key presses to type the given string in C++## Introduction:You can use of key presses to type a given string in C++. The idea is to build a table where each entry minimum number represents the minimum number of key presses required to type the dp[i][j] The table is filled in a bottom-up manner.substring s[i..j].## Program:
Enter the string: abcdefghij Minimum number of keypresses: 10
**Initialization:**
In this example, the code starts by including necessary The required to type a given string.minimum keypressesA 2D vector named dp is created to store the results. The function takes a string as input: **Dynamic Programming Table:**
The diagonal elements of the dp table are initialized to 1. This is because a substring of length 1 requires only one keypress. The function then iterates over all possible substring lengths (2 or more) and considers all possible substrings within the given string. **Palindrome Check:**
For each substring, it checks if the endpoints are the same (i.e., the substring is a palindrome). If so, it sets the keypresses to If the endpoints are the same, the result is the same as for the substring without those endpoints. **Partitioning:**
The code then tries different partition points within the substring and updates the minimum keypresses accordingly. It checks all possible partitions and keeps track of the minimum keypresses required. **Result:**
The final result is stored in **Main Function:**
The main function takes user input for the string. It calls the
The code uses a bottom-up dynamic programming approach to fill in the dp table. The two nested loops iterate over all possible substring lengths and positions within the string. The third loop iterates over possible partition points. Therefore, the time complexity is O(n^3), where 'n' is the length of the input string.
The 2D vector dp determine The size of the dp table is is the length of the input string.'n'Therefore, the space complexity is ## Approach 1: Using Greedy AlgorithmIterate through the string and count consecutive characters. If a character is repeated consecutively, you can consider typing the character and the number of times it repeats in a single keypress. ## Program:
Enter the string: JavaTpoint Minimum number of keypresses: 10
**Input:**
Takes a string s as input. **Edge Case:**
Checks if the string is empty. If it is, the function returns 0 as there are no characters to type. **Initialization:**
Initializes totalKeypresses to **Iterating through the String:**
Uses a loop to go through each character in the string. For each character, it counts how many consecutive characters are the same using a while loop. While counting, it moves to the next character until a different one is encountered. **Calculating Keypresses:**
Adds to If the count is **Result:**
It returns the final value of totalKeypresses, representing the minimum keypresses needed to type the given string.
**Input:**
It takes user input for the string. **Function Call:**
Calls the **Printing Result:**
Prints the minimum number of keypresses obtained from the function.
The for loop iterates through each character in the string once. The while loop inside it counts consecutive characters. In the worst case, each character might be visited twice (once in the outer loop and once in the while loop). Therefore, the time complexity for the iteration is
The calculations inside the loop take constant time per character. The overall time complexity remains
The main function takes input and calls the O(n).The overall time complexity of the code is
The code uses a constant amount of additional space regardless of the input size. Variables like variables don't depend on the input size.loopThe space complexity is O(1), indicating constant space usage. ## Approach 2: Using Manacher's Algorithm for Longest Palindromic SubstringFind the longest palindromic substring in the given string. The length of the remaining string minus the length of the palindromic substring gives the minimum keypresses needed. ## Program:
Enter the string: abcde Minimum number of keypresses: 4
**Input:**
It takes a string s as input. **Process:**
Inserts a
**Input:**
Takes the preprocessed string as input. **Initialization:**
Initializes variables for the center and right boundary of the rightmost palindrome found so far. **Palindrome Lengths Array:**
Creates an array to store the lengths of palindromes centered at each index. **Algorithm Execution:**
Iterates through each character in the preprocessed string. Utilizes symmetry properties to avoid unnecessary comparisons and efficiently finds the length of palindromes centered at each index. **Update Center and Right Boundary:**
Updates the center and right boundary based on the palindromes found. **Finding Maximum Length:**
Finds the maximum length in the palindrome lengths array. **Result:**
Returns the length of the longest palindromic substring.
**Input:**
Takes the original string as input. **Longest Palindromic Substring Length:**
Calls the **Calculating Minimum Keypresses:**
Calculate the minimum keypresses by subtracting the length of the longest palindromic substring from the total length of the original string. **Result:**
Returns the
**Input:**
Takes user input for the **Function Call:**
Calls the **Printing Result:**
It prints the minimum number of keypresses obtained from the function.
Iterates through each character in the original string once to insert The time complexity of this function is
It iterates through each character in the preprocessed string once. The time complexity of Manacher's algorithm is
It calls the .O(n)Performs simple arithmetic operations, which take constant time. The overall time complexity of the code is
Creates a new string with length 2n (each character in the original string plus the inserted '#'). The space complexity of this function is
Uses additional space for the palindrome lengths array, which has length n. The space complexity of Manacher's algorithm is
It uses The space complexity is The overall space complexity of the code is O(n). ## Approach 3: Using Count Unique CharactersCount the number of The minimum keypresses would be the length of the string minus the count of unique characters. ## Program:
Enter the string: jtp Minimum number of keypresses: 0
**Input:**
The **Set for Unique Characters:**
Initializes an **Traverse and Insert:**
It iterates through each character in the string and inserts it into the set. The set automatically ensures uniqueness. **Minimum Keypresses Calculation:**
It calculates the minimum keypresses using the formula: length of string - number of unique characters. **Result:**
Returns the minimum keypresses as the result.
Iterates through each character in the string once and inserts it into the unordered set. The time complexity of this operation is is the length of the input string.'n'
Calculating the minimum keypresses involves finding the size of the unordered set. The time complexity of finding the size of an unordered set is The overall time complexity of the code
The space complexity is influenced by the unordered set, which stores unique characters. In the worst case, the space required is is the length of the input string and 'm' is the number of unique characters.'n'
Other than the unordered set, the code uses a constant amount of additional space for variables like n, result, and the loop variables. The overall space complexity is is the length of the input string and 'm' is the number of unique characters.'n'## Approach 4: Using Divide and ConquerThe ## Program:
Enter the string: world Minimum number of keypresses: 5
Represents a function that calculates the minimum keypresses for a given segment. This function can be customized based on characteristics of the segments.
It takes the input string s and the indices start and end to define the segment.
as typing a single character requires one keypress.return 1Divide the segment into two halves (start to mid and Recursively calculate the minimum keypresses for each half. Combine the results based on the characteristics of the segments. In this example, we add the keypresses for each half. Return the combined result.
Takes user input for the Calls the
In each recursive call, the function divides the segment into two halves. The number of recursive calls is proportional to the depth of the recursion tree, which is logarithmic in the length of the input string. Each level of the recursion tree takes The overall time complexity is
The time complexity of this function depends on the custom logic implemented to calculate keypresses for a segment. In the provided example, it simply returns the length of the segment, which takes If more complex logic is implemented, the time complexity may vary. The recursive calls dominate the overall time complexity and are
The space complexity is determined by the maximum depth of the recursion stack. In the worst case, the maximum depth is logarithmic in the length of the input string. Therefore, the space complexity due to recursion is
The code uses a constant amount of additional space for variables like mid, The overall space complexity is |

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