Optimal Partition of String

class Solution {
public:
    int countUniquePartitions(string str) {
        vector<int> lastPosition(26, -1);
        int partitionCount = 1, startIdx = 0;
    
        for (int index = 0; index < str.size(); index++) {
            if (lastPosition[str[index] - 'a'] >= startIdx) {
                partitionCount++;
                startIdx = index;
            }
            lastPosition[str[index] - 'a'] = index;
        }
    
        return partitionCount;
    }
};

The given C++ program provides a solution to the problem of partitioning a string into the maximum number of substrings, such that each substring contains unique characters. Here's a breakdown of how the solution works:

• lastPosition vector keeps track of the last position of each character in the alphabet relative to the string being analyzed.
• partitionCount starts at 1, representing at least one partition.
• startIdx represents the starting index of the current partition.

For each character in the string:

• If the character has appeared before and its last known position is within the boundary of the current partition (lastPosition[str[index] - 'a'] >= startIdx), then start a new partition and update startIdx to the current character's index.
• Update the lastPosition of the current character to the current index (lastPosition[str[index] - 'a'] = index).

At the end of the loop, partitionCount will contain the number of unique partitions possible. This approach ensures that you compute the maximum number of substrings where no character repeats within each substring, effectively parsing the string for the optimal partition. The solution works efficiently with a linear pass over the string and constant space relative to the charset involved (i.e., 26 for the English alphabet).

class Solution {
    public int splitString(String input) {
        int[] charIndex = new int[26];
        Arrays.fill(charIndex, -1);
        int partitions = 1, start = 0;
    
        for (int i = 0; i < input.length(); i++) {
            if (charIndex[input.charAt(i) - 'a'] >= start) {
                partitions++;
                start = i;
            }
            charIndex[input.charAt(i) - 'a'] = i;
        }
    
        return partitions;
    }
}

The provided Java solution aims to determine the optimal number of partitions a string can be divided into so that no letter appears in more than one partition. It works by tracking the last occurrence of each character and adjusting the start of a new partition whenever a character is found that would otherwise conflict with the integrity of the current partition.

Here's a concise explanation of the algorithm:

1. Initialize an array charIndex to store the most recent index of each character encountered in the string. The array is initially filled with -1 to indicate no occurrence.
2. Set the number of partitions to 1 as the string will have at least one partition.
3. Iterate through every character in the input string using a for loop:
   • Calculate the array index for each character by subtracting 'a' from the character itself.
   • Check if the current character was seen after the start of the current partition (start):
     • If yes, increment the partition count (partitions).
     • Update start to the current index i.
   • Update the latest occurrence of the current character in the charIndex array.

Ultimately, return the total number of partitions (partitions) that were calculated. The logic ensures each character only appears in one partition, leading to the optimal subdivision of the string.

class Solution:
    def countSubstrings(self, input_string: str) -> int:
        last_position = [-1] * 26
        partition_count = 1
        start_index = 0
    
        for index in range(len(input_string)):
            char_index = ord(input_string[index]) - ord('a')
            if last_position[char_index] >= start_index:
                partition_count += 1
                start_index = index
            last_position[char_index] = index
    
        return partition_count

The given Python script defines a solution class method named countSubstrings which aims to analyze optimal partitions of a string such that no letter appears in more than one partition. Here's a breakdown of how the method works:

• Instantiate an array last_position with a size of 26 (representing each letter of the alphabet) initialized to -1. This array tracks the most recent index where each character was encountered.
• Set partition_count to 1, starting as the first partition.
• Initialize start_index at 0 to mark the starting index of the current partition.

The method then iterates through each character in the input string:

• Calculate char_index by determining the position of the character in the alphabet using its Unicode code point offset by that of 'a'.
• If the character has been seen after start_index, increment partition_count as this requires starting a new partition. Adjust start_index to the current character's index.
• Update the last_position array with the current index for the character.

After iterating through the string, the method returns partition_count which reflects the number of distinct partitions needed in order to ensure that no character gets repeated across partitions.

This structured approach ensures an efficient partition strategy by keeping track of character positions dynamically, making the complexity of this solution largely depend on the length of the string, O(n).