DI String Match

class Solution {
public:
    vector<int> diStringMatch(string pattern) {
        int length = pattern.size();
        int minVal = 0, maxVal = length;
        vector<int> result(length + 1);

        for (int i = 0; i < length; ++i) {
            if (pattern[i] == 'I') {
                result[i] = minVal++;
            } else {
                result[i] = maxVal--;
            }
        }

        result[length] = minVal;
        return result;
    }
};

The provided C++ solution solves the problem of generating a sequence of integers based on a given pattern of 'I' (increasing) and 'D' (decreasing) characters. Here's a breakdown of how the solution works:

• A string pattern, consisting of characters 'I' and 'D', defines the order.
• Two variables, minVal and maxVal, are initialized to represent the minimum and maximum possible values in the resulting sequence, set initially to 0 and the length of pattern respectively.
• A result vector is created of size equal to pattern.length() + 1 to store the resultant sequence.
• Loop through each character in pattern:
  • If the character is 'I', assign minVal to the current index in result and increment minVal.
  • If the character is 'D', assign maxVal to the current index in result and decrement maxVal.
• After the loop, assign minVal to the last element in result to handle the last index due to 'I' or the smallest remaining number.
• Return the result vector which contains the desired sequence of integers.

This solution ensures that the sequence correctly follows the 'increase' or 'decrease' order dictated by the pattern string, leveraging the properties of integers and looping control flow to construct the result efficiently.

class Solution {

    public int[] createPermutation(String pattern) {
        int size = pattern.length();
        int min = 0, max = size;
        int[] result = new int[size + 1];
        for (int i = 0; i < size; i++) {
            if (pattern.charAt(i) == 'I') result[i] = min++;
            else result[i] = max--;
        }

        result[size] = min;
        return result;
    }
}

The given Java solution addresses the problem of generating an array that matches a specific "DI string" pattern. The method createPermutation takes a string pattern consisting of the characters 'D' and 'I'. Based on this pattern, the method generates an array of integers where 'I' indicates that the sequence must increase and 'D' indicates a decrease. Here's a breakdown of how the solution implements this:

• Initialize size to the length of the pattern to determine the output array’s size, which will be size + 1.
• Set min to 0 and max to size to represent the smallest and largest numbers to be used in the array.
• Create an integer array result of length size + 1.
• Loop through each character of the pattern:
  • If the character is 'I', assign result[i] to min and then increment min.
  • If the character is 'D', assign result[i] to max and then decrement max.
• After exiting the loop, set the last element of result to min to ensure all numbers from 0 to size are utilized.

This implementation efficiently maps the increase/decrease pattern to a concrete number sequence, effectively solving the "DI String Match" problem within the constraints assured by using single-pass logic and direct assignments.

class Solution:
    def orderString(self, input_str: str) -> List[int]:
        low, high = 0, len(input_str)
        result = []
        for char in input_str:
            if char == "I":
                result.append(low)
                low += 1
            else:
                result.append(high)
                high -= 1

        return result + [low]

This Python solution is designed to solve the DI String Match problem by mapping the input string comprised of 'I' and 'D' characters to an array of integers that satisfy certain conditions. The approach utilizes two pointers, low and high, which represent the smallest and largest values yet to be assigned, initialized as 0 and the length of the input string, respectively. As you iterate over the input string:

• Append low to the result list and increment low by 1 for every 'I'.
• Append high to the result list and decrement high by 1 for every 'D'.

After processing all characters, append the current value of low to the result list to account for the final value. This method ensures no missed or duplicated elements, producing a sequence where each 'I' represents an increment and each 'D' represents a decrement in relation to the previous element. The solution effectively translates the pattern in the input string into a sequence of integers meeting the described criteria.