package com.fishercoder.solutions;

/**
 * 484. Find Permutation
 *
 * By now, you are given a secret signature consisting of character 'D' and 'I'.
 * 'D' represents a decreasing relationship between two numbers, 'I' represents an increasing relationship between two numbers.
 * And our secret signature was constructed by a special integer array, which contains uniquely all the different number from 1 to n (n is the length of the secret signature plus 1).
 * For example, the secret signature "DI" can be constructed by array [2,1,3] or [3,1,2],
 * but won't be constructed by array [3,2,4] or [2,1,3,4], which are both illegal constructing special string that can't represent the "DI" secret signature.

 On the other hand, now your job is to find the lexicographically smallest permutation of [1, 2, ... n] could refer to the given secret signature in the input.

 Example 1:
 Input: "I"
 Output: [1,2]
 Explanation: [1,2] is the only legal initial spectial string can construct secret signature "I", where the number 1 and 2 construct an increasing relationship.

 Example 2:
 Input: "DI"
 Output: [2,1,3]
 Explanation: Both [2,1,3] and [3,1,2] can construct the secret signature "DI",
 but since we want to find the one with the smallest lexicographical permutation, you need to output [2,1,3]

 Note:
 The input string will only contain the character 'D' and 'I'.
 The length of input string is a positive integer and will not exceed 10,000
 */
public class _484 {
    public static class Solution1 {

        /**
         * credit:https://discuss.leetcode.com/topic/76221/java-o-n-clean-solution-easy-to-understand
         *
         * For example, given IDIIDD we start with sorted sequence 1234567
         * Then for each k continuous D starting at index i we need to reverse [i, i+k] portion of the sorted sequence.
         *
         * e.g.
         * IDIIDD
         *
         * 1234567 // sorted
         * 1324765 // answer
         */
        public int[] findPermutation(String s) {
            int[] result = new int[s.length() + 1];
            for (int i = 0; i <= s.length(); i++) {
                result[i] = i + 1;
            }
            for (int i = 0; i < s.length(); i++) {
                if (s.charAt(i) == 'D') {
                    int left = i;
                    while (i < s.length() && s.charAt(i) == 'D') {
                        i++;
                    }
                    reverse(result, left, i);
                }
            }
            return result;
        }

        private void reverse(int[] result, int left, int i) {
            while (left < i) {
                int temp = result[left];
                result[left] = result[i];
                result[i] = temp;
                left++;
                i--;
            }
        }
    }

}

package com.fishercoder.solutions;

public class _484 {
    public static class Solution1 {

        /**
         * credit:https://discuss.leetcode.com/topic/76221/java-o-n-clean-solution-easy-to-understand
         * <p>
         * For example, given IDIIDD we start with sorted sequence 1234567
         * Then for each k continuous D starting at index i we need to reverse [i, i+k] portion of the sorted sequence.
         * <p>
         * e.g.
         * IDIIDD
         * <p>
         * 1234567 // sorted
         * 1324765 // answer
         */
        public int[] findPermutation(String s) {
            int[] result = new int[s.length() + 1];
            for (int i = 0; i <= s.length(); i++) {
                result[i] = i + 1;
            }
            for (int i = 0; i < s.length(); i++) {
                if (s.charAt(i) == 'D') {
                    int left = i;
                    while (i < s.length() && s.charAt(i) == 'D') {
                        i++;
                    }
                    reverse(result, left, i);
                }
            }
            return result;
        }

        private void reverse(int[] result, int left, int i) {
            while (left < i) {
                int temp = result[left];
                result[left] = result[i];
                result[i] = temp;
                left++;
                i--;
            }
        }
    }

}

refactor 484