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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0901_1000.s0942_di_string_match;

// #Easy #Array #String #Math #Greedy #Two_Pointers
// #2022_03_30_Time_4_ms_(33.74%)_Space_48.7_MB_(20.18%)

/**
 * 942 - DI String Match\.
 *
 * Easy
 *
 * A permutation `perm` of `n + 1` integers of all the integers in the range `[0, n]` can be represented as a string `s` of length `n` where:
 *
 * *   `s[i] == 'I'` if `perm[i] < perm[i + 1]`, and
 * *   `s[i] == 'D'` if `perm[i] > perm[i + 1]`.
 *
 * Given a string `s`, reconstruct the permutation `perm` and return it. If there are multiple valid permutations perm, return **any of them**.
 *
 * **Example 1:**
 *
 * **Input:** s = "IDID"
 *
 * **Output:** [0,4,1,3,2]
 *
 * **Example 2:**
 *
 * **Input:** s = "III"
 *
 * **Output:** [0,1,2,3]
 *
 * **Example 3:**
 *
 * **Input:** s = "DDI"
 *
 * **Output:** [3,2,0,1]
 *
 * **Constraints:**
 *
 * *   1 <= s.length <= 10⁵
 * *   `s[i]` is either `'I'` or `'D'`.
**/
public class Solution {
    public int[] diStringMatch(String s) {
        int[] arr = new int[s.length() + 1];
        int max = s.length();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == 'D') {
                arr[i] = max;
                max--;
            }
        }
        for (int i = s.length() - 1; i >= 0 && max > 0; i--) {
            if (s.charAt(i) == 'I' && arr[i + 1] == 0) {
                arr[i + 1] = max;
                max--;
            }
        }
        for (int i = 0; i < arr.length && max > 0; i++) {
            if (arr[i] == 0) {
                arr[i] = max;
                max--;
            }
        }

        return arr;
    }
}