Given a string S that only contains “I” (increase) or “D” (decrease), let N = S.length.

Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1:

• If S[i] == "I", then A[i] < A[i+1]
• If S[i] == "D", then A[i] > A[i+1]

Example 1:

Input: "IDID"
Output: [0,4,1,3,2]

Example 2:

Input: "III"
Output: [0,1,2,3]

Example 3:

Input: "DDI"
Output: [3,2,0,1]

Note:

1. 1 <= S.length <= 10000
2. S only contains characters "I" or "D".

Solution: Greedy

“I” -> use the smallest possible number

“D” -> use the largest possible number

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua, 44 ms
class Solution {
public:
  vector<int> diStringMatch(string S) {
    const int n = S.length();
    vector<int> ans;
    int lo = 0;
    int hi = n;
    for (char c : S) {
      if (c == 'I')
        ans.push_back(lo++);
      else
        ans.push_back(hi--);
    }
    ans.push_back(lo);
    return ans;
  }
};

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