LeetCode Solutions

423. Reconstruct Original Digits from English

Time: $O(n)$

Space: $O(n)$

			

class Solution {
 public:
  string originalDigits(string s) {
    string ans;
    vector<int> count(10);

    for (const char c : s) {
      if (c == 'z')
        ++count[0];
      if (c == 'o')
        ++count[1];
      if (c == 'w')
        ++count[2];
      if (c == 'h')
        ++count[3];
      if (c == 'u')
        ++count[4];
      if (c == 'f')
        ++count[5];
      if (c == 'x')
        ++count[6];
      if (c == 's')
        ++count[7];
      if (c == 'g')
        ++count[8];
      if (c == 'i')
        ++count[9];
    }

    count[1] -= count[0] + count[2] + count[4];
    count[3] -= count[8];
    count[5] -= count[4];
    count[7] -= count[6];
    count[9] -= count[5] + count[6] + count[8];

    for (int i = 0; i < 10; ++i)
      for (int j = 0; j < count[i]; ++j)
        ans += i + '0';

    return ans;
  }
};

			

class Solution {
  public String originalDigits(String s) {
    StringBuilder sb = new StringBuilder();
    int[] count = new int[10];

    for (final char c : s.toCharArray()) {
      if (c == 'z')
        ++count[0];
      if (c == 'o')
        ++count[1];
      if (c == 'w')
        ++count[2];
      if (c == 'h')
        ++count[3];
      if (c == 'u')
        ++count[4];
      if (c == 'f')
        ++count[5];
      if (c == 'x')
        ++count[6];
      if (c == 's')
        ++count[7];
      if (c == 'g')
        ++count[8];
      if (c == 'i')
        ++count[9];
    }

    count[1] -= count[0] + count[2] + count[4];
    count[3] -= count[8];
    count[5] -= count[4];
    count[7] -= count[6];
    count[9] -= count[5] + count[6] + count[8];

    for (int i = 0; i < 10; ++i)
      for (int j = 0; j < count[i]; ++j)
        sb.append(i);

    return sb.toString();
  }
}

			

class Solution:
  def originalDigits(self, s: str) -> str:
    count = [0] * 10

    for c in s:
      if c == 'z':
        count[0] += 1
      if c == 'o':
        count[1] += 1
      if c == 'w':
        count[2] += 1
      if c == 'h':
        count[3] += 1
      if c == 'u':
        count[4] += 1
      if c == 'f':
        count[5] += 1
      if c == 'x':
        count[6] += 1
      if c == 's':
        count[7] += 1
      if c == 'g':
        count[8] += 1
      if c == 'i':
        count[9] += 1

    count[1] -= count[0] + count[2] + count[4]
    count[3] -= count[8]
    count[5] -= count[4]
    count[7] -= count[6]
    count[9] -= count[5] + count[6] + count[8]

    return ''.join(chr(i + ord('0')) for i, c in enumerate(count) for j in range(c))