Check If String Is Transformable With Substring Sort Operations LeetCode solution in Java

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class Solution {
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  public boolean isTransformable(String s, String t) {
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    if (!equal(s, t)) return false;
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    int[] countS = new int[10];
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    int[] countT = new int[10];
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    int[][] prev = new int[10][10];
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    int[][] after = new int[10][10];
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    for (int i = 0; i < s.length(); i++) {
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      int s1 = s.charAt(i) - '0';
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      int t1 = t.charAt(i) - '0';
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      countS[s1]++;
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      countT[t1]++;
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      // This step is to calculate how many digit less than s1/t1 occur before time i
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      // Store the frequency pair into 2-d array
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      for (int j = 0; j < s1; j++) {
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        if (countS[j] == 0) continue;
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        prev[j][s1] += countS[j];
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      }
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      for (int j = 0; j < t1; j++) {
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        if (countT[j] == 0) continue;
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        after[j][t1] += countT[j];
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      }
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    }
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    for (int i = 0; i <= 8; i++) {
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      for (int j = i + 1; j <= 9; j++) {
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        if (prev[i][j] == 0) continue;
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        // Check if any ascending pair's frequency has been reduced after modified.
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        if (after[i][j] < prev[i][j]) return false;
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      }
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    }
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    return true;
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  }
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  // Judge whether the two strings has the same digits
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  public boolean equal(String s, String t) {
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    char[] sc = s.toCharArray();
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    char[] tc = t.toCharArray();
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    Arrays.sort(sc);
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    Arrays.sort(tc);
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    for (int i = 0; i < s.length(); i++) {
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      if (sc[i] != tc[i]) return false;
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    }
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    return true;
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  }
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}