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Problem solution · Java

Count Stepping Numbers in Range

Count Stepping Numbers in Range: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
41 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Count Stepping Numbers in Range, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 41 lines of Java from the credited upstream file 2801.java.
  • The implementation visibly relies on sequence storage.
  • 1 loop block detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCount Stepping Numbers in Range · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int countSteppingNumbers(String low, String high) {    final String lowWithLeadingZeros =        String.valueOf('0').repeat(high.length() - low.length()) + low;    Integer[][][][] mem = new Integer[high.length()][11][2][2];    return count(lowWithLeadingZeros, high, 0, 10, /*isLeadingZero=*/true, true, true, mem);  }   private static final int MOD = 1_000_000_007;   // Returns the number of valid integers, considering the i-th digit, where  // `prevDigit` is the previous digit, `tight1` indicates if the current  // digit is tightly bound for `low`, and `tight2` indicates if the current  // digit is tightly bound for `high`.  private int count(final String low, final String high, int i, int prevDigit,                    boolean isLeadingZero, boolean tight1, boolean tight2, Integer[][][][] mem) {    if (i == high.length())      return 1;    if (mem[i][prevDigit][tight1 ? 1 : 0][tight2 ? 1 : 0] != null)      return mem[i][prevDigit][tight1 ? 1 : 0][tight2 ? 1 : 0];     int res = 0;    final int minDigit = tight1 ? low.charAt(i) - '0' : 0;    final int maxDigit = tight2 ? high.charAt(i) - '0' : 9;     for (int d = minDigit; d <= maxDigit; ++d) {      final boolean nextTight1 = tight1 && (d == minDigit);      final boolean nextTight2 = tight2 && (d == maxDigit);      if (isLeadingZero)        // Can place any digit in [minDigit, maxDigit].        res += count(low, high, i + 1, d, isLeadingZero && d == 0, nextTight1, nextTight2, mem);      else if (Math.abs(d - prevDigit) == 1)        // Can only place prevDigit - 1 or prevDigit + 1.        res += count(low, high, i + 1, d, false, nextTight1, nextTight2, mem);      res %= MOD;    }     return mem[i][prevDigit][tight1 ? 1 : 0][tight2 ? 1 : 0] = res;  }}

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