LeetCode-in-Java
2801. Count Stepping Numbers in Range
Hard
Given two positive integers low and high represented as strings, find the count of stepping numbers in the inclusive range [low, high].
A stepping number is an integer such that all of its adjacent digits have an absolute difference of exactly 1.
Return an integer denoting the count of stepping numbers in the inclusive range [low, high].
Since the answer may be very large, return it modulo 109 + 7.
Note: A stepping number should not have a leading zero.
Example 1:
Input: low = “1”, high = “11”
Output: 10
Explanation: The stepping numbers in the range [1,11] are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. There are a total of 10 stepping numbers in the range. Hence, the output is 10.
Example 2:
Input: low = “90”, high = “101”
Output: 2
Explanation: The stepping numbers in the range [90,101] are 98 and 101. There are a total of 2 stepping numbers in the range. Hence, the output is 2.
Constraints:
1 <= int(low) <= int(high) < 101001 <= low.length, high.length <= 100lowandhighconsist of only digits.lowandhighdon’t have any leading zeros.
Solution
public class Solution {
private static final int MOD = (int) (1e9 + 7);
private Integer[][][][] dp;
public int countSteppingNumbers(String low, String high) {
dp = new Integer[low.length() + 1][10][2][2];
int count1 = solve(low, 0, 0, 1, 1);
dp = new Integer[high.length() + 1][10][2][2];
int count2 = solve(high, 0, 0, 1, 1);
return (count2 - count1 + isStep(low) + MOD) % MOD;
}
private int solve(String s, int i, int prevDigit, int hasBound, int curIsZero) {
if (i >= s.length()) {
if (curIsZero == 1) {
return 0;
}
return 1;
}
if (dp[i][prevDigit][hasBound][curIsZero] != null) {
return dp[i][prevDigit][hasBound][curIsZero];
}
int count = 0;
int limit = 9;
if (hasBound == 1) {
limit = s.charAt(i) - '0';
}
for (int digit = 0; digit <= limit; digit++) {
int nextIsZero = (curIsZero == 1 && digit == 0) ? 1 : 0;
int nextHasBound = (hasBound == 1 && digit == limit) ? 1 : 0;
if (curIsZero == 1 || Math.abs(digit - prevDigit) == 1) {
count = (count + solve(s, i + 1, digit, nextHasBound, nextIsZero)) % MOD;
}
}
dp[i][prevDigit][hasBound][curIsZero] = count;
return dp[i][prevDigit][hasBound][curIsZero];
}
private int isStep(String s) {
boolean isValid = true;
for (int i = 0; i < s.length() - 1; i++) {
if (Math.abs(s.charAt(i + 1) - s.charAt(i)) != 1) {
isValid = false;
break;
}
}
if (isValid) {
return 1;
}
return 0;
}
}