Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
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 <= 100low and high consist of only digits.low and high don't have any leading zeros.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
"1" "11"
"90" "101"
stepping-numbers)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2801: Count Stepping Numbers in Range
import java.math.BigInteger;
class Solution {
private final int mod = (int) 1e9 + 7;
private String num;
private Integer[][] f;
public int countSteppingNumbers(String low, String high) {
f = new Integer[high.length() + 1][10];
num = high;
int a = dfs(0, -1, true, true);
f = new Integer[high.length() + 1][10];
num = new BigInteger(low).subtract(BigInteger.ONE).toString();
int b = dfs(0, -1, true, true);
return (a - b + mod) % mod;
}
private int dfs(int pos, int pre, boolean lead, boolean limit) {
if (pos >= num.length()) {
return lead ? 0 : 1;
}
if (!lead && !limit && f[pos][pre] != null) {
return f[pos][pre];
}
int ans = 0;
int up = limit ? num.charAt(pos) - '0' : 9;
for (int i = 0; i <= up; ++i) {
if (i == 0 && lead) {
ans += dfs(pos + 1, pre, true, limit && i == up);
} else if (pre == -1 || Math.abs(pre - i) == 1) {
ans += dfs(pos + 1, i, false, limit && i == up);
}
ans %= mod;
}
if (!lead && !limit) {
f[pos][pre] = ans;
}
return ans;
}
}
// Accepted solution for LeetCode #2801: Count Stepping Numbers in Range
func countSteppingNumbers(low string, high string) int {
const mod = 1e9 + 7
f := [110][10]int{}
for i := range f {
for j := range f[i] {
f[i][j] = -1
}
}
num := high
var dfs func(int, int, bool, bool) int
dfs = func(pos, pre int, lead bool, limit bool) int {
if pos >= len(num) {
if lead {
return 0
}
return 1
}
if !lead && !limit && f[pos][pre] != -1 {
return f[pos][pre]
}
var ans int
up := 9
if limit {
up = int(num[pos] - '0')
}
for i := 0; i <= up; i++ {
if i == 0 && lead {
ans += dfs(pos+1, pre, true, limit && i == up)
} else if pre == -1 || abs(pre-i) == 1 {
ans += dfs(pos+1, i, false, limit && i == up)
}
ans %= mod
}
if !lead && !limit {
f[pos][pre] = ans
}
return ans
}
a := dfs(0, -1, true, true)
t := []byte(low)
for i := len(t) - 1; i >= 0; i-- {
if t[i] != '0' {
t[i]--
break
}
t[i] = '9'
}
num = string(t)
f = [110][10]int{}
for i := range f {
for j := range f[i] {
f[i][j] = -1
}
}
b := dfs(0, -1, true, true)
return (a - b + mod) % mod
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
# Accepted solution for LeetCode #2801: Count Stepping Numbers in Range
class Solution:
def countSteppingNumbers(self, low: str, high: str) -> int:
@cache
def dfs(pos: int, pre: int, lead: bool, limit: bool) -> int:
if pos >= len(num):
return int(not lead)
up = int(num[pos]) if limit else 9
ans = 0
for i in range(up + 1):
if i == 0 and lead:
ans += dfs(pos + 1, pre, True, limit and i == up)
elif pre == -1 or abs(i - pre) == 1:
ans += dfs(pos + 1, i, False, limit and i == up)
return ans % mod
mod = 10**9 + 7
num = high
a = dfs(0, -1, True, True)
dfs.cache_clear()
num = str(int(low) - 1)
b = dfs(0, -1, True, True)
return (a - b) % mod
// Accepted solution for LeetCode #2801: Count Stepping Numbers in Range
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2801: Count Stepping Numbers in Range
// import java.math.BigInteger;
//
// class Solution {
// private final int mod = (int) 1e9 + 7;
// private String num;
// private Integer[][] f;
//
// public int countSteppingNumbers(String low, String high) {
// f = new Integer[high.length() + 1][10];
// num = high;
// int a = dfs(0, -1, true, true);
// f = new Integer[high.length() + 1][10];
// num = new BigInteger(low).subtract(BigInteger.ONE).toString();
// int b = dfs(0, -1, true, true);
// return (a - b + mod) % mod;
// }
//
// private int dfs(int pos, int pre, boolean lead, boolean limit) {
// if (pos >= num.length()) {
// return lead ? 0 : 1;
// }
// if (!lead && !limit && f[pos][pre] != null) {
// return f[pos][pre];
// }
// int ans = 0;
// int up = limit ? num.charAt(pos) - '0' : 9;
// for (int i = 0; i <= up; ++i) {
// if (i == 0 && lead) {
// ans += dfs(pos + 1, pre, true, limit && i == up);
// } else if (pre == -1 || Math.abs(pre - i) == 1) {
// ans += dfs(pos + 1, i, false, limit && i == up);
// }
// ans %= mod;
// }
// if (!lead && !limit) {
// f[pos][pre] = ans;
// }
// return ans;
// }
// }
// Accepted solution for LeetCode #2801: Count Stepping Numbers in Range
function countSteppingNumbers(low: string, high: string): number {
const mod = 1e9 + 7;
const m = high.length;
let f: number[][] = Array(m + 1)
.fill(0)
.map(() => Array(10).fill(-1));
let num = high;
const dfs = (pos: number, pre: number, lead: boolean, limit: boolean): number => {
if (pos >= num.length) {
return lead ? 0 : 1;
}
if (!lead && !limit && f[pos][pre] !== -1) {
return f[pos][pre];
}
let ans = 0;
const up = limit ? +num[pos] : 9;
for (let i = 0; i <= up; i++) {
if (i == 0 && lead) {
ans += dfs(pos + 1, pre, true, limit && i == up);
} else if (pre == -1 || Math.abs(pre - i) == 1) {
ans += dfs(pos + 1, i, false, limit && i == up);
}
ans %= mod;
}
if (!lead && !limit) {
f[pos][pre] = ans;
}
return ans;
};
const a = dfs(0, -1, true, true);
num = (BigInt(low) - 1n).toString();
f = Array(m + 1)
.fill(0)
.map(() => Array(10).fill(-1));
const b = dfs(0, -1, true, true);
return (a - b + mod) % mod;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.