Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You are given two integers low and high.
An integer is called balanced if it satisfies both of the following conditions:
Return an integer representing the number of balanced integers in the range [low, high] (both inclusive).
Example 1:
Input: low = 1, high = 100
Output: 9
Explanation:
The 9 balanced numbers between 1 and 100 are 11, 22, 33, 44, 55, 66, 77, 88, and 99.
Example 2:
Input: low = 120, high = 129
Output: 1
Explanation:
Only 121 is balanced because the sum of digits at even and odd positions are both 2.
Example 3:
Input: low = 1234, high = 1234
Output: 0
Explanation:
1234 is not balanced because the sum of digits at odd positions (1 + 3 = 4) does not equal the sum at even positions (2 + 4 = 6).
Constraints:
1 <= low <= high <= 1015Problem summary: You are given two integers low and high. An integer is called balanced if it satisfies both of the following conditions: It contains at least two digits. The sum of digits at even positions is equal to the sum of digits at odd positions (the leftmost digit has position 1). Return an integer representing the number of balanced integers in the range [low, high] (both inclusive).
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
1 100
120 129
1234 1234
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3791: Number of Balanced Integers in a Range
class Solution {
private char[] num;
private Long[][] f;
private final int base = 90;
public long countBalanced(long low, long high) {
if (high < 11) {
return 0;
}
low = Math.max(low, 11);
num = String.valueOf(low - 1).toCharArray();
f = new Long[num.length][base << 1 | 1];
long a = dfs(0, 0, true);
num = String.valueOf(high).toCharArray();
f = new Long[num.length][base << 1 | 1];
long b = dfs(0, 0, true);
return b - a;
}
private long dfs(int pos, int diff, boolean lim) {
if (pos >= num.length) {
return diff == 0 ? 1 : 0;
}
if (!lim && f[pos][diff + base] != null) {
return f[pos][diff + base];
}
int up = lim ? num[pos] - '0' : 9;
long res = 0;
for (int i = 0; i <= up; ++i) {
res += dfs(pos + 1, diff + i * (pos % 2 == 0 ? 1 : -1), lim && i == up);
}
if (!lim) {
f[pos][diff + base] = res;
}
return res;
}
}
// Accepted solution for LeetCode #3791: Number of Balanced Integers in a Range
func countBalanced(low int64, high int64) int64 {
if high < 11 {
return 0
}
if low < 11 {
low = 11
}
const base = 90
var num []byte
var f [20][181]int64
var dfs func(pos int, diff int, lim bool) int64
dfs = func(pos int, diff int, lim bool) int64 {
if pos >= len(num) {
if diff == 0 {
return 1
}
return 0
}
if !lim && f[pos][diff+base] != -1 {
return f[pos][diff+base]
}
up := 9
if lim {
up = int(num[pos] - '0')
}
var res int64 = 0
for i := 0; i <= up; i++ {
if pos%2 == 0 {
res += dfs(pos+1, diff+i, lim && i == up)
} else {
res += dfs(pos+1, diff-i, lim && i == up)
}
}
if !lim {
f[pos][diff+base] = res
}
return res
}
num = []byte(fmt.Sprint(low - 1))
for i := range f {
for j := range f[i] {
f[i][j] = -1
}
}
a := dfs(0, 0, true)
num = []byte(fmt.Sprint(high))
for i := range f {
for j := range f[i] {
f[i][j] = -1
}
}
b := dfs(0, 0, true)
return b - a
}
# Accepted solution for LeetCode #3791: Number of Balanced Integers in a Range
class Solution:
def countBalanced(self, low: int, high: int) -> int:
@cache
def dfs(pos: int, diff: int, lim: int) -> int:
if pos >= len(num):
return 1 if diff == 0 else 0
res = 0
up = int(num[pos]) if lim else 9
for i in range(up + 1):
res += dfs(
pos + 1, diff + i * (1 if pos % 2 == 0 else -1), lim and i == up
)
return res
if high < 11:
return 0
low = max(low, 11)
num = str(low - 1)
a = dfs(0, 0, True)
dfs.cache_clear()
num = str(high)
b = dfs(0, 0, True)
return b - a
// Accepted solution for LeetCode #3791: Number of Balanced Integers in a 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 #3791: Number of Balanced Integers in a Range
// class Solution {
// private char[] num;
// private Long[][] f;
// private final int base = 90;
//
// public long countBalanced(long low, long high) {
// if (high < 11) {
// return 0;
// }
// low = Math.max(low, 11);
// num = String.valueOf(low - 1).toCharArray();
// f = new Long[num.length][base << 1 | 1];
// long a = dfs(0, 0, true);
// num = String.valueOf(high).toCharArray();
// f = new Long[num.length][base << 1 | 1];
// long b = dfs(0, 0, true);
// return b - a;
// }
//
// private long dfs(int pos, int diff, boolean lim) {
// if (pos >= num.length) {
// return diff == 0 ? 1 : 0;
// }
// if (!lim && f[pos][diff + base] != null) {
// return f[pos][diff + base];
// }
// int up = lim ? num[pos] - '0' : 9;
// long res = 0;
// for (int i = 0; i <= up; ++i) {
// res += dfs(pos + 1, diff + i * (pos % 2 == 0 ? 1 : -1), lim && i == up);
// }
// if (!lim) {
// f[pos][diff + base] = res;
// }
// return res;
// }
// }
// Accepted solution for LeetCode #3791: Number of Balanced Integers in a Range
function countBalanced(low: number, high: number): number {
if (high < 11) {
return 0;
}
if (low < 11) {
low = 11;
}
const base = 90;
let num: string;
let f: number[][];
function dfs(pos: number, diff: number, lim: boolean): number {
if (pos >= num.length) {
return diff === 0 ? 1 : 0;
}
if (!lim && f[pos][diff + base] !== -1) {
return f[pos][diff + base];
}
const up = lim ? num.charCodeAt(pos) - 48 : 9;
let res = 0;
for (let i = 0; i <= up; ++i) {
res += dfs(pos + 1, diff + i * (pos % 2 === 0 ? 1 : -1), lim && i === up);
}
if (!lim) {
f[pos][diff + base] = res;
}
return res;
}
num = String(low - 1);
f = Array.from({ length: num.length }, () => Array((base << 1) | 1).fill(-1));
const a = dfs(0, 0, true);
num = String(high);
f = Array.from({ length: num.length }, () => Array((base << 1) | 1).fill(-1));
const b = dfs(0, 0, true);
return b - a;
}
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.