LeetCode-in-Java
3791. Number of Balanced Integers in a Range
Hard
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).
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 <= 1015
Solution
import java.util.Arrays;
public class Solution {
private long solve(String number, int idx, int diff, int tight, long[][][] dp) {
if (idx == number.length()) {
return diff == 0 ? 1 : 0;
}
if (dp[idx][diff + dp[0].length / 2][tight] != -1) {
return dp[idx][diff + dp[0].length / 2][tight];
}
long ans = 0;
int ub = tight == 1 ? number.charAt(idx) - '0' : 9;
for (int i = 0; i <= ub; i++) {
int newTight = (tight == 1 && i == ub) ? 1 : 0;
if (idx % 2 == 0) {
ans += solve(number, idx + 1, diff - i, newTight, dp);
} else {
ans += solve(number, idx + 1, diff + i, newTight, dp);
}
}
dp[idx][diff + dp[0].length / 2][tight] = ans;
return dp[idx][diff + dp[0].length / 2][tight];
}
private long solve(long num) {
if (num == 0) {
return 1;
}
String number = Long.toString(num);
int len = number.length();
long[][][] dp =
len % 2 == 1 ? new long[len][(len + 1) * 9 + 1][2] : new long[len][len * 9 + 1][2];
for (long[][] tbl : dp) {
for (long[] row : tbl) {
Arrays.fill(row, -1);
}
}
return solve(number, 0, 0, 1, dp);
}
public long countBalanced(long low, long high) {
return solve(high) - solve(low - 1);
}
}