Leetcode 3791. Number of Balanced Integers in a Range

• [Leetcode 3791. Number of Balanced Integers in a Range](#Leetcode 3791. Number of Balanced Integers in a Range)
  • [1. 解题思路](#1. 解题思路)
  • [2. 代码实现](#2. 代码实现)

• 题目链接：3791. Number of Balanced Integers in a Range

1. 解题思路

这一题是leetcode周赛482的第四题，是一道hard的题目。

这一题其实算是一个标准的模板题了，就是转换为求不超过 n n n的平衡数，然后使用动态规划进行解答即可。

2. 代码实现

给出python代码实现如下：

class Solution:
    def countBalanced(self, low: int, high: int) -> int:
        
        def count_balanced(num):
            if num <= 10:
                return 0
            digits = [int(digit) for digit in str(num)]
            n = len(digits)

            @lru_cache(None)
            def dp(idx, is_zero, allow_bigger, flag, delta):
                if idx == n-1:
                    if is_zero:
                        return 0
                    need = -delta if flag == 1 else delta
                    if allow_bigger and (0 <= need <= 9):
                        return 1
                    elif not allow_bigger and (0 <= need <= digits[-1]):
                        return 1
                    else:
                        return 0
                if flag == 1 and ((n-idx+1)//2) * 9 + delta < 0:
                    return 0
                elif flag == -1 and -((n-idx+1)//2) * 9 + delta > 0:
                    return 0
                
                if is_zero:
                    ans = dp(idx+1, True, True, 1, 0)
                    if allow_bigger:
                        for i in range(1, 10):
                            ans += dp(idx+1, False, True, -1, i)
                    else:
                        for i in range(1, digits[idx]):
                            ans += dp(idx+1, False, True, -1, i)
                        ans += dp(idx+1, False, False, -1, digits[idx])
                else:
                    ans = 0
                    if allow_bigger:
                        for i in range(10):
                            ans += dp(idx+1, False, True, -flag, delta + flag*i)
                    else:
                        for i in range(digits[idx]):
                            ans += dp(idx+1, False, True, -flag, delta + flag*i)
                        ans += dp(idx+1, False, False, -flag, delta + flag*digits[idx])
                return ans

            ans = dp(0, True, False, 1, 0)
            return ans

        return count_balanced(high) - count_balanced(low-1)

提交代码评测得到：耗时1129ms，占用内存25.79MB。