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Problem solution · C++

Sum of K Mirror Numbers

Sum of K Mirror Numbers: a C++ solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Binary search
Source
Kamyu LeetCode Solutions
Length
98 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Binary search

For Sum of K Mirror Numbers, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 98 lines of C++ from the credited upstream file sum-of-k-mirror-numbers.cpp.
  • The implementation visibly relies on sequence storage.
  • 9 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeSum of K Mirror Numbers · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(10^6), the most times of finding x is 665502 (k = 7, n = 30)// Space: O(1) class Solution {public:    long long kMirror(int k, int n) {        const int base1 = k, base2 = 10;  // (10, k) is slower        int64_t result = 0;        vector<int> prefix_num(2, 1), total(2, base1);        uint8_t odd = 1;        while (n--) {            int64_t x;            do {                x = mirror(prefix_num[odd], base1, odd);                if (++prefix_num[odd] == total[odd]) {                    total[odd] *= base1;                    odd ^= 1;                }            } while (x != reverse(x, base2));            result += x;        }        return result;    } private:    int64_t mirror(int n, int base, bool odd) {        int64_t result = n;        if (odd) {            n /= base;        }        for (; n; n /= base) {            result = result * base + (n % base);        }        return result;    }     int64_t reverse(int64_t n, int base) {        int64_t result = 0;        for (; n; n /= base) {            result = result * base + n % base;        }        return result;    }}; // Time:  O(10^6), the most times of finding x is 665502 (k = 7, n = 30)// Space: O(1)class Solution2 {public:    long long kMirror(int k, int n) {        string s = "0";        int64_t result = 0;        while (n--) {            int64_t x;            do {                x = next_num_in_base_k(k, &s);            } while (!is_mirror(to_string(x)));            result += x;        }        return result;    } private:    int64_t next_num_in_base_k(int k, string *s) {        int result = 0;        for (int i = size(*s) / 2; i < size(*s); ++i) {            if ((*s)[i] + 1 - k < '0') {                (*s)[i] = (*s)[size(*s) - 1 - i] = (*s)[i] + 1;                break;            }            (*s)[i] = (*s)[size(*s) - 1 - i] = '0';        }        if ((*s)[0] == '0') {            s->back() = '1';            s->insert(begin(*s), '1');        }        return to_int_from_base_k(*s, k);    }     int64_t to_int_from_base_k(const string& s, int k) {        int64_t result = 0;        for (int64_t i = size(s) - 1, base = 1; i >= 0; --i, base *= k) {            result += (s[i] - '0') * base;        }        return result;    }     bool is_mirror(const string& s) {        int left = 0, right = size(s) - 1;        while (left < right) {            if (s[left++] != s[right--]) {                return false;            }        }        return true;    }};

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