CPMagical CP Library

Problem solution · C++

Sum of Compatible Numbers in Range I

Sum of Compatible Numbers in Range I: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Direct simulation
Source
Kamyu LeetCode Solutions
Length
61 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

Direct simulation

For Sum of Compatible Numbers in Range I, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 61 lines of C++ from the credited upstream file sum-of-compatible-numbers-in-range-i.cpp.
  • The implementation keeps its working state in language-native values and containers.
  • 3 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 Compatible Numbers in Range I · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(log(n + k))// Space: O(1) // bitmasks, combinatoricsclass Solution {public:    int sumOfGoodIntegers(int n, int k) {        const auto& count = [&](int x) {            if (x <= 0) {                return 0;            }            const auto& l = bit_width(static_cast<uint32_t>(x));            int total = 0, cnt = 1;            for (int i = 0; i < l; ++i) {                if (n & (1 << i)) {                    continue;                }                total = total * 2 + (1 << i) * cnt;                cnt *= 2;            }            int result = 0, prefix = 0;            for (int i = l - 1; i >= 0; --i) {                if ((n & (1 << i)) == 0) {                    if (!(n & (1 << i))) {                        cnt /= 2;                        total = (total - (1 << i) * cnt) / 2;                    }                }                if (!(x & (1 << i))) {                    continue;                }                result += prefix * cnt + total;                if (n & (1 << i)) {                    return result;                }                prefix |= 1 << i;            }            result += prefix;            return result;        };         return count(n + k) - count((n - k) - 1);    }}; // Time:  O(k)// Space: O(1)// simulationclass Solution2 {public:    int sumOfGoodIntegers(int n, int k) {        int result = 0;        for (int i = max(n - k, 1); i <= n + k; ++i) {            if ((n & i) == 0) {                result += i;            }        }        return result;    }};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗