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Problem solution · C++

Minimum Operations to Make the Integer Zero

Minimum Operations to Make the Integer Zero: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
20 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 Minimum Operations to Make the Integer Zero, 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

  • 20 lines of C++ from the credited upstream file 2749.cpp.
  • The implementation keeps its working state in language-native values and containers.
  • 1 loop block 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 walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Operations to Make the Integer Zero · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int makeTheIntegerZero(int num1, int num2) {    // If k operations are used, num1 - [(num2 + 2^{i_1}) + (num2 + 2^{i_2}) +    // ... + (num2 + 2^{i_k})] = 0. So, num1 - k * num2 = (2^{i_1} + 2^{i_2} +    // ... + 2^{i_k}), where i_1, i_2, ..., i_k are in the range [0, 60].    // Note that for any number x, we can use "x's bit count" operations to make    // x equal to 0. Additionally, we can also use x operations to deduct x by    // 2^0 (x times), which also results in 0.     for (long ops = 0; ops <= 60; ++ops) {      const long target = num1 - ops * num2;      if (__builtin_popcountl(target) <= ops && ops <= target)        return ops;    }     return -1;  }};

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