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Problem solution · C++

Minimum Moves to Pick K Ones

Minimum Moves to Pick K Ones: a C++ solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
55 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

Sliding window or two pointers

For Minimum Moves to Pick K Ones, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 55 lines of C++ from the credited upstream file 3086.cpp.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 Moves to Pick K Ones · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  long long minimumMoves(vector<int>& nums, int k, int maxChanges) {    // Dylan has two actions for collecting '1's in a sequence:    // Action 1: Put a '1' next to him and pick it up.    //           The cost is 2.    // Action 2: Swap a '1' towards him and collect it.    //           The cost equals the distance to the '1'.    //    // To minimize the swapping cost, Dylan can use a sliding window strategy,    // selecting the optimal position (middle '1' in the window) for efficient    // collection. The window's size is crucial:     // The minimum window size: min(0, k - maxChanges), ensuring the window    // isn't too small.    // The maximum window size: min(k, minOnesByTwo + 3, the number of ones),    // preventing overly ambitious swaps.    //    // Note that if needing to move a '1' beyond `minOnesByTwo + 3`, it's    // cheaper to use Action 1.     // At most three indices, (dylanIndex - 1, dylanIndex, dylanIndex + 1), have    // a distance <= 1 from dylanIndex, implying that we'll be taking at most    // `maxOnesByTwo + 3` using Action 2. Any more Action 2 is not optimal and    // should be replaced with Action 1.    constexpr int kNumOfIndicesWithinOneDistance = 3;    long ans = LONG_MAX;    vector<int> oneIndices;  // the indices of 1s    vector<long> prefix{0};  // the accumulated indices of 1s     for (int i = 0; i < nums.size(); ++i)      if (nums[i] == 1)        oneIndices.push_back(i);     for (const int oneIndex : oneIndices)      prefix.push_back(prefix.back() + oneIndex);     const int minOnesByTwo = max(0, k - maxChanges);    const int maxOnesByTwo =        min({k, minOnesByTwo + kNumOfIndicesWithinOneDistance,             static_cast<int>(oneIndices.size())});     for (int onesByTwo = minOnesByTwo; onesByTwo <= maxOnesByTwo; ++onesByTwo)      for (int l = 0; l + onesByTwo < prefix.size(); ++l) {        const int r = l + onesByTwo;  // Collect 1s in oneIndices[l - 1..r - 1].        const long cost1 = (k - onesByTwo) * 2;        const long cost2 = (prefix[r] - prefix[(l + r) / 2]) -                           (prefix[(l + r + 1) / 2] - prefix[l]);        ans = min(ans, cost1 + cost2);      }     return ans;  }};

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