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Problem solution · C++

Next Greater Element IV

Next Greater Element IV: a C++ solution using stack-based processing. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Stack-based processing
Source
walkccc LeetCode Solutions
Length
28 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

Stack-based processing

For Next Greater Element IV, the implementation keeps unresolved items in last-in, first-out order, often to match boundaries, parse structure, or maintain monotonic candidates.

  1. Define what every stack entry represents.
  2. Pop entries once the current item resolves or invalidates them.
  3. Push the current item with only the information later steps need.

Code notes

  • 28 lines of C++ from the credited upstream file 2454.cpp.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

If each item is pushed and popped at most once, the stack work is linear.

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 codeNext Greater Element IV · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  vector<int> secondGreaterElement(vector<int>& nums) {    vector<int> ans(nums.size(), -1);    // a decreasing stack that stores indices that met the first greater number    stack<int> prevStack;    // a decreasing stack that stores indices    stack<int> currStack;     for (int i = 0; i < nums.size(); ++i) {      // Indices in `prevStack` meet the second greater number.      while (!prevStack.empty() && nums[prevStack.top()] < nums[i])        ans[prevStack.top()] = nums[i], prevStack.pop();      // Push indices that meet the first greater number from `currStack` to      // `prevStack`. We need a temporary array to make the indices in the      // `prevStack` increasing.      stack<int> decreasingIndices;      while (!currStack.empty() && nums[currStack.top()] < nums[i])        decreasingIndices.push(currStack.top()), currStack.pop();      while (!decreasingIndices.empty())        prevStack.push(decreasingIndices.top()), decreasingIndices.pop();      currStack.push(i);    }     return ans;  }};

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