Problem solution · C++

Longest Non-decreasing Subarray From Two Arrays

Longest Non-decreasing Subarray From Two Arrays: 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
21 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 Longest Non-decreasing Subarray From Two Arrays, 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

  • 21 lines of C++ from the credited upstream file 2771.cpp.
  • The implementation visibly relies on sequence storage.
  • 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.

Full codeLongest Non-decreasing Subarray From Two Arrays · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int maxNonDecreasingLength(vector<int>& nums1, vector<int>& nums2) {    int ans = 1;    int dp1 = 1;  // the longest subarray that ends in nums1[i] so far    int dp2 = 1;  // the longest subarray that ends in nums2[i] so far     for (int i = 1; i < nums1.size(); ++i) {      const int dp11 = nums1[i - 1] <= nums1[i] ? dp1 + 1 : 1;      const int dp21 = nums2[i - 1] <= nums1[i] ? dp2 + 1 : 1;      const int dp12 = nums1[i - 1] <= nums2[i] ? dp1 + 1 : 1;      const int dp22 = nums2[i - 1] <= nums2[i] ? dp2 + 1 : 1;      dp1 = max(dp11, dp21);      dp2 = max(dp12, dp22);      ans = max({ans, dp1, dp2});    }     return ans;  }}; 

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