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Problem solution · C++

Find the Count of Monotonic Pairs II

Find the Count of Monotonic Pairs II: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
42 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

Dynamic programming

For Find the Count of Monotonic Pairs II, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 42 lines of C++ from the credited upstream file 3251.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 4 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 codeFind the Count of Monotonic Pairs II · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  // Same as 3250. Find the Count of Monotonic Pairs I  int countOfPairs(vector<int>& nums) {    constexpr int kMod = 1'000'000'007;    constexpr int kMax = 1000;    const int n = nums.size();    int ans = 0;    // dp[i][num] := the number of valid ways to fill the arrays up to index i    // with arr1[i] = num    vector<vector<int>> dp(n, vector<int>(kMax + 1));     for (int num = 0; num <= nums[0]; ++num)      dp[0][num] = 1;     for (int i = 1; i < n; ++i) {      int ways = 0;      int prevNum = 0;      // To satisfy arr1, prevNum <= num.      // To satisfy arr2, nums[i - 1] - prevNum >= nums[i] - num.      //               => prevNum <= min(num, num - (nums[i] - nums[i - 1])).      // As we move from `num` to `num + 1`, the range of valid `prevNum` values      // becomes prevNum <= min(num + 1, num + 1 - (nums[i] - nums[i - 1])).      // Since the range of `prevNum` can only increase by at most 1, there's      // no need to iterate through all possible values of `prevNum`. We can      // simply increment `prevNum` by 1 if it meets the condition.      for (int num = 0; num <= nums[i]; ++num) {        if (prevNum <= min(num, num - (nums[i] - nums[i - 1]))) {          ways = (ways + dp[i - 1][prevNum]) % kMod;          ++prevNum;        }        dp[i][num] = ways;      }    }     for (int i = 0; i <= kMax; ++i)      ans = (ans + dp[n - 1][i]) % kMod;     return ans;  }};

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