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Problem solution · Java

Find Sorted Submatrices With Maximum Element at Most K

Find Sorted Submatrices With Maximum Element at Most K: a Java 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
46 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 Sorted Submatrices With Maximum Element at Most K, 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

  • 46 lines of Java from the credited upstream file 3359.java.
  • The implementation visibly relies on sequence storage, work queue, 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 Sorted Submatrices With Maximum Element at Most K · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public long countSubmatrices(int[][] grid, int k) {    record T(int subarrayWidth, int rowIndex, int accumulatedSubmatrices) {}    final int m = grid.length;    final int n = grid[0].length;    long ans = 0;    // dp[i][j] := the number of valid subarrays ending in grid[i][j]    int[][] dp = new int[m][n];    // stacks[j] := the stack of valid    // (subarray width, row index, number of accumulated submatrices) ending in    // column j    List<Deque<T>> stacks = new ArrayList<>(n);     for (int j = 0; j < n; ++j) {      Deque<T> stack = new ArrayDeque<>();      stack.push(new T(0, -1, 0));      stacks.add(stack);    }     for (int i = 0; i < m; ++i)      for (int j = 0; j < n; ++j)        if (grid[i][j] > k) {          stacks.get(j).clear();          stacks.get(j).push(new T(0, i, 0));        } else {          dp[i][j] = 1;          if (j > 0 && grid[i][j - 1] <= k && grid[i][j - 1] >= grid[i][j])            // Extend the valid subarrays to the current number.            dp[i][j] += dp[i][j - 1];          final int width = dp[i][j];          // Remove subarray widths greater than the current width since they          // will become invalid.          Deque<T> stack = stacks.get(j);          while (!stack.isEmpty() && width < stack.peek().subarrayWidth)            stack.pop();          final int height = i - stack.peek().rowIndex;          final int newSubmatrices = width * height;          final int accumulatedSubmatrices = stack.peek().accumulatedSubmatrices + newSubmatrices;          ans += accumulatedSubmatrices;          stack.push(new T(width, i, accumulatedSubmatrices));        }     return ans;  }}

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