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Problem solution · C++

Minimum Number of Visited Cells in a Grid

Minimum Number of Visited Cells in a Grid: a C++ solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Segment tree or range structure
Source
walkccc LeetCode Solutions
Length
78 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

Segment tree or range structure

For Minimum Number of Visited Cells in a Grid, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

  • 78 lines of C++ from the credited upstream file 2617.cpp.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of 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 Number of Visited Cells in a Grid · C++C++
Use this to learn the idea, then write your own version.
class SegmentTree { public:  explicit SegmentTree(int n, int kInf) : kInf(kInf), n(n), tree(4 * n, kInf) {}   // Updates nums[i] to val.  void update(int i, int val) {    update(0, 0, n - 1, i, val);  }   // Returns min(nums[i..j]).  int query(int i, int j) const {    return query(0, 0, n - 1, i, j);  }  private:  const int kInf;    // the invalid value  const int n;       // the size of the input array  vector<int> tree;  // the segment tree   void update(int treeIndex, int lo, int hi, int i, int val) {    if (lo == hi) {      tree[treeIndex] = val;      return;    }    const int mid = (lo + hi) / 2;    if (i <= mid)      update(2 * treeIndex + 1, lo, mid, i, val);    else      update(2 * treeIndex + 2, mid + 1, hi, i, val);    tree[treeIndex] = merge(tree[2 * treeIndex + 1], tree[2 * treeIndex + 2]);  }   int query(int treeIndex, int lo, int hi, int i, int j) const {    if (i <= lo && hi <= j)  // [lo, hi] lies completely inside [i, j].      return tree[treeIndex];    if (j < lo || hi < i)  // [lo, hi] lies completely outside [i, j].      return kInf;    const int mid = (lo + hi) / 2;    return merge(query(treeIndex * 2 + 1, lo, mid, i, j),                 query(treeIndex * 2 + 2, mid + 1, hi, i, j));  }   int merge(int left, int right) const {    return min(left, right);  }}; class Solution { public:  int minimumVisitedCells(vector<vector<int>>& grid) {    const int m = grid.size();    const int n = grid[0].size();    const int kInf = (m + n) * 2 - 1;    vector<SegmentTree> rows(m, SegmentTree(n, kInf));    vector<SegmentTree> cols(n, SegmentTree(m, kInf));     // The min # cells to visit (m - 1, n - 1) from (m - 1, n - 1) is 1.    rows[m - 1].update(n - 1, 1);    cols[n - 1].update(m - 1, 1);     for (int i = m - 1; i >= 0; --i)      for (int j = n - 1; j >= 0; --j) {        // There's no k s.t. j < k <= 0 + j.        // There's no k s.t. i < k <= 0 + i.        if (grid[i][j] == 0)          continue;        const int moveRight = rows[i].query(j + 1, grid[i][j] + j);        const int moveDown = cols[j].query(i + 1, grid[i][j] + i);        const int minMove = min(kInf, min(moveRight, moveDown) + 1);        rows[i].update(j, minMove);        cols[j].update(i, minMove);      }     const int res = rows[0].query(0, 0);    return res == kInf ? -1 : res;  }};

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