Number of People That Can Be Seen in a Grid

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Problem Highlights

• 🔗 Leetcode Link: Number of People That Can Be Seen in a Grid
• 💡 Problem Difficulty: Medium
• ⏰ Time to complete: 15 mins
• 🛠️ Topics: Stack
• 🗒️ Similar Questions: Number of Visible People in a Queue

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• What happens when a newcomer encounters one with equal height? Can you provide an example?
  • When a newcomer encounters one with equal height, the newcomer replace the existing one and stops there. Eg. two persons in stack [height == 4, height == 2], a third one with height == 2 shows up, first person cannot see person 3, because his view of person 3 blocked by person 2.
• How is this different from Number of Visible People in a Queue?
  • One difference is that we can have duplicates here.

HAPPY CASE
Input: heights = [[3,1,4,2,5]]
Output: [[2,1,2,1,0]]


EDGE CASE
Input: heights = [[5,1],[3,1],[4,1]]
Output: [[3,1],[2,1],[1,0]]

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

• Stack: What could we push onto a stack to make this problem easier? Stacks don’t allow us to keep track of data based on keys.
• Queue: Queues fall into the same category as Stacks, do we need to maintain any sense of ordering to solve this problem?
• HashMap: HashMaps allow us to store data for quick access. What could we store in a HashMap to make this problem easier?
• Heap: Do we need some sort of ordering to our data that a Heap could provide?

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

Maintain a stack and pop an element off 
Increment answer by 1 when the current height is more than the top element on the stack.

⚠️ Common Mistakes

• There is one thing to watch out for - when the current element is equal to the top element on the stack, do not push it onto the stack, just skip it. If we push onto the stack, it will lead to future bigger elements to over count.

4: I-mplement

Implement the code to solve the algorithm.

class Solution:
    def seePeople(self, heights: List[List[int]]) -> List[List[int]]:
        m, n = len(heights), len(heights[0])
        ans = [[0] * n for _ in range(m)]
        s = collections.deque()                   # a deque behave like mono-stack
        for i in range(m):                        # look right
            for j in range(n-1, -1, -1):
                num = heights[i][j]
                idx = bisect.bisect_left(s, num)  # binary search on an increasing order sequence
                ans[i][j] += idx + (idx < len(s)) # if `idx` is not out of bound, meaning the next element in `s` is the first one large than `num`, we can count it too
                while s and s[0] <= num:          # keep a mono-descreasing stack
                    s.popleft()
                s.appendleft(num)    
            s.clear()
        for j in range(n):                        # look below
            for i in range(m-1, -1, -1):
                num = heights[i][j]
                idx = bisect.bisect_left(s, num)
                ans[i][j] += idx + (idx < len(s))
                while s and s[0] <= num:
                    s.popleft()
                s.appendleft(num)    
            s.clear()
        return ans

class Solution {
    public int[][] seePeople(int[][] heights) {
        int m = heights.length, n = heights[0].length;
        int[][] ans = new int[m][n];
        for (int i = 0; i < n; i++){ // DOWN
            Deque<Integer> stack = new ArrayDeque<>();
            for (int j = m - 1; j >= 0; j--){
                while(!stack.isEmpty() && heights[j][i] > stack.peek()){
                    ans[j][i]++;
                    stack.pop();
                }
                if (!stack.isEmpty()){
                    ans[j][i]++;
                }
                if (stack.isEmpty() || heights[j][i] != stack.peek()){
                    stack.push(heights[j][i]);
                }
            }
        }

        for (int i = 0; i < m; i++){ // RIGHT
            Deque<Integer> stack = new ArrayDeque<>();
            for (int j = n - 1; j >= 0; j--){
                while(!stack.isEmpty() && heights[i][j] > stack.peek()){
                    ans[i][j]++;
                    stack.pop();
                }
                if (!stack.isEmpty()){
                    ans[i][j]++;
                }
                if (stack.isEmpty() || heights[i][j] != stack.peek()){
                    stack.push(heights[i][j]);
                }
            }
        }

        return ans;
    }
}

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Trace through your code with an input to check for the expected output
• Catch possible edge cases and off-by-one errors

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

• Time Complexity: O(MN), where M and N are the sizes of its two inputs
• Space Complexity: O(N) since a stack is used