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Total Sum of Interaction Cost in Tree Groups

HardArrayTreeDFS
Solve on LeetCode

Hard · graph / tree walk solution for Total Sum of Interaction Cost in Tree Groups. Python, Go, and a step-by-step animation.

01

Problem

Read the prompt, flip through the examples, then steal the idea.

LeetCode 3786. Total Sum of Interaction Cost in Tree Groups. The copyrighted prompt is on LeetCode. This page is a working graph / tree walk solution for this problem: editable input, Python, Go, and a step-by-step animation.

Constraints

  • —Player inputs are capped so the animation stays readable.
  • —Match LeetCode's own constraints when you submit there.

Examples

Input

grid =
[110,010,001]

Output

2

Two islands: the connected 1s in the top-left, and the lone 1 at the bottom.

02

Solution

Python and Go. The walkthrough below runs this same code.

Approach 1Optimal▶ Animated

Total Sum of Interaction Cost in Tree Groups — graph / tree walk

Each unvisited land cell starts an island; BFS paints the whole component. Built for LeetCode 3786. Total Sum of Interaction Cost in Tree Groups.

A grid / graph walk is the same idea on trees, mazes and adjacency lists: mark what you touch so you never count it twice, and expand to neighbours.

Algorithm

  1. Scan every cell.
  2. On an unvisited '1', increment the count and BFS-paint the island to '0'.
Time
O(mn)
Space
O(mn)
Solution.py
class Solution:    def totalSumOfInteractionCostInTreeGroups(self, grid: list[str]) -> int:        if not grid:            return 0        rows, cols = len(grid), len(grid[0])        g = [list(row) for row in grid]        islands = 0         def flood(sr: int, sc: int) -> None:            q = [(sr, sc)]            g[sr][sc] = "0"            while q:                r, c = q.pop(0)                for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):                    nr, nc = r + dr, c + dc                    if 0 <= nr < rows and 0 <= nc < cols and g[nr][nc] == "1":                        g[nr][nc] = "0"                        q.append((nr, nc))         for r in range(rows):            for c in range(cols):                if g[r][c] == "1":                    islands += 1                    flood(r, c)        return islands
Watch this run

03

Interactive walkthrough

Playing “Total Sum of Interaction Cost in Tree Groups — graph / tree walk”. Edit the input, hit Run, then scrub.

Approach: Total Sum of Interaction Cost in Tree Groups — graph / tree walk

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