Skip to content
LEETFLIX

Lexicographically Smallest String After Applying Operations

MediumStringDFSBFS
Solve on LeetCode

Medium · graph / tree walk solution for Lexicographically Smallest String After Applying Operations. Python, Go, and a step-by-step animation.

01

Problem

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

LeetCode 1625. Lexicographically Smallest String After Applying Operations. 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

Lexicographically Smallest String After Applying Operations — graph / tree walk

Each unvisited land cell starts an island; BFS paints the whole component. Built for LeetCode 1625. Lexicographically Smallest String After Applying Operations.

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 lexicographicallySmallestStringAfterApplyingOperations(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 “Lexicographically Smallest String After Applying Operations — graph / tree walk”. Edit the input, hit Run, then scrub.

Approach: Lexicographically Smallest String After Applying Operations — graph / tree walk

Loading animation…