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花花酱 LeetCode 733. Flood Fill

By zxi on November 28, 2017

Problem:

An image is represented by a 2-D array of integers, each integer representing the pixel value of the image (from 0 to 65535).

Given a coordinate (sr, sc) representing the starting pixel (row and column) of the flood fill, and a pixel value newColor, “flood fill” the image.

To perform a “flood fill”, consider the starting pixel, plus any pixels connected 4-directionally to the starting pixel of the same color as the starting pixel, plus any pixels connected 4-directionally to those pixels (also with the same color as the starting pixel), and so on. Replace the color of all of the aforementioned pixels with the newColor.

At the end, return the modified image.

Example 1:

Input:

image = [[1,1,1],[1,1,0],[1,0,1]]

sr = 1, sc = 1, newColor = 2

Output: [[2,2,2],[2,2,0],[2,0,1]]

Explanation:

From the center of the image (with position (sr, sc) = (1, 1)), all pixels connected

by a path of the same color as the starting pixel are colored with the new color.

Note the bottom corner is not colored 2, because it is not 4-directionally connected

to the starting pixel.

Note:

The length of image and image[0] will be in the range [1, 50].
The given starting pixel will satisfy 0 <= sr < image.length and 0 <= sc < image[0].length.
The value of each color in image[i][j] and newColor will be an integer in [0, 65535].

Idea

DFS

Solution:

C++ / DFS

Time complexity: O(mn)

Space complexity: O(1)

// Author: Huahua

// Runtime: 33 ms

class Solution {

public:

vector<vector<int>> floodFill(vector<vector<int>>& image,

int sr, int sc, int newColor) {

if (image[sr][sc] == newColor) return image;

int m = image.size();

int n = image[0].size();

floodFill(image, sc, sr, n, m, image[sr][sc], newColor);

return image;

}

private:

void floodFill(vector<vector<int>>& image,

int x, int y, int n, int m, int orgColor, int newColor) {

if (x < 0 || x >= n || y < 0 || y >= m) return;

if (image[y][x] != orgColor) return;

image[y][x] = newColor;

floodFill(image, x + 1, y, n, m, orgColor, newColor);

floodFill(image, x - 1, y, n, m, orgColor, newColor);

floodFill(image, x, y + 1, n, m, orgColor, newColor);

floodFill(image, x, y - 1, n, m, orgColor, newColor);

}

};

Related Problems:

花花酱 LeetCode 64. Minimum Path Sum

By zxi on November 27, 2017

Problem:

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example 1:

[[1,3,1],

[1,5,1],

[4,2,1]]

Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum.

Idea:

Solution 1:

C++ / Recursion with memoization

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

int minPathSum(vector<vector<int>>& grid) {

int m = grid.size();

if (m == 0) return 0;

int n = grid[0].size();

s_ = vector<vector<int>>(m, vector<int>(n, 0));

return minPathSum(grid, n - 1, m - 1, n, m);

}

private:

long minPathSum(const vector<vector<int>>& grid,

int x, int y, int n, int m) {

if (x == 0 && y == 0) return grid[y][x];

if (x < 0 || y < 0) return INT_MAX;

if (s_[y][x] > 0) return s_[y][x];

int ans = grid[y][x] + min(minPathSum(grid, x - 1, y, n, m),

minPathSum(grid, x, y - 1, n, m));

return s_[y][x] = ans;

}

vector<vector<int>> s_;

};

Solution 2:

C++ / DP

Time complexity: O(mn)

Space complexity: O(1)

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

int minPathSum(vector<vector<int>>& grid) {

int m = grid.size();

if (m == 0) return 0;

int n = grid[0].size();

for (int i = 0; i < m; ++i)

for (int j = 0; j < n; ++j) {

if (i == 0 && j == 0) continue;

if (i == 0)

grid[i][j] += grid[i][j - 1];

else if (j == 0)

grid[i][j] += grid[i - 1][j];

else

grid[i][j] += min(grid[i][j - 1], grid[i - 1][j]);

}

return grid[m - 1][n - 1];

}

};

Related Problems:

[解题报告] LeetCode 113. Path Sum II

花花酱 LeetCode 113. Path Sum II

By zxi on November 27, 2017

Problem:

Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

/ \

4 8

/ / \

11 13 4

/ \ / \

7 2 5 1

return

[

[5,4,11,2],

[5,8,4,5]

]

Idea:

Recursion

Solution:

C++

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

vector<vector<int>> pathSum(TreeNode* root, int sum) {

vector<vector<int>> ans;

vector<int> curr;

pathSum(root, sum, curr, ans);

return ans;

}

private:

void pathSum(TreeNode* root, int sum, vector<int>& curr, vector<vector<int>>& ans) {

if(root==nullptr) return;

if(root->left==nullptr && root->right==nullptr) {

if(root->val == sum) {

ans.push_back(curr);

ans.back().push_back(root->val);

}

return;

}

curr.push_back(root->val);

int new_sum = sum - root->val;

pathSum(root->left, new_sum, curr, ans);

pathSum(root->right, new_sum, curr, ans);

curr.pop_back();

}

};

Related Problems:

花花酱 LeetCode 53. Maximum Subarray

By zxi on November 27, 2017

Problem:

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.

Idea:

Solution:

C++

// Author: Huahua

// Runtime: 6 ms (better than 98.66%)

class Solution {

public:

int maxSubArray(vector<int>& nums) {

vector<int> f(nums.size());

f[0] = nums[0];

for (int i = 1; i < nums.size(); ++i)

f[i] = max(f[i - 1] + nums[i], nums[i]);

return *std::max_element(f.begin(), f.end());

}

};

花花酱 LeetCode 657. Judge Route Circle

By zxi on November 27, 2017

Problem:

Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place.

The move sequence is represented by a string. And each move is represent by a character. The valid robot moves are R (Right), L (Left), U (Up) and D (down). The output should be true or false representing whether the robot makes a circle.

Example 1:

1 2	Input: "UD" Output: true

Example 2:

1 2	Input: "LL" Output: false

Idea:

Simulation

Solution:

C++

// Author: Huahua

// Runtime: 19 ms

class Solution {

public:

bool judgeCircle(string moves) {

int cx{0};

int cy{0};

for (const char move : moves) {

switch(move) {

case 'U':

cy--;

break;

case 'D':

cy++;

break;

case 'R':

cx++;

break;

case 'L':

cx--;

break;

}

return cx==0 && cy==0;

}

};

Posts published in “Difficulty”

花花酱 LeetCode 733. Flood Fill

花花酱 LeetCode 64. Minimum Path Sum

花花酱 LeetCode 113. Path Sum II

花花酱 LeetCode 53. Maximum Subarray

花花酱 LeetCode 657. Judge Route Circle