Posts tagged as “O(n)”

花花酱 LeetCode 1253. Reconstruct a 2-Row Binary Matrix

By zxi on November 9, 2019

Given the following details of a matrix with n columns and 2 rows :

The matrix is a binary matrix, which means each element in the matrix can be 0 or 1.
The sum of elements of the 0-th(upper) row is given as upper.
The sum of elements of the 1-st(lower) row is given as lower.
The sum of elements in the i-th column(0-indexed) is colsum[i], where colsum is given as an integer array with length n.

Your task is to reconstruct the matrix with upper, lower and colsum.

Return it as a 2-D integer array.

If there are more than one valid solution, any of them will be accepted.

If no valid solution exists, return an empty 2-D array.

Example 1:

Input: upper = 2, lower = 1, colsum = [1,1,1]
Output: [[1,1,0],[0,0,1]]
Explanation: [[1,0,1],[0,1,0]], and [[0,1,1],[1,0,0]] are also correct answers.

Example 2:

Input: upper = 2, lower = 3, colsum = [2,2,1,1]
Output: []

Example 3:

Input: upper = 5, lower = 5, colsum = [2,1,2,0,1,0,1,2,0,1]
Output: [[1,1,1,0,1,0,0,1,0,0],[1,0,1,0,0,0,1,1,0,1]]

Constraints:

1 <= colsum.length <= 10^5
0 <= upper, lower <= colsum.length
0 <= colsum[i] <= 2

Solution: Greedy?

Two passes:
first pass, only process sum = 2, upper = 1, lower = 1
second pass, only process sum = 1, whoever has more leftover, assign 1 to that row.

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

vector<vector<int>> reconstructMatrix(int upper, int lower, vector<int>& colsum) {

const int n = colsum.size();

vector<vector<int>> a(2, vector<int>(n));

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

if (colsum[i] == 2) {

a[0][i] = a[1][i] = 1;

--upper;

--lower;

}

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

if (colsum[i] == 1)

if (upper > lower) {

a[0][i] = 1;

--upper;

} else {

a[1][i] = 1;

--lower;

}

if (upper != 0 || lower != 0) return {};

return a;

}

};

花花酱 LeetCode 1247. Minimum Swaps to Make Strings Equal

By zxi on November 7, 2019

You are given two strings s1 and s2 of equal length consisting of letters "x" and "y" only. Your task is to make these two strings equal to each other. You can swap any two characters that belong to different strings, which means: swap s1[i] and s2[j].

Return the minimum number of swaps required to make s1 and s2 equal, or return -1 if it is impossible to do so.

Example 1:

Input: s1 = "xx", s2 = "yy"
Output: 1
Explanation: 
Swap s1[0] and s2[1], s1 = "yx", s2 = "yx".

Example 2:

Input: s1 = "xy", s2 = "yx"
Output: 2
Explanation: 
Swap s1[0] and s2[0], s1 = "yy", s2 = "xx".
Swap s1[0] and s2[1], s1 = "xy", s2 = "xy".
Note that you can't swap s1[0] and s1[1] to make s1 equal to "yx", cause we can only swap chars in different strings.

Example 3:

Input: s1 = "xx", s2 = "xy"
Output: -1

Example 4:

Input: s1 = "xxyyxyxyxx", s2 = "xyyxyxxxyx"
Output: 4

Constraints:

1 <= s1.length, s2.length <= 1000
s1, s2 only contain 'x' or 'y'.

Solution: Math

if s1[i] == s2[i] than no need to swap, so we can only look for
case1. s1[i] = x, s2[i] = y, xy
case2. s1[i] = y, s2[i] = x, yx

If case1 + case2 is odd, then there’s no solution.

Otherwise we can use one swap to fix two xys (or two yxs)
xx, yy => xy, yx

One special case is there an extra xy and and extra yx, which takes two swaps
xy, yx => yy, xx => xy, xy

Finally,
ans = (case1 + 1) / 2 + (case2 + 1) / 2

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int minimumSwap(string s1, string s2) {

int xy = 0;

int yx = 0;

for (int i = 0; i < s1.length(); ++i) {

if (s1[i] == 'x' && s2[i] == 'y') ++xy;

if (s1[i] == 'y' && s2[i] == 'x') ++yx;

}

if ((xy + yx) % 2) return -1;

return (xy + 1) / 2 + (yx + 1) / 2;

}

};

花花酱 LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal

By zxi on October 3, 2019

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

Solution: Recursion

Similar to LC 105

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {

unordered_map<int, int> pos;

for (int i = 0; i < inorder.size(); i++)

pos[inorder[i]] = i;

function<TreeNode*(int, int, int, int)> buildTree = [&](int is, int ie, int ps, int pe) {

if (ps > pe) return (TreeNode*)nullptr;

int im = pos[postorder[pe]];

int pm = ps + (im - is) - 1;

auto root = new TreeNode(postorder[pe]);

root->left = buildTree(is, im - 1, ps, pm);

root->right = buildTree(im + 1, ie, pm + 1, pe - 1);

return root;

};

return buildTree(0, inorder.size() - 1, 0, postorder.size() - 1);

}

};

花花酱 LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal

By zxi on October 3, 2019

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

Solution: Recursion

Preprocessing: use a hashtable to store the index of element in preorder array.

For an element in inorder array, find the pos of it in preorder array in O(1), anything to the left will be the leftchild and anything to the right will be the right child.

e.g.
buildTree([9, 3, 15, 20, 7], [3, 9, 20, 15, 7]):
root = TreeNode(9) # inorder[0] = 9
root.left = buildTree([3], [3])
root.right = buildTree([15, 20, 7], [20, 15, 7])
return root

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {

unordered_map<int, int> pos;

for (int i = 0; i < inorder.size(); i++)

pos[inorder[i]] = i;

function<TreeNode*(int, int, int, int)> buildTree = [&](int is, int ie, int ps, int pe) {

if (ps > pe) return (TreeNode*)nullptr;

int im = pos[preorder[ps]];

int pm = ps + (im - is);

auto root = new TreeNode(preorder[ps]);

root->left = buildTree(is, im - 1, ps + 1, pm);

root->right = buildTree(im + 1, ie, pm + 1, pe);

return root;

};

return buildTree(0, inorder.size() - 1, 0, preorder.size() - 1);

}

};

花花酱 LeetCode 46. Permutations

By zxi on October 2, 2019

Given a collection of distinct integers, return all possible permutations.

Example:

Input: [1,2,3]
Output:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

Solution: DFS

Time complexity: O(n!)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

vector<vector<int>> permute(vector<int>& nums) {

const int n = nums.size();

vector<vector<int>> ans;

vector<int> used(n);

vector<int> path;

function<void(int)> dfs = [&](int d) {

if (d == n) {

ans.push_back(path);

return;

}

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

if (used[i]) continue;

used[i] = 1;

path.push_back(nums[i]);

dfs(d + 1);

path.pop_back();

used[i] = 0;

}

};

dfs(0);

return ans;

}

};

Posts tagged as “O(n)”

花花酱 LeetCode 1253. Reconstruct a 2-Row Binary Matrix

Solution: Greedy?

C++

花花酱 LeetCode 1247. Minimum Swaps to Make Strings Equal

Solution: Math

C++

花花酱 LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal

Solution: Recursion

C++

Related Problems

花花酱 LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal

Solution: Recursion

C++

Related Problems

花花酱 LeetCode 46. Permutations

Solution: DFS

C++

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