Posts tagged as “sum”

花花酱 LeetCode 563. Binary Tree Tilt

By zxi on March 20, 2018

Problem

https://leetcode.com/problems/binary-tree-tilt/description/

题目大意：返回树的总tilt值。对于一个节点的tilt值定义为左右子树元素和的绝对之差。

Given a binary tree, return the tilt of the whole tree.

The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.

The tilt of the whole tree is defined as the sum of all nodes’ tilt.

Example:

Input: 
         1
       /   \
      2     3
Output: 1
Explanation: 
Tilt of node 2 : 0
Tilt of node 3 : 0
Tilt of node 1 : |2-3| = 1
Tilt of binary tree : 0 + 0 + 1 = 1

Note:

The sum of node values in any subtree won’t exceed the range of 32-bit integer.
All the tilt values won’t exceed the range of 32-bit integer.

Solution: Recursion

Time complexity: O(n)

Space complexity: O(h)

C++

// Author: Huahua

// Running time: 17 ms

class Solution {

public:

int findTilt(TreeNode* root) {

int sum;

return findTilt(root, sum);

}

private:

// return the total tilt and sum of the root.

int findTilt(TreeNode* root, int& sum) {

if (root == nullptr) return 0;

int ls = 0;

int rs = 0;

int tl = findTilt(root->left, ls);

int tr = findTilt(root->right, rs);

sum = root->val + ls + rs;

return tl + tr + abs(ls - rs);

}

};

v1-2

// Author: Huahua

// Running time: 17 ms

class Solution {

public:

int findTilt(TreeNode* root) {

return findTiltSum(root).first;

}

private:

pair<int, int> findTiltSum(TreeNode* root) {

if (root == nullptr) return {0, 0};

auto l = findTiltSum(root->left);

auto r = findTiltSum(root->right);

return { l.first + r.first + abs(l.second - r.second),

root->val + l.second + r.second };

}

};

// Author: Huahua

// Running time: 16 ms

class Solution {

public:

int findTilt(TreeNode* root) {

ans_ = 0;

sum(root);

return ans_;

}

private:

int ans_;

int sum(TreeNode* root) {

if (root == nullptr) return 0;

int l = sum(root->left);

int r = sum(root->right);

ans_ += abs(l - r);

return root->val + l + r;

}

};

Python3

"""

Author: Huahua

Running time: 112 ms

"""

class Solution:

def findTilt(self, root):

def findTilt(root):

if not root: return 0, 0

tl, sl = findTilt(root.left)

tr, sr = findTilt(root.right)

return tl + tr + abs(sl - sr), root.val + sl + sr

return findTilt(root)[0]

花花酱 LeetCode 454. 4Sum II

By zxi on March 5, 2018

题目大意：给你4个组数：A, B, C, D。问有多少组组合的A[i] + B[j] + C[k] + D[l] 的和为0。

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l]is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -2²⁸ to 2²⁸ – 1 and the result is guaranteed to be at most 2³¹ – 1.

Example:

Input:

A = [ 1, 2]

B = [-2,-1]

C = [-1, 2]

D = [ 0, 2]

Output:

Explanation:

The two tuples are:

1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0

2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

Solution 1: HashTable

Split the arrays into two groups: (A, B), (C, D)

Create the Minkowski sum of two groups and count the occurrence of each element.

e.g. A = [1, 2, 3], B = [0, -1, 1]

Minkowski sum(A, B) SAB= [0, 1, 2, 3, 4] => [0:1, 1:2, 2:3, 3:2, 4:1]

for each element s in SAB, check whether -s is in Minkowski sum(C, D) SCD

ans = sum(SAB[s] * SCD[-s]), for all s, that s in SAB and -s in SCD

Time complexity: O(n^2)

Space complexity: O(n^2)

C++

// Author: Huahua

// Running time: 217 ms

class Solution {

public:

int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {

unordered_map<int, int> sums;

for (int a : A)

for (int b : B)

++sums[a + b];

int ans = 0;

for (int c : C)

for (int d : D) {

auto it = sums.find(- c - d);

if (it != sums.end()) ans += it->second;

}

return ans;

}

};

Python3

"""

Author: Huahua

Running time: 672 ms

"""

class Solution:

def fourSumCount(self, A, B, C, D):

s = collections.Counter([a + b for a in A for b in B])

return sum(s[-c - d] for c in C for d in D)

Related Problems:

[解题报告] LeetCode 1. Two Sum – 花花酱

花花酱 LeetCode 508. Most Frequent Subtree Sum

By zxi on March 2, 2018

Given the root of a tree, you are asked to find the most frequent subtree sum. The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself). So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order.

Examples 1
Input:

/ \

2 -3

return [2, -3, 4], since all the values happen only once, return all of them in any order.

Examples 2
Input:

/ \

2 -5

return [2], since 2 happens twice, however -5 only occur once.

Note: You may assume the sum of values in any subtree is in the range of 32-bit signed integer.

// Author: Huahua

// Running time: 18 ms

class Solution {

public:

vector<int> findFrequentTreeSum(TreeNode* root) {

unordered_map<int, int> freqs;

int max_freq = -1;

vector<int> ans;

(void)treeSum(root, freqs, max_freq, ans);

return ans;

}

private:

int treeSum(TreeNode* root, unordered_map<int, int>& freqs, int& max_freq, vector<int>& ans) {

if (!root) return 0;

int sum = root->val +

treeSum(root->left, freqs, max_freq, ans) +

treeSum(root->right, freqs, max_freq, ans);

int freq = ++freqs[sum];

if (freq > max_freq) {

max_freq = freq;

ans.clear();

}

if (freq == max_freq)

ans.push_back(sum);

return sum;

}

};

Python

"""

Author: Huahua

Running time: 72 ms (beats 100%)

"""

class Solution:

def findFrequentTreeSum(self, root):

if not root: return []

def treeSum(root):

if not root: return 0

s = root.val + treeSum(root.left) + treeSum(root.right)

f[s] += 1

return s

f = collections.Counter()

treeSum(root)

max_freq = max(f.values())

return [s for s in f.keys() if f[s] == max_freq]

花花酱 LeetCode 494. Target Sum

By zxi on January 9, 2018

题目大意：给你一串数字，你可以在每个数字前放置+或-，问有多少种方法可以使得表达式的值等于target。You are given a list of non-negative integers, a1, a2, …, an, and a target, S. Now you have 2 symbols + and -. For each integer, you should choose one from + and - as its new symbol.

Find out how many ways to assign symbols to make sum of integers equal to target S.

Example 1:

Input: nums is [1, 1, 1, 1, 1], S is 3.

Output: 5

Explanation:

-1+1+1+1+1 = 3

+1-1+1+1+1 = 3

+1+1-1+1+1 = 3

+1+1+1-1+1 = 3

+1+1+1+1-1 = 3

There are 5 ways to assign symbols to make the sum of nums be target 3.

Note:

The length of the given array is positive and will not exceed 20.
The sum of elements in the given array will not exceed 1000.
Your output answer is guaranteed to be fitted in a 32-bit integer.

Idea: DP

Solution 1: DP

Time complexity: O(n*sum)

Space complexity: O(n*sum)

C++

// Author: Huahua

// Running time: 16 ms

class Solution {

public:

int findTargetSumWays(vector<int>& nums, int S) {

const int n = nums.size();

const int sum = std::accumulate(nums.begin(), nums.end(), 0);

if (sum < S) return 0;

const int offset = sum;

// ways[i][j] means total ways to sum up to (j - offset) using nums[0] ~ nums[i - 1].

vector<vector<int>> ways(n + 1, vector<int>(sum + offset + 1, 0));

ways[0][offset] = 1;

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

for (int j = nums[i]; j < 2 * sum + 1 - nums[i]; ++j)

if (ways[i][j]) {

ways[i + 1][j + nums[i]] += ways[i][j];

ways[i + 1][j - nums[i]] += ways[i][j];

}

return ways.back()[S + offset];

}

};

C++ SC O(n)

// Author: Huahua

// Running time: 12 ms

class Solution {

public:

int findTargetSumWays(vector<int>& nums, int S) {

const int n = nums.size();

const int sum = std::accumulate(nums.begin(), nums.end(), 0);

if (sum < std::abs(S)) return 0;

const int kOffset = sum;

const int kMaxN = sum * 2 + 1;

vector<int> ways(kMaxN, 0);

ways[kOffset] = 1;

for (int num : nums) {

vector<int> tmp(kMaxN, 0);

for (int i = num; i < kMaxN - num; ++i)

if (ways[i]) {

tmp[i + num] += ways[i];

tmp[i - num] += ways[i];

}

std::swap(ways, tmp);

}

return ways[S + kOffset];

}

};

Java

// Author: Huahua

// Running time: 28 ms

class Solution {

public int findTargetSumWays(int[] nums, int S) {

int sum = 0;

for (final int num : nums)

sum += num;

if (sum < S) return 0;

final int kOffset = sum;

final int kMaxN = sum * 2 + 1;

int[] ways = new int[kMaxN];

ways[kOffset] = 1;

for (final int num : nums) {

int[] tmp = new int[kMaxN];

for (int i = num; i < kMaxN - num; ++i) {

tmp[i + num] += ways[i];

tmp[i - num] += ways[i];

}

ways = tmp;

}

return ways[S + kOffset];

}

C++ / V2

// Author: Huahua

// Running time: 17 ms

class Solution {

public:

int findTargetSumWays(vector<int>& nums, int S) {

const int n = nums.size();

const int sum = std::accumulate(nums.begin(), nums.end(), 0);

if (sum < std::abs(S)) return 0;

const int kOffset = sum;

const int kMaxN = sum * 2 + 1;

vector<int> ways(kMaxN, 0);

ways[kOffset] = 1;

for (int num : nums) {

vector<int> tmp(kMaxN, 0);

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

if (i + num < kMaxN) tmp[i] += ways[i + num];

if (i - num >= 0) tmp[i] += ways[i - num];

}

std::swap(ways, tmp);

}

return ways[S + kOffset];

}

};

Solution 2: DFS

Time complexity: O(2^n)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 422 ms

class Solution {

public:

int findTargetSumWays(vector<int>& nums, int S) {

const int sum = std::accumulate(nums.begin(), nums.end(), 0);

if (sum < std::abs(S)) return 0;

int ans = 0;

dfs(nums, 0, S, ans);

return ans;

}

private:

void dfs(const vector<int>& nums, int d, int S, int& ans) {

if (d == nums.size()) {

if (S == 0) ++ans;

return;

}

dfs(nums, d + 1, S - nums[d], ans);

dfs(nums, d + 1, S + nums[d], ans);

}

};

Java

// Author: Huahua

// Running time: 615 ms

class Solution {

private int ans;

public int findTargetSumWays(int[] nums, int S) {

int sum = 0;

for (final int num : nums)

sum += num;

if (sum < Math.abs(S)) return 0;

ans = 0;

dfs(nums, 0, S);

return ans;

}

private void dfs(int[] nums, int d, int S) {

if (d == nums.length) {

if (S == 0) ++ans;

return;

}

dfs(nums, d + 1, S - nums[d]);

dfs(nums, d + 1, S + nums[d]);

}

Solution 3: Subset sum

Time complexity: O(n*sum)

Space complexity: O(sum)

C++ w/ copy

// Author: Huahua

// Running time: 7 ms

class Solution {

public:

int findTargetSumWays(vector<int>& nums, int S) {

S = std::abs(S);

const int sum = std::accumulate(nums.begin(), nums.end(), 0);

if (sum < S || (S + sum) % 2 != 0) return 0;

const int target = (S + sum) / 2;

vector<int> dp(target + 1, 0);

dp[0] = 1;

for (int num : nums) {

vector<int> tmp(dp);

for (int j = 0; j <= target - num; ++j)

tmp[j + num] += dp[j];

std::swap(dp, tmp);

}

return dp[target];

}

};

C++ w/o copy

// Author: Huahua

// Running time: 6 ms

class Solution {

public:

int findTargetSumWays(vector<int>& nums, int S) {

S = std::abs(S);

const int n = nums.size();

const int sum = std::accumulate(nums.begin(), nums.end(), 0);

if (sum < S || (S + sum) % 2 != 0) return 0;

const int target = (S + sum) / 2;

vector<int> dp(target + 1, 0);

dp[0] = 1;

for (int num : nums)

for (int j = target; j >= num; --j)

dp[j] += dp[j - num];

return dp[target];

}

};

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: