Berliner FC Preussen vs Hertha 03 Zehlendorf

Expert Analysis: Berliner FC Preussen vs Hertha 03 Zehlendorf

The upcoming match between Berliner FC Preussen and Hertha 03 Zehlendorf is poised to be an exciting encounter, with several betting predictions suggesting a dynamic game. Both teams have shown varying levels of offensive and defensive prowess throughout the season, which will play a crucial role in determining the outcome of this match. With an average total goals forecasted at 4.18, it indicates a high-scoring affair might be on the cards. The odds for over 0.5 goals in the first half are particularly favorable, suggesting early action could be expected.

Prediction Analysis

• Over 0.5 Goals HT (85.20): The high probability of more than half a goal being scored in the first half suggests both teams will likely start aggressively.
• Over 1.5 Goals (78.20): A strong likelihood exists for this match to feature more than one and a half goals overall, pointing towards an open game.
• Away Team Not To Score In 2nd Half (80.70): This prediction hints at a potential strong start from Berliner FC Preussen or a defensive tightening up by Hertha 03 Zehlendorf in the latter stages.
• Both Teams Not To Score In 1st Half (68.70): Despite high expectations for goals, there’s still a significant chance neither team will score early on.
• Home Team To Score In 1st Half (67.30): Berliner FC Preussen is expected to capitalize on their home advantage by scoring in the opening period.
• Away Team Not To Score In 1st Half (64.80): Hertha 03 Zehlendorf may face challenges breaking through early, as indicated by these odds.
• Average Total Goals (4.18): This statistic supports the idea of an entertaining match with plenty of scoring opportunities for both sides.
• Average Conceded Goals (1.82) & Average Goals Scored (2.26): These averages suggest that while both teams are capable of scoring, they also have vulnerabilities that could be exploited by their opponents.
• Red Cards (0.56): The relatively low expectation for red cards indicates that while tempers may flare, outright dismissals are not highly anticipated.

Betting Insights

• Both Teams To Score: 58.90: This is one of the more balanced bets, suggesting both teams have enough firepower to find the back of the net during this encounter.
• Home Team To Win: 60.90: Given their home ground advantage and current form metrics, Berliner FC Preussen appears slightly favored to take all three points.
• First Goal Between Minute 0-29: 54.80: Early goals seem probable based on these odds, indicating aggressive tactics from kickoff.
• Home Team To Score In 2nd Half: 69.20: Berliner FC Preussen might use their familiarity with home conditions to push forward later in the game as well.
• Away Team Not To Score In First Half: 64.80:</stronzhaoyongkun/leetcode/src/main/java/com/leetcode/solution/P_0039.java package com.leetcode.solution; import java.util.ArrayList; import java.util.Arrays; import java.util.List; /** * Created by zhaoyongkun on Nov/17/2017. */ public class P_0039 { public List<List> combinationSum(int[] candidates, int target) { List<List> result = new ArrayList(); if(candidates == null || candidates.length ==0){ return result; } Arrays.sort(candidates); backtrack(result,new ArrayList(),candidates,target,0); return result; } private void backtrack(List<List> result,List list,int[] candidates,int remain,int start){ if(remain<0){ return; } else if(remain==0){ result.add(new ArrayList(list)); return; } for(int i=start;i<candidates.length;i++){ list.add(candidates[i]); backtrack(result,list,candidates,remain-candidates[i],i); list.remove(list.size()-1); } } public static void main(String[] args) { // int [] candidates = {10}; // int target =8; int [] candidates = {2,3}; int target =4; System.out.println(new P_0039().combinationSum(candidates,target)); // System.out.println(new P_0039().combinationSum(new int[]{},target)); // System.out.println(new P_0039().combinationSum(null,target)); // int [] candidates = {8}; // int target =8; // // // System.out.println(new P_0039().combinationSum(candidates,target)); // // int [] candidates = {3}; // int target =3; // // // System.out.println(new P_0039().combinationSum(candidates,target)); // // // //// [5] //// [3] //// [3] //// [2] //// //// [5] //// [3] //// [2] //// //// [5] //// [2] //// //// [3] //// [3] //// //// [3] //// [2] // // // [[5], //[3], //[2], //[3], //[2]] // // //[5] +[3]+[2] +[3]+[2] +[5] +[3]+[2] +[5] +[2] +[3]+[3] +[3]+[2] //[[8],[8],[8],[8],[8],[8],[8],[8]] // // //// //// //// //// //// //// //} # LeetCode LeetCode Solution ## Solution ### Easy * Two Sum : https://github.com/zhaoyongkun/leetcode/blob/master/src/main/java/com/leetcode/easy/P_0001.java * Add Two Numbers : https://github.com/zhaoyongkun/leetcode/blob/master/src/main/java/com/leetcode/easy/P_0002.java * Longest Substring Without Repeating Characters : https://github.com/zhaoyongkun/leetcode/blob/master/src/main/java/com/leetcode/easy/P_0003.java * Roman to Integer : https://github.com/zhaoyongkun/leetcode/blob/master/src/main/java/com/leetcode/easy/P_0007.java * Reverse Integer : https://github.com/zhaoyongkun/leetcode/blob/master/src/main/java/com/leetcode/easy/P_0007.java * Palindrome Number : https://github.com/zhaoyongkun/leetcode/blob/master/src/main/java/com/leetcode/easy/P_0009.java * String to Integer(Atoi) : https://github.com/zhaoyongkun/leetcode/blob/master/src/main/java/com/leetcode/easy/P_0084.java * Valid Parentheses : https://github.com/zhaoyongkun/leetcode/blob/master/src/main/java/com/leetcode/easy/P_0026.java ### Medium * Best Time to Buy and Sell Stock II :https://github.com/zhaoyongkun/leetcode/blob/master/src/main/java/com/leetcode/solution/P_0014.java * Best Time to Buy and Sell Stock III :https://github.com/zhaoyongkun/leetcode/blob/master/src/main/java/com/leetcode/solution/P_0016.java * Best Time to Buy and Sell Stock IV :https://github.com/zhaoyongkun/leetcode/blob/master/src/main/java/com/leetcode/solution/P_0017.java * Climbing Stairs :https://github.com/zhaoyongkun/leetcode/blob/master/src/main/java/com/algorithms/dynamicprogramming/p0200_climbing_stairs/Solution.java * Longest Increasing Subsequence :https://github.com/zhaoyongkun/algorithms/tree/master/dynamic-programming/p0300_longest_increasing_subsequence/Solution.py ### Hard #### Tree Traversal ##### Recursive traversal Pre-order traversal: void preorderTraversal(TreeNode root) { if(root == null) return ; visit(root); // Visit root. preorderTraversal(root.left); preorderTraversal(root.right); } In-order traversal: void inorderTraversal(TreeNode root) { if(root == null) return ; inorderTraversal(root.left); visit(root); // Visit root. inorderTraversal(root.right); } Post-order traversal: void postorderTraversal(TreeNode root) { if(root == null) return ; postorderTraversal(root.left); postorderTraversal(root.right); visit(root); // Visit root. } ##### Iterative traversal using stack Pre-order traversal: Stack stack; stack.push(root); while(!stack.empty()) { TreeNode *node = stack.top(); stack.pop(); visit(node); if(node->right != NULL) stack.push(node->right); if(node->left != NULL) stack.push(node->left); } In-order traversal: Stack stack; TreeNode *current = root; while(current != NULL || !stack.empty()) { while(current != NULL){ // Reach left most Node of current Node stack.push(current); current = current->left; } current = stack.top(); stack.pop(); visit(current); // Visit node current = current->right; // Process right subtree } Post-order traversal: The key idea here is that we need process nodes from bottom up according to their depth level. For example, given binary tree `{1,#,2,3}` , its post-order traversal is `[3,2,1]`. If we use iterative method with single stack , we can only get `[1,3,2]` , because when we pop `root` from top of `stack` , then process its right child `root.right` . But what we want is processing its left child `root.left` first . So we need another data structure , e.g., set , to record whether left child has been processed . So our algorithm becomes : Push root into `stack`. While `stack` is not empty: Pop node out from top of `stack`, call it `root`. If left child has been processed or does not exist: Visit it. If right child does not exist: Push it into `set`. Else : If right child has been processed : Visit it . Push it into set . Push left child into `stack`. Repeat until `stack` is empty . Here’s code implementation : set s; void postorderTraversal(TreeNode *root){ Stack st; st.push(root); while(!st.empty()){ TreeNode *node=st.top(); if((node->left==NULL && s.find(node)==s.end()) || node->right==NULL){ visit(node); s.insert(node); if(node->right!=NULL){ st.push(node->right);} else{ st.pop();}} else{ if(s.find(node->right)==s.end()){ visit(node); s.insert(node);} st.push(node->left);} } } #### Divide & Conquer Divide & Conquer algorithm consists of four steps : Divide: Divide original problem into smaller subproblems recursively. Conquer: Solve subproblem recursively until subproblem becomes small enough. Combine: Combine results from subproblems together. #### Dynamic Programming Dynamic programming problems usually contain two key elements : Optimal substructure property. Overlapping subproblems property. Typically , dynamic programming algorithms follow below steps : Characterize structure of optimal solution. Define recursive function based on above structure . Bottom-up approach – solve problem iteratively. Top-down approach – solve problem recursively using memoization . Bottom-up approach usually uses iteration whereas top-down approach uses recursion. ## Reference http://www.lintcode.com/en/problem/binary-tree-inorder-traversal/ http://www.lintcode.com/en/problem/binary-tree-preorder-traversal/ http://www.lintcode.com/en/problem/binary-tree-postorder-traversal/ http://www.geeksforgeeks.org/dynamic-programming-set-32-longest-common-subsequence/ zhaoyongkun/leetcode<|file_sep#!/usr/bin/env python class Solution(object): def lengthOfLongestSubstring(self,s): """ Use dictionary instead of list so that lookup time is O(1). """ n=len(s) max_length=0 # length longest substring without duplicate characters. dict={} # i marks starting index+1 of window. # j marks ending index+1 of window. i,j=0,len(s) while i<j: if j<n and s[j] not in dict: dict[s[j]]=j # Insert new character into dictionary. j+=1 # Expand window size by expanding end pointer 'j'. else: max_length=max(max_length,j-i) dict[s[i]]=None # Remove character out side window from dictionary. i+=1 # Shrink window size by moving start pointer 'i'. return max_length if __name__=='__main__': s='abcabcbb' print(Solution().lengthOfLongestSubstring(s))zhaoyongkun/algorithms<|file_sep[](https://www.youtube.com/watch?v=dQw4w9WgXcQ) # Algorithms ## Reference ### Books #### Designing Data Intensive Applications Data Modeling – Chapter 6 Consistency – Chapter7&Chapter11&Chapter12&Chapter13&Chapter14&Chapter15&Chapter16&Chapter17&Chapter18 Availability & Partition Tolerance – Chapter19&Chapter20&Chapter21&Chapter22&Chaper23&Chaper24 #### Clean Code Clean Code – Chapter01-06 Code Smells – Chapter07-09 #### Refactoring Refactoring – Chapter01-05 #### Effective Java Effective Java Item01-02 Effective Java Item16-17 #### Clean Architecture Clean Architecture – Part01-02 ### Online Courses #### Coursera Algorithms Specialization I-IV Algorithms Specialization I Problem Set Solutions ### Blogs / Websites / Videos / Etc… Google Interview Questions YouTube Channel "BackToBackSWE" zhaoyongkun/algorithms<|file_sep` class Solution(object): def twoSum(self,numbers,target): i,j=0,len(numbers)-1 while(itarget: j-=1 elif numbers[i]+numbers[j]<target: i+=1 else:# If sum equals target, return i+1,j+1 return None,None if __name__=='__main__': numbers=[-101,-100,-99,-98,-97,-96,-95,-94,-93,-92,-91, -90,-89,-88,-87,-86,-85,-84, -83, -82, -81, -80, -79, -78, -77, -76, -75, ] target=-165 print(Solution().twoSum(numbers,target)) class Solution(object): def twoSum(self,numbers,target): i,j=0,len(numbers)-1 while(itarget: j-=1 elif numbers[i]+numbers[j]<target: i+=1 else:# If sum equals target, return i+1,j+1 return None,None if __name__=='__main__': numbers=[-10100,-10000,-9900,…99999] target=-165 print(Solution().twoSum(numbers,target)) class Solution(object): def twoSum(self,numbers,target): i,j=0,len(numbers)-1 while(itarget: j-=j elif numbers[i]+numbers[j]<target: i+=i else:# If sum equals target, return i+1,j+1 return None,None if __name__=='__main__': numbers=[-10100,…99999] target=-165 print(Solution().twoSum(numbers,target)) class Solution(object): def twoSum(self,numbers,target): i,j=0,len(numbers)-1 while(itarget: j-=j elif numbers[i]+numbers[j]<target: i+=i else:# If sum equals target, return i+1,j+1 return None,None if __name__=='__main__': numbers=[-10100,…99999] target=-165 print(Solution().twoSum(numbers,target)) # ===================== # # === Final Version === # # ===================== # class Solution(object): def twoSum(self,numbers,target): i,j=0,len(numbers)-i while(itarget: j-=j else: i+=i else:# If sum equals target, return i+1,j+i return None,None if __name__==’__main__’: numbers=[-10100,…99999] target=-165 print(Solution().twoSum(numbers,target)) def test(): s=self.twoSum([-10100,…99999],-165) assert s==(x,y),str(s)+’:’+str((x,y)) test()