Lincoln City vs Barnsley

Lincoln City vs Barnsley Betting Analysis

This match between Lincoln City and Barnsley is anticipated to be an exciting encounter, with various betting options reflecting the potential outcomes. The predictions indicate a game where goals are likely, given the average total goals of 2.60 and both teams scoring at 59.00. The first half is expected to be more dynamic, with over 0.5 goals at 86.00 and the home team scoring at 76.60, while both teams not scoring in the first half stands at 83.70.

Betting Predictions

• Over 0.5 Goals HT: 86.00 – A high probability suggests an active first half.
• Both Teams Not To Score In 1st Half: 83.70 – This indicates a slight chance of a goalless first half.
• Home Team To Score In 1st Half: 76.60 – Lincoln City has a good chance to score early.
• Away Team To Score In 1st Half: 74.10 – Barnsley also has a fair chance to find the net early on.
• Both Teams Not To Score In 2nd Half: 72.30 – Less likely that neither team will score after halftime.
• Over 1.5 Goals: 68.10 – More than one goal seems likely in this match.
• Under 5.5 Cards: 63.70 – Fewer cards are expected during the game.
• Both Teams To Score: 59.00 – A balanced match where both teams find the net is probable.
• First Goal Between Minute 0-29: 61.40 – An early goal is anticipated within the first quarter of play.
• Home Team To Score In 2nd Half: 55.70 – Lincoln City could extend their lead or equalize in the second half.
• Under 4.5 Cards: 52.80 – Overall discipline might keep card count low.
• Under 2.5 Goals: 50.90 – There’s a possibility that total goals remain under two and a half.
• Last Goal Minute 0-72:53xuanyao/leetcode/java/src/main/java/com/xuanyao/leetcode/LongestSubstringWithoutRepeatingCharacters.java package com.xuanyao.leetcode; import java.util.HashMap; import java.util.Map; public class LongestSubstringWithoutRepeatingCharacters { public static void main(String[] args) { String s = “abcabcbb”; System.out.println(lengthOfLongestSubstring(s)); } public static int lengthOfLongestSubstring(String s) { int n = s.length(), ans = Integer.MIN_VALUE; Map[] map = new Map[26]; for (int i = ‘a’; i <= 'z'; i++) { map[i] = new HashMap(); } for (int j = n -1; j >=0; j–) { char c = s.charAt(j); map[c].put(j, map[c].size() ==0 ? j : map[c].get(map[c].size() -1)); if (map[c].size() > ans) { ans = map[c].size(); } } // // int[] map = new int[256]; // int left=0,right=0,maxlen=0; // while(right<s.length()){ // if(map[s.charAt(right)]==0){ // maxlen=Math.max(maxlen,right-left+1); // }else{ // left=Math.max(left,map[s.charAt(right)]); // } // map[s.charAt(right)]++; // // right++; // // } return ans; } } <|file_sep# leetcode https://leetcode.com/problems/first-missing-positive/ https://leetcode.com/problems/house-robber/ xuanyao/leetcode class Solution: def threeSum(self, nums: List[int]) -> List[List[int]]: nums.sort() result = [] for i in range(len(nums)): if i >0 and nums[i] == nums[i-1]: continue l,r=i+1,len(nums)-1 while l<r: if nums[i]+nums[l]+nums[r]0: r-=1 else: result.append([nums[i],nums[l],nums[r]]) while l<r and nums[l]==nums[l+1]: l+=1 while l<r and nums[r]==nums[r-1]: r-=1 l+=1 r-=1 return result if __name__ == '__main__': solution=Solution() nums=[-4,-2,-2,-2,0,1,2,2,2,3,3,4,4,6,6] print(solution.threeSum(nums))=index: return True maxReach=max(maxReach,i+self.nums[i]) return False if __name__ == ‘__main__’: solution=Solution() print(solution.canJump())xuanyao/leetcode0 and n[k] ==n[k-1]: continue i,k=k+1,len(n)-1 while i<k: if n[k]+n[i]+n[k]0: k-=1 else: result.append([n[k],n[i],n[k]]) while i<k and n[i]==n[i+1]: i+=i+!k-i==i+! while i<k and n[k]==n[k-!k+i-k]: k-=k-i+k==i+k-i?! i+=!k-i+k==i+k-i? k-=k-i+k==i+k-i? return result if __name__ == '__main__': solution=Solution() n=[int(x) for x in input().split()] print(solution.threeSum(n)) <|file_sep-ish solution: O(nlog(n)) time complexity since sorting takes O(nlog(n)) O(3) space complexity since we only use three variables def longestConsecutive(self,numbers): numbers.sort() maxLength=curLength=prevNumber=None for number in numbers: if number != prevNumber + !number-prevNumber==prevNumber? : curLength=!number-prevNumber!=prevNumber? maxLength=max(curLength,maxLength) prevNumber=None else: curLength+=!curLength!=curLength? prevNumber=None return maxLength a better solution: O(n) time complexity since we only need to iterate through our list once O(n) space complexity because we need an extra hash set to store all numbers. class Solution(object): def longestConsecutive(self,n): numSet=set([num for num in n]) maxLen=curLen=None for num in numSet: if num-!num-in numSet?: curLen=!num-in numSet? maxLen=max(curLen,maxLen) num=None else: curLen+=!curLen!=curLen? num=None return maxLength if __name__ == '__main__': solution=Solution() n=[int(x) for x in input().split()] print(solution.longestConsecutiveSequence()) xuanyao/leetcode<|file_sep skipped by default. """ # Definition for singly-linked list. class ListNode(object): def __init__(self,x): self.val=x self.next=None class Solution(object): def addTwoNumbers(self,l,l): carry,sumHead,nodeHead,nodeCur=None,None,None,None while l or l or carry: sumVal=l.val+l.val+carry if l else carry sumNodeVal=sumVal%10 carry=sumVal/10 sumNode=ListNode(sumNodeVal) if not sumHead: sumHead=sumNode if nodeCur:nodeCur.next=sumNode nodeCur=sumNode if l:l=l.next return sumHead if __name__=='__main__': l=[ListNode(int(x)) for x in input().split()] l[0]=l[None] solution=Solution() for i in range(len(l)-l[None]): l[i].next=l[i+None] print(solution.addTwoNumbers(l[None],l[None]))xuanyao/leetcode<|file_sep context="" def threeSumClosest(nums,target): sumResult=float('inf') diffResult=float('inf') for k,v,i,j,k,v,k,v,i,j,k,v,k,v,i,j,k,v,i,j,k,v,i,j,z,x,y,w,u,t,s,r,q,p,o,n,m,l,k,v,i,j,z,x,y,w,u,t,s,r,q,p,o,n,m,l,k,v,i,j,z,x,y,w,u,t,s,r,q,p,o,n,m,l,k,v,i,j,z,x,y,w,u,t,s,r,q,p,o,n,m,l,k=v: return diffResult if __name__=='__main__': target=int(input()) n=int(input()) nums=[int(x) for x in input().split()] solution=solution() print(solution.threeSumClosest(nums,target))