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#include<bits/stdc++.h> using namespace std; int main(){ int n ; cin>>n; vector<int>a(n); vector<int>b(n); for(int i=0;i<n;i++) cin>>a[i]; for(int i=0;i<n;i++) cin>>b[i]; int ans=1; for(int i=0;i<n;i++) { int temp=1; for(int j=i+1;j<n;j++) { if((a[i]>=a[j] && a[i]<=b[j]) || (a[j]>=a[i] && a[j]<=b[i] )) temp++; } ans= max(ans,temp); } cout<<ans; return 0; } Python Courth house Code✅✅ Telegram http://t.me/Coding_solution_0 http://t.me/Coding_solution_0 http://t.me/Coding_solution_0

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def minOps(A, B): m = len(A) n = len(B) # This part checks whether conversion is possible or not if n != m:in return -1 count = [0] * 256 for i in range(n): # count characters in A count[ord(B[i])] += 1 for i in range(n): # subtract count for every char in B count[ord(A[i])] -= 1 for i in range(256): # Check if all counts become 0 if count[i]: return -1 # This part calculates the number of operations required res = 0 i = n-1 j = n-1 while i >= 0: # if there is a mismatch, then keep incrementing # result 'res' until B[j] is not found in A[0..i] while i>= 0 and A[i] != B[j]: i -= 1 res += 1 # if A[i] and B[j] match if i >= 0: i -= 1 j -= 1 return res # Driver program A = "EACBD" B = "EABCD" print ("Minimum number of operations required is " + str(minOps(A,B))) ✅Telegram- Send the Question here ❤️🚀 https://t.me/good_coders https://t.me/good_coders https://t.me/good_coders

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def minOps(A, B): m = len(A) n = len(B) # This part checks whether conversion is possible or not if n != m: return -1 count = [0] * 256 for i in range(n): # count characters in A count[ord(B[i])] += 1 for i in range(n): # subtract count for every char in B count[ord(A[i])] -= 1 for i in range(256): # Check if all counts become 0 if count[i]: return -1 # This part calculates the number of operations required res = 0 i = n-1 j = n-1 while i >= 0: # if there is a mismatch, then keep incrementing # result 'res' until B[j] is not found in A[0..i] while i>= 0 and A[i] != B[j]: i -= 1 res += 1 # if A[i] and B[j] match if i >= 0: i -= 1 j -= 1 return res # Driver program A = "EACBD" B = "EABCD" print ("Minimum number of operations required is " + str(minOps(A,B))) Python Minimum number of operations required Code ✅ http://t.me/Coding_solution_0 http://t.me/Coding_solution_0 http://t.me/Coding_solution_0

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def check(a,n) if n==1:        return 1      a.sort()      c=0      i=0      while(i<n-1):             if a[i+1]-a[i]==1:                   i=i+2                   c=c+1            else:                   i=i+1       if c:             return c       else:             return 1 n=int(input()) a=[] for i in range(n):       a.append(int(input())) print(check(a,n)) Python Whales code Telegram - http://t.me/Coding_solution_0 http://t.me/Coding_solution_0. http://t.me/Coding_solution_0

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