#User function Template for python3
class Solution:
def minOperations(self, N):
# Code here
if N == 1:
return 0
return N * N // 4
#{
# Driver Code Starts
#Initial Template for Python 3
if __name__ == '__main__':
T = int(input())
while T > 0:
N =int(input())
ob = Solution()
print(ob.minOperations(N))
T -= 1
#User function Template for python3
class Solution():
def lexicographicallyLargest(self, a, n):
#your code here
ans = []
j = 0
for i in range(len(a)):
if (a[i] - a[j]) % 2:
t = a[j:i]
t.sort(reverse=True)
ans.extend(t)
j = i
t = a[j:len(a)]
t.sort(reverse=True)
ans.extend(t)
return ans
You are given an array A of size N. Let us denote S as the sum of all integers present in the array. Among all integers present in the array, find the minimum integer X such that S ≤ N*X.
Example 1:
Input:
N = 3,
A = { 1, 3, 2}
Output:
2
Explanation:
Sum of integers in the array is 6.
since 6 ≤ 3*2, therefore 2 is the answer.
Example 2:
Input:
N = 1,
A = { 3 }
Output:
3
Explanation:
3 is the only possible answer
Your Task:
The task is to complete the function minimumInteger() which takes an integer N and an integer array A as input parameters and returns the minimum integer which satisfies the condition.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 105
1 ≤ Ai ≤ 109
SOLUTION IN C++
from typing import List
class Solution:
def minimumInteger(self, N : int, A : List[int]) -> int:
# code here
s = sum(A)
ans = max(A)
for a in A:
if N * a >= s and a < ans:
ans = a
return ans
#{
# Driver Code Starts
class IntArray:
def __init__(self) -> None:
pass
def Input(self,n):
arr=[int(i) for i in input().strip().split()]#array input
return arr
def Print(self,arr):
for i in arr:
print(i,end=" ")
print()
if __name__=="__main__":
t = int(input())
for _ in range(t):
N = int(input())
A=IntArray().Input(N)
obj = Solution()
res = obj.minimumInteger(N, A)
print(res)
# } Driver Code Ends
Count the Substrings.py
Python Code !!
#{
# Driver Code Starts
#Initial Template for Python 3
from collections import Counter
class Solution:
def countSubstring(self, S):
#code here
n = len(S)
d = Counter()
d[0] = 1
cnt = 0
ans = 0
for c in S:
if c.isupper():
cnt += 1
else:
cnt -= 1
ans += d[cnt];
d[cnt] += 1
return ans
#{
# Driver Code Starts.
if __name__ == '__main__':
t = int(input())
for _ in range(t):
S = input()
ob = Solution()
ans = ob.countSubstring(S)
print(ans)
# } Driver Code Ends
