1976D - Invertible Bracket Sequences - CodeForces Solution

binary search combinatorics data structures divide and conquer implementation two pointers *2000

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Python Code: 

from collections import Counter
for i in range(int(input())):
  s = input()
  res=cur=0
  cnt = Counter({0:1})
  for c in s:
    cur += 1 if c == '(' else -1
    res += cnt[cur]
    cnt[cur] += 1
    cnt[(cur-1)//2] = 0
  print(res)


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