Given that the length of the provided string is 10, the approach involves iterating through the string and comparing each character at position i with the corresponding character at position i in "codeforces."

t = int(input())
c = "codeforces"
for _ in range(t):
    s = input()
    res = 0
    for i in range(10):
        if s[i] != c[i]:
             res += 1
    print(res)	

Consider the first character in each pair.

Take every other character starting from the first one. Since the last character can not be a starting letter for a pair, add it to 'a' last.

t = int(input())
for _ in range(t):
    b = input()


    a = [] # Storing the characters in a list and appending gives a better time complexity compared to string concatenation
    for index in range(0, len(b), 2):
        a.append(b[index])
   
    a.append(b[-1])
    a = "".join(a)
    print(a)

Can you rewrite the formula?

The equation $$$a_j - a_i = j - i$$$ can be transformed into $$$a_j - j = a_i - i$$$, and let $$$b_i = a_i - i$$$. Then we need to count the number of pairs $$$(i, j)$$$ where $$$b_i = b_j$$$. We can use a dictionary(hash map) to count occurrences of each difference $$$(b_i)$$$.

T = int(input())
for _ in range(T):
    N = int(input())
    a = list(map(int,input().split()))
    difference_count = {}
    pair_count = 0  

    for i in range(N):
        difference = a[i] - i
        pair_count += difference_count.get(difference, 0)
        difference_count[difference] = difference_count.get(difference, 0) + 1
    print(pair_count)