Find the minimum k where valid(k) = 1 using binary search. The valid(k) function uses a set intersection of all segments of the string with length k. Then if the final set isn't empty, then that means there is at least one character common in those segments.

Time Complexity: Binary search of length of the string takes --> O(log(n)) operations

Checking if valid(k) takes --> O((n-k)*k) ~ O(n*k) operations Reference

Total Time Complexity = O(k*nlog(n)) operations

def isValid(n, s, k): # Check if **k** is valid on the string **s** of length **n**
    ans = set(s[:k])
    for i in range(1, n-k+1):
        ans = ans.intersection(set(s[i:i+k]))
    if len(ans) >= 1: return 1
    return 0

def solve(s):
    n = len(s)
    l, h = 0, n
    mid = 0
    while l < h-1:
        mid = (l + h) // 2
        # print(l, mid, h, isValid(n, s, mid))
        if isValid(n, s, mid):
            h = mid
        else:
            l = mid
    return mid

s = list(input())
print(solve(s))