We maintain two hash maps for bi-directional mapping between characters of s and t. Alternatively, we can use a single map and a set to ensure no two characters map to the same target.

1. If len(s) != len(t), return False.
2. Initialize mapping_s_t = {}, mapped_t_chars = set().
3. For each pair of characters (c1, c2) in zip(s, t):
   • If c1 in mapping_s_t:
     • If mapping_s_t[c1] != c2, return False.
   • Else:
     • If c2 in mapped_t_chars, return False.
     • mapping_s_t[c1] = c2, mapped_t_chars.add(c2).
4. Return True.

• Time Complexity: O(N).
• Space Complexity: O(1) (size of character set).

def is_isomorphic(s, t):
    if len(s) != len(t): return False
    
    map_s = {}
    map_t = {}
    
    for c1, c2 in zip(s, t):
        if c1 in map_s:
            if map_s[c1] != c2: return False
        elif c2 in map_t:
            return False
        else:
            map_s[c1] = c2
            map_t[c2] = c1
            
    return True