We use a stack to keep track of previous strings and repetition multipliers. When we see an opening bracket, we push the current context. When we see a closing bracket, we pop and reconstruct.

1. stack = [], curr_str = "", curr_num = 0.
2. For each character c in s:
   • If c is digit: curr_num = curr_num * 10 + int(c).
   • If c == '[':
     • Push (curr_str, curr_num) to stack.
     • Reset curr_str = "", curr_num = 0.
   • If c == ']':
     • prev_str, num = stack.pop().
     • curr_str = prev_str + num * curr_str.
   • Else: curr_str += c.
3. Return curr_str.

• Time Complexity: O(total_length_of_output_string).
• Space Complexity: O(N) for the stack.

def decode_string(s):
    stack = []
    curr_num = 0
    curr_str = ""
    
    for char in s:
        if char == '[':
            stack.append((curr_str, curr_num))
            curr_str = ""
            curr_num = 0
        elif char == ']':
            prev_str, num = stack.pop()
            curr_str = prev_str + num * curr_str
        elif char.isdigit():
            curr_num = curr_num * 10 + int(char)
        else:
            curr_str += char
            
    return curr_str