CF 282A - Bit++

CF 282A - Bit++

Rating: 800
Tags: implementation
Solve time: 1m 16s
Verified: yes

Solution

Problem Understanding

The problem presents a toy programming language with only one variable, x, initially set to zero. Each line of the program is a statement that either increments or decrements this variable. The operation can appear before or after the variable, so the statement could be ++X, X++, --X, or X--. Our task is to determine the final value of x after executing all statements in order.

The input starts with the number of statements n, followed by n lines, each a valid statement. The output is a single integer representing the final value of x.

Given the constraints, n is at most 150, which is extremely small for modern computers. This means even a naive approach that examines every statement individually will run instantly. There are no concerns about large input sizes, integer overflows, or performance optimizations.

A non-obvious edge case arises from the order of symbols. Since the operation can appear before or after X, a naive check for statement == "++X" or statement == "X++" might miss a decrement operation like X--. We must detect increment and decrement by checking for the presence of ++ or -- anywhere in the statement, not just by position. For example, the input:

3
++X
X--
--X

should result in a final value of 0, because the first increments, the second decrements, and the third decrements again.

Approaches

The brute-force approach is straightforward. Initialize a counter x to zero. For each statement, examine the string. If it contains ++, increment x; if it contains --, decrement x. This works because every statement contains exactly one operation and one variable. Since each statement is examined individually and we perform only a constant-time string search per statement, the total operation count is proportional to n. For n = 150, this is trivial.

There is no meaningful "faster" algorithm here because the brute-force method already runs in linear time relative to the number of statements. The key insight is that we do not need to parse the statement positionally or tokenize it; the presence of the two-character operator ++ or -- is sufficient. This observation simplifies the implementation and eliminates edge cases related to the order of characters.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n)	O(1)	Accepted
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Initialize a variable x to zero. This represents the current value of the only variable in the program.
2. Read the number of statements, n.
3. For each statement in the program, check whether it contains the substring ++. If it does, increment x.
4. If the statement contains --, decrement x.
5. After processing all statements, print the value of x.

Why it works: The algorithm maintains the invariant that x always reflects the net effect of all statements processed so far. Each statement modifies x by exactly one, either up or down, and every statement is guaranteed to contain exactly one operation. Therefore, by scanning sequentially and updating x according to the operation detected, the final value is guaranteed to be correct.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
x = 0

for _ in range(n):
    statement = input().strip()
    if "++" in statement:
        x += 1
    elif "--" in statement:
        x -= 1

print(x)

The solution first reads the number of statements and initializes x. For each statement, it removes any trailing newline or spaces and checks for the presence of the increment or decrement operator. We use in rather than a positional check because the operator may appear before or after X. This ensures correctness for all permutations like ++X or X++. Finally, the result is printed.

Worked Examples

Trace through the sample input:

1
++X

Step	Statement	x before	x after
1	++X	0	1

This confirms that a single increment works correctly.

Another example:

3
++X
X--
--X

Step	Statement	x before	x after
1	++X	0	1
2	X--	1	0
3	--X	0	-1

This demonstrates handling a mix of increments and decrements in any order.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each statement is checked exactly once, with a fixed-length substring search.
Space	O(1)	Only a single integer x is stored; no additional data structures are needed.

Given n <= 150, the solution runs in negligible time and uses minimal memory.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n = int(input())
    x = 0
    for _ in range(n):
        statement = input().strip()
        if "++" in statement:
            x += 1
        elif "--" in statement:
            x -= 1
    return str(x)

# provided samples
assert run("1\n++X\n") == "1", "sample 1"

# custom cases
assert run("3\n++X\nX--\n--X\n") == "-1", "mixed increments and decrements"
assert run("5\n++X\nX++\n++X\nX++\n++X\n") == "5", "all increments"
assert run("4\n--X\nX--\n--X\nX--\n") == "-4", "all decrements"
assert run("1\nX--\n") == "-1", "single decrement"
assert run("2\nX++\n--X\n") == "0", "increment followed by decrement"

Test input	Expected output	What it validates
3 increments and decrements mixed	-1	Correctly handles multiple operations in different orders
5 increments all	5	Correctly handles all increments
4 decrements all	-4	Correctly handles all decrements
Single decrement	-1	Minimal input handling
Increment then decrement	0	Operations cancel out correctly

Edge Cases

The main edge case is the order of symbols in the statement. For example, X++ should be treated exactly the same as ++X. Our algorithm handles this because it checks for the substring ++ or -- anywhere in the statement.

Input:

2
X++
--X

Execution trace:

Step	Statement	x before	x after
1	X++	0	1
2	--X	1	0

Output is 0, confirming the algorithm correctly interprets operator positions. This ensures no off-by-one errors due to the position of the operator relative to the variable.