Fixing Runtime Error in Codeforces 2190A | Generated by AI

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Question: Why does this code get a Runtime Error on test 1 for Codeforces problem 2190A?

Answer:

Your code gets Runtime Error on test 1 mainly because of this critical logical error:

for (i = 0; i < n; i++) {
    char target;
    if (i < zero) {
        target = '0';
    } else {
        target = '1';
    }
    if (s.charAt(i) != target) {
        positions.add(i);
    }
}

This logic assumes that after sorting, the string should become:

• first zero characters = ‘0’
• remaining one characters = ‘1’

But this is not what the problem asks for.

The problem requires the lexicographically smallest string you can obtain by at most one swap of two characters.

That means the target sorted string is always:

All ‘0’s come before all ‘1’s
→ 000...0111...1 (exactly zero zeros followed by one ones)

But your code is comparing the current string against this ideal sorted string and collecting all mismatch positions.

This is completely wrong for the purpose of the problem, because:

• The problem allows at most one swap (0 or 1 swap)
• You are supposed to output either:
  • “Bob” (if already sorted / no swap needed or impossible with 1 swap)
  • “Alice” + exactly two positions to swap (1-based)

Your code almost always outputs more than two positions (or zero), which leads to:

• Wrong output format → judge expects either:
  • single line “Bob”
  • or two lines: “Alice” followed by exactly two numbers
• When it prints many numbers or wrong count → Runtime Error or Wrong Answer

Real reasons for Runtime Error on test 1

Most probably test 1 is one of these simple cases:

1
3
010

or

1
4
1100

or

1
1
0

In such cases your code:

• Counts zero and one
• If all same → prints “Bob” (ok)
• Otherwise → goes into the positions collection logic
• Prints “Alice” and then many positions (or none)
• Judge expects either “Bob” or “Alice\nx y” → your output format is invalid → Runtime Error

Correct idea (very brief)

You only need to find at most one misplaced ‘1’ before a ‘0’.

The string is already good (Bob) if it is non-decreasing (all 0s before 1s).

Otherwise, you usually need to swap one ‘1’ that appears too early with one ‘0’ that appears too late.

Most common pattern: find the leftmost ‘1’ that should be ‘0’, and the rightmost ‘0’ that should be ‘1’, and swap them.

But you must check if after this one swap the string becomes sorted.

Quick correct approach

Read t
For each test:
    Read n, s
    Count zeros = count of '0'
    
    // Check if already sorted (non-decreasing)
    boolean sorted = true;
    for (int i = 1; i < n; i++) {
        if (s.charAt(i) < s.charAt(i-1)) {
            sorted = false;
            break;
        }
    }
    
    if (sorted) {
        out.println("Bob");
        continue;
    }
    
    // Find leftmost position where we have '1' but should be '0'
    int left = -1;
    for (int i = 0; i < zeros; i++) {
        if (s.charAt(i) == '1') {
            left = i;
            break;
        }
    }
    
    // Find rightmost position where we have '0' but should be '1'
    int right = -1;
    for (int i = n-1; i >= zeros; i--) {
        if (s.charAt(i) == '0') {
            right = i;
            break;
        }
    }
    
    // If we found both → one swap fixes it
    if (left != -1 && right != -1) {
        out.println("Alice");
        out.println((left+1) + " " + (right+1));
    } else {
        out.println("Bob");  // impossible with one swap
    }

This is not 100% complete (some edge cases need careful check), but it explains the direction.

Your current code is trying to do something completely different from what the problem asks, which is why it fails very early (test 1).

References:

• https://codeforces.com/contest/2190/problem/A

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