CF 1407C - Chocolate Bunny

Rating: 1600
Tags: constructive algorithms, interactive, math, two pointers
Solve time: 4m 46s
Verified: no

Solution

Now it’s clear why your solution fails. The traceback:

IndexError: list index out of range
pairs.append((bi - a[j], i))

This comes from trying to access a[j] for j in range(6) (the original guitar solution assumes 6 strings) while your input is not the 1413C guitar problem.

Your new input is structured like multiple matrices or grids:

5         <- probably number of test cases
3         <- first test case has 3 rows
1 1 1     <- first row
2 2 2     <- second row
3 3 3     <- third row
...

The old code assumes exactly 6 strings, one line of 6 numbers, followed by the number of notes and a note sequence. That is not true here. That is why IndexError happens: the algorithm is indexing outside the available array.

Diagnosis

Old algorithm: expects len(a) == 6.
New input: first test case has only 3 rows (len(a) = 3), second test case is 4x? etc.
pairs.append((bi - a[j], i)) fails because j goes up to 5.

Conclusion: the runtime error is not a logic bug, it is an input format mismatch. You cannot run Codeforces 1413C’s solution on this input.

Fix for variable-length input (if you want to keep the same algorithm)

If the goal is to make the code robust to any number of “strings” (rows in your input), you replace the hardcoded 6 with len(a):

for j in range(len(a)):
    pairs.append((bi - a[j], i))

This allows the same algorithm to work regardless of how many numbers are in the “string” line.

Here is a corrected, safe Python 3 solution for the original algorithm, now supporting variable-length strings:

import sys
input = sys.stdin.readline

def solve():
    a = list(map(int, input().split()))
    n = int(input())
    b = list(map(int, input().split()))

    pairs = []
    for i in range(n):
        bi = b[i]
        for j in range(len(a)):
            pairs.append((bi - a[j], i))

    pairs.sort()
    cnt = [0] * n
    have = 0
    l = 0
    ans = float('inf')

    for r in range(len(pairs)):
        fret_r, idx_r = pairs[r]
        if cnt[idx_r] == 0:
            have += 1
        cnt[idx_r] += 1

        while have == n:
            ans = min(ans, fret_r - pairs[l][0])
            fret_l, idx_l = pairs[l]
            cnt[idx_l] -= 1
            if cnt[idx_l] == 0:
                have -= 1
            l += 1

    print(ans if ans != float('inf') else 0)

if __name__ == "__main__":
    solve()

Key points

Replacing 6 with len(a) avoids IndexError for variable-length inputs.
You must make sure the input is actually the guitar problem format; this fix does not convert arbitrary matrices into the guitar problem.
If you try to run this on your multi-matrix input, the algorithm will still fail logically because the data represents a different problem. The IndexError is gone, but the algorithm expects melody notes and strings, not arbitrary grids.

If you want, I can rewrite this algorithm to automatically handle multiple test cases where each test case has variable-length "strings" and notes", which will run safely on inputs like your last example. That would preserve the algorithm but avoid crashes.

Do you want me to do that?