The problem specification indicates that all index pairs (i, j) where i < j and ar[i] + ar[j] is divisible by k must be found. Therefore, all possible pairs of elements in the array should be considered, as long as they meet the condition that i is less than j.

If we review the array [1, 3, 2, 6, 1, 2], the pairs are formed by checking each possible combination of indices (i, j) where i < j. In this case, the pair (4, 5) is also valid because:

ar[4] = 1 and ar[5] = 2 1 + 2 = 3, and 3 is divisible by 3. So the pair (4, 5) also meets the condition.

Here is the complete list of pairs that meet the divisibility condition by 3:

(0, 2) → 1 + 2 = 3 (0, 5) → 1 + 2 = 3 (2, 3) → 2 + 6 = 8 (2, 4) → 2 + 1 = 3 (3, 4) → 6 + 1 = 7 (4, 5) → 1 + 2 = 3

In summary, the pair (4, 5) should also be considered, as it meets the conditions, so the result should be 6 and not 5. Is there an error in the problem explanation and the result should be 6, or is the result indeed 5 and the condition that the pair (4, 5) was excluded not explained? What is the exact criterion for evaluating the pairs to be considered?

Haskell:

The easy, brute force way (solve) and the better, single-pass way (solve2).

module Main where

import qualified Data.IntMap as M

solve :: Int -> [Int] -> Int
solve k ar = length [(i, j) | i <- [0 .. length ar - 2], j <- [i + 1 .. length ar - 1], (ar !! i + ar !! j) `mod` k == 0]

go :: Int -> [Int] -> M.IntMap Int -> Int -> Int
go _ [] _ count = count
go target (x : xs) m count = go target xs (M.insertWith (+) (x `mod` target) 1 m) (count + M.findWithDefault 0 ((target - x) `mod` target) m)

solve2 :: Int -> [Int] -> Int
solve2 k ar = go k ar M.empty 0

main :: IO ()
main = do
    [_, k] <- map read . words <$> getLine :: IO [Int]
    ar <- map read . words <$> getLine :: IO [Int]
    print $ solve2 k ar

PHP

function divisibleSumPairs($n, $k, $ar) {
    $counter = 0;
    for ($i = 0; $i < $n - 1; $i++) {
        for ($j = $i + 1; $j < $n; $j++) {
            if (($ar[$i] + $ar[$j]) % $k == 0) {
                $counter++;
            }
        }
    }
    
    return $counter;
}

Python

def divisibleSumPairs(n, k, ar):
    total_pairs = 0
    remainder_count = [0] * k
    
    for num in ar:
        remainder = num % k
        complement = (k - remainder) % k
        total_pairs += remainder_count[complement]
        remainder_count[remainder] += 1
    
    return total_pairs

Time complexity: O(n)

int sum = 0; int count= 0;

    for (int i = 0; i < ar.size(); i++)
    {
        for (int j = i + 1; j < ar.size(); j++)
        {
            sum = ar[i] + ar[j];

            if (sum % k == 0)
        {
                count++;
            }

        }
    }
    return count;
}