Problem Description

You have an integer array nums of length n and an integer array queries.

The problem asks you to:

1. Generate all possible GCD pairs: Calculate the GCD (Greatest Common Divisor) for every possible pair (nums[i], nums[j]) where 0 <= i < j < n. This means you need to find the GCD for all combinations of two different elements from the array.

2. Create a sorted GCD array: Take all these GCD values and sort them in ascending order. This sorted array is called gcdPairs.

3. Answer queries: For each value queries[i], return the element at index queries[i] in the gcdPairs array.

The function should return an array answer where answer[i] contains the value at position queries[i] in the sorted gcdPairs array.

Example to illustrate:

• If nums = [2, 4, 6], the pairs would be: (2,4), (2,6), (4,6)
• The GCDs would be: gcd(2,4) = 2, gcd(2,6) = 2, gcd(4,6) = 2
• So gcdPairs = [2, 2, 2] after sorting
• If queries = [0, 1, 2], the answer would be [2, 2, 2]

The key challenge is efficiently computing and storing the GCD values for all pairs, especially when dealing with large arrays, since the number of pairs grows quadratically with the size of the input array.

Intuition

The naive approach would be to compute the GCD for all pairs, sort them, and answer queries directly. However, with n elements, we'd have n*(n-1)/2 pairs, which could be extremely large and inefficient.

The key insight is to think about the problem from a different angle: instead of computing individual GCDs, we can count how many pairs have each possible GCD value.

Why is this helpful? Consider that:

1. The GCD of any pair must be a divisor of both numbers in the pair
2. The GCD values are limited by the maximum value in the array (at most max(nums) different values)
3. If we know the count of pairs for each GCD value, we can build the sorted gcdPairs array conceptually without storing all individual values

Here's the clever observation: if we want to count pairs with GCD equal to g, we need to:

• First, count all pairs where both numbers are divisible by g (these pairs have GCD that's at least g)
• Then subtract pairs whose GCD is actually 2g, 3g, 4g, ... (multiples of g)

For example, if we want pairs with GCD = 2:

• Count all pairs where both numbers are divisible by 2
• Subtract pairs with GCD = 4, 6, 8, ... (which we've already computed)

This suggests a bottom-up approach: start from the largest possible GCD and work downwards. For each value i:

1. Count how many numbers in nums are multiples of i. If there are v such numbers, they can form v*(v-1)/2 pairs
2. These pairs have GCD that's a multiple of i, but some might have larger GCDs
3. Subtract the counts of pairs with GCD equal to multiples of i (which we've already calculated)

Once we have counts for each GCD value, we can use prefix sums to quickly determine which GCD value corresponds to each query index, using binary search to find the answer efficiently.

This transforms a potentially quadratic problem into one that's linear in the range of values, making it much more efficient.

Learn more about Math, Binary Search, Combinatorics and Prefix Sum patterns.

Solution Approach

Let's walk through the implementation step by step:

Step 1: Initialize data structures

• Find mx = max(nums) - the maximum value in the array
• Create cnt = Counter(nums) to store the frequency of each number
• Initialize cnt_g = [0] * (mx + 1) to store the count of pairs with each GCD value

Step 2: Calculate GCD counts (traverse from large to small)

We iterate i from mx down to 1. For each value i:

1. Count multiples of i:

   v = 0
   for j in range(i, mx + 1, i):
       v += cnt[j]

   This counts how many numbers in nums are multiples of i (i.e., i, 2i, 3i, ...)

2. Calculate pairs with GCD being a multiple of i:

   cnt_g[i] += v * (v - 1) // 2

   If there are v numbers that are multiples of i, they can form v*(v-1)/2 pairs, all having GCD that's at least i.

3. Subtract over-counted pairs:

   for j in range(i, mx + 1, i):
       cnt_g[i] -= cnt_g[j]

   We need to subtract pairs whose GCD is actually 2i, 3i, 4i, ... (not exactly i). Since we're iterating from large to small, these values have already been computed.

Step 3: Build prefix sum array

s = list(accumulate(cnt_g))

The prefix sum array s[i] tells us how many pairs have GCD less than or equal to i. This essentially gives us the sorted gcdPairs array in compressed form.

Step 4: Answer queries using binary search

return [bisect_right(s, q) for q in queries]

For each query q, we use bisect_right to find the smallest index i where s[i] > q. This index i is the GCD value at position q in the conceptual sorted gcdPairs array.

Example walkthrough: If nums = [4, 6, 8] and mx = 8:

• For i = 8: One number (8) is divisible by 8, so 0 pairs
• For i = 4: Two numbers (4, 8) divisible by 4, giving 1 pair with GCD ≥ 4
• For i = 2: All three numbers divisible by 2, giving 3 pairs, minus 1 pair with GCD=4, equals 2 pairs with GCD=2
• For i = 1: Would count all pairs minus those already counted

The prefix sum array then allows us to map query indices to their corresponding GCD values efficiently.

Example Walkthrough

Let's walk through a concrete example with nums = [4, 6, 8] and queries = [0, 1, 2].

Step 1: Setup

• Maximum value: mx = 8
• Frequency counter: cnt = {4: 1, 6: 1, 8: 1}
• Initialize cnt_g = [0, 0, 0, 0, 0, 0, 0, 0, 0] (indices 0 to 8)

Step 2: Calculate GCD counts (from largest to smallest)

For i = 8:

• Multiples of 8 in nums: just 8 → v = 1
• Pairs with GCD ≥ 8: 1 * (1-1) / 2 = 0
• No multiples to subtract
• cnt_g[8] = 0

For i = 7:

• No multiples of 7 in nums → v = 0
• cnt_g[7] = 0

For i = 6:

• Multiples of 6: just 6 → v = 1
• Pairs with GCD ≥ 6: 1 * 0 / 2 = 0
• cnt_g[6] = 0

For i = 5:

• No multiples of 5 → cnt_g[5] = 0

For i = 4:

• Multiples of 4: 4 and 8 → v = 2
• Pairs with GCD ≥ 4: 2 * 1 / 2 = 1
• Subtract cnt_g[8] = 0 (since 8 is a multiple of 4)
• cnt_g[4] = 1 - 0 = 1
• This represents the pair (4, 8) with GCD = 4

For i = 3:

• Multiples of 3: just 6 → v = 1
• cnt_g[3] = 0

For i = 2:

• Multiples of 2: 4, 6, and 8 → v = 3
• Pairs with GCD ≥ 2: 3 * 2 / 2 = 3
• Subtract: cnt_g[4] = 1, cnt_g[6] = 0, cnt_g[8] = 0
• cnt_g[2] = 3 - 1 = 2
• This represents pairs (4, 6) and (6, 8), both with GCD = 2

For i = 1:

• All numbers are multiples of 1 → v = 3
• Pairs: 3 * 2 / 2 = 3
• Subtract all other GCDs: 0 + 2 + 0 + 1 + 0 + 0 + 0 + 0 = 3
• cnt_g[1] = 3 - 3 = 0

Step 3: Build prefix sum array

• cnt_g = [0, 0, 2, 0, 1, 0, 0, 0, 0]
• Prefix sums: s = [0, 0, 2, 2, 3, 3, 3, 3, 3]

This means:

• Indices 0-1 in gcdPairs have GCD = 2 (2 pairs)
• Index 2 in gcdPairs has GCD = 4 (1 pair)

Step 4: Answer queries

• queries[0] = 0: bisect_right(s, 0) = 2 → answer is 2
• queries[1] = 1: bisect_right(s, 1) = 2 → answer is 2
• queries[2] = 2: bisect_right(s, 2) = 4 → answer is 4

Verification: The actual pairs and their GCDs:

• gcd(4, 6) = 2
• gcd(4, 8) = 4
• gcd(6, 8) = 2

Sorted gcdPairs = [2, 2, 4]

So queries[0] = 2, queries[1] = 2, queries[2] = 4 ✓

The algorithm efficiently computes these values without explicitly generating all pairs, using the counting technique to handle large inputs effectively.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n + M * log M + q * log M)

• Creating the Counter from nums takes O(n) time
• The nested loop iterates through all divisors:
  • The outer loop runs from mx down to 1, which is O(M) iterations
  • For each value i, the inner loop iterates through multiples of i up to mx, which takes O(M/i) time
  • The total time for all iterations is O(M/1 + M/2 + M/3 + ... + M/M) = O(M * log M) (harmonic series)
• Building the prefix sum array with accumulate takes O(M) time
• For each of the q queries, binary search on the prefix sum array takes O(log M) time
• Total: O(n + M * log M + q * log M) which simplifies to O(n + (M + q) * log M)

Space Complexity: O(M)

• The Counter object stores at most n elements but uses keys up to value M, effectively O(M) in worst case
• The cnt_g array has size M + 1, which is O(M)
• The prefix sum array s also has size M + 1, which is O(M)
• Total space complexity is O(M)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Order of Subtraction in Inclusion-Exclusion

The Pitfall: A common mistake is subtracting the over-counted pairs BEFORE calculating the total pairs with GCD being a multiple of i. This leads to incorrect results because you're subtracting from a value that hasn't been computed yet.

Incorrect Implementation:

# WRONG: Subtracting before calculating total pairs
for gcd in range(max_value, 0, -1):
    multiples_count = 0
  
    # Subtract first (WRONG!)
    for multiple in range(gcd, max_value + 1, gcd):
        gcd_pair_count[gcd] -= gcd_pair_count[multiple]
  
    # Then calculate pairs
    for multiple in range(gcd, max_value + 1, gcd):
        multiples_count += frequency_count[multiple]
  
    gcd_pair_count[gcd] += multiples_count * (multiples_count - 1) // 2

Correct Implementation:

# CORRECT: Calculate total pairs first, then subtract over-counted
for gcd in range(max_value, 0, -1):
    multiples_count = 0
  
    # First, count all multiples
    for multiple in range(gcd, max_value + 1, gcd):
        multiples_count += frequency_count[multiple]
  
    # Calculate total pairs with GCD being a multiple of current value
    gcd_pair_count[gcd] = multiples_count * (multiples_count - 1) // 2
  
    # Then subtract pairs with GCD that is a proper multiple (2*gcd, 3*gcd, etc.)
    for multiple in range(2 * gcd, max_value + 1, gcd):
        gcd_pair_count[gcd] -= gcd_pair_count[multiple]

2. Starting Subtraction from Wrong Multiple

The Pitfall: Another subtle error is starting the subtraction loop from gcd instead of 2 * gcd. This causes the algorithm to subtract gcd_pair_count[gcd] from itself, which zeros out the value incorrectly.

Incorrect Implementation:

# WRONG: Starting from gcd itself
for multiple in range(gcd, max_value + 1, gcd):  # Starts at gcd
    gcd_pair_count[gcd] -= gcd_pair_count[multiple]

Correct Implementation:

# CORRECT: Starting from 2*gcd to avoid self-subtraction
for multiple in range(2 * gcd, max_value + 1, gcd):  # Starts at 2*gcd
    gcd_pair_count[gcd] -= gcd_pair_count[multiple]

3. Off-by-One Error with Query Indices

The Pitfall: Using bisect_left instead of bisect_right (or vice versa) without proper adjustment can lead to incorrect query answers, especially at boundary cases.

Why it matters:

• bisect_right(s, q) returns the smallest index where all values to the left are ≤ q
• This directly gives us the GCD value at position q in our conceptual sorted array
• Using bisect_left would require additional logic to handle edge cases

Example to illustrate the difference: If prefix_sum = [0, 2, 5, 8] and query = 2:

• bisect_right([0, 2, 5, 8], 2) returns 2 (correct GCD value)
• bisect_left([0, 2, 5, 8], 2) returns 1 (would need adjustment)