Skip to content

4010. Maximize Pair Strength Using GCD

Description

You are given an integer array nums.

Choose exactly one pair of distinct indices i and j. The strength of the pair is defined as (nums[i] * nums[j]) / gcd(nums[i], nums[j])2.

Return the maximum strength over all possible pairs.

The term gcd(a, b) denotes the greatest common divisor of a and b.

Β 

Example 1:

Input: nums = [2,3,5]

Output: 15

Explanation:

Choosing i = 1 and j = 2 gives strength (3 * 5) / gcd(3, 5)2 = 15 / 1 = 15, which is the maximum over all pairs.

Example 2:

Input: nums = [4,6,8]

Output: 12

Explanation:

Choosing i = 1 and j = 2 gives strength (6 * 8) / gcd(6, 8)2 = 48 / 4 = 12, which is the maximum over all pairs.

Example 3:

Input: nums = [3,3]

Output: 1

Explanation:

Choosing i = 0 and j = 1 gives strength (3 * 3) / gcd(3, 3)2 = 9 / 9 = 1, the maximum over all pairs.

Β 

Constraints:

  • 2 <= nums.length <= 2000
  • 1 <= nums[i] <= 105

Solutions

Solution 1: Enumeration

We directly enumerate all pairs \((i, j)\) where \(i < j\), calculate the strength of each pair \(\frac{\textit{nums}[i] \times \textit{nums}[j]}{\gcd(\textit{nums}[i], \textit{nums}[j])^2}\), and take the maximum.

The greatest common divisor \(\gcd\) can be computed using the Euclidean algorithm.

The time complexity is \(O(n^2 \times \log M)\), where \(n\) is the length of the array \(\textit{nums}\) and \(M\) is the maximum value in the array. The space complexity is \(O(1)\).

1
2
3
4
5
6
7
8
9
class Solution:
    def maxPairStrength(self, nums: list[int]) -> int:
        n = len(nums)
        ans = 0
        for i in range(n):
            for j in range(i + 1, n):
                x = nums[i] * nums[j] // gcd(nums[i], nums[j]) ** 2
                ans = max(ans, x)
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
class Solution {
    public long maxPairStrength(int[] nums) {
        int n = nums.length;
        long ans = 0;

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                long g = gcd(nums[i], nums[j]);
                long x = (long) nums[i] * nums[j] / (g * g);
                ans = Math.max(ans, x);
            }
        }

        return ans;
    }

    private long gcd(long a, long b) {
        while (b != 0) {
            long t = a % b;
            a = b;
            b = t;
        }
        return a;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
class Solution {
public:
    long long maxPairStrength(vector<int>& nums) {
        int n = nums.size();
        long long ans = 0;

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                long long g = gcd(nums[i], nums[j]);
                long long x = 1LL * nums[i] * nums[j] / (g * g);
                ans = max(ans, x);
            }
        }

        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
func maxPairStrength(nums []int) int64 {
    n := len(nums)
    var ans int64 = 0

    for i := 0; i < n; i++ {
        for j := i + 1; j < n; j++ {
            g := gcd(int64(nums[i]), int64(nums[j]))
            x := int64(nums[i]) * int64(nums[j]) / (g * g)
            ans = max(ans, x)
        }
    }

    return ans
}

func gcd(a, b int64) int64 {
    for b != 0 {
        a, b = b, a%b
    }
    return a
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
function maxPairStrength(nums: number[]): number {
    const n = nums.length;
    let ans = 0;

    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            const g = gcd(nums[i], nums[j]);
            const x = Math.floor((nums[i] * nums[j]) / (g * g));
            ans = Math.max(ans, x);
        }
    }

    return ans;
}

function gcd(a: number, b: number): number {
    while (b !== 0) {
        const t = a % b;
        a = b;
        b = t;
    }
    return a;
}

Comments