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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.

 

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.

 

Constraints:

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

Solutions

Solution 1: Enumeration

Thinking

Pair strength depends only on the two values and their $\gcd$. For $n\le 2000$ there are about $2\times 10^6$ unordered pairs.

We enumerate $i<j$, compute $\frac{\textit{nums}[i]\cdot\textit{nums}[j]}{\gcd^2}$ by Euclidean algorithm, and keep the maximum.

$O(n^2\log M)$ already fits the limits, so grouping the array by common divisors is unnecessary.

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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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
  
  

• 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
  }
  
  

• 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;
  }
  
  

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