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2862. Maximum Element-Sum of a Complete Subset of Indices

Description

You are given a 1-indexed array nums. Your task is to select a complete subset from nums where every pair of selected indices multiplied is a perfect square,. i. e. if you select ai and aj, i * j must be a perfect square.

Return the sum of the complete subset with the maximum sum.

 

Example 1:

Input: nums = [8,7,3,5,7,2,4,9]

Output: 16

Explanation:

We select elements at indices 2 and 8 and 2 * 8 is a perfect square.

Example 2:

Input: nums = [8,10,3,8,1,13,7,9,4]

Output: 20

Explanation:

We select elements at indices 1, 4, and 9. 1 * 4, 1 * 9, 4 * 9 are perfect squares.

 

Constraints:

  • 1 <= n == nums.length <= 104
  • 1 <= nums[i] <= 109

Solutions

Solution 1: Enumeration

We note that if a number can be expressed in the form of \(k \times j^2\), then all numbers of this form have the same \(k\).

Therefore, we can enumerate \(k\) in the range \([1,..n]\), and then start enumerating \(j\) from \(1\), each time adding the value of \(nums[k \times j^2 - 1]\) to \(t\), until \(k \times j^2 > n\). At this point, update the answer to \(ans = \max(ans, t)\).

Finally, return the answer \(ans\).

The time complexity is \(O(n)\), where \(n\) is the length of the array. The space complexity is \(O(1)\).

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class Solution:
    def maximumSum(self, nums: List[int]) -> int:
        n = len(nums)
        ans = 0
        for k in range(1, n + 1):
            t = 0
            j = 1
            while k * j * j <= n:
                t += nums[k * j * j - 1]
                j += 1
            ans = max(ans, t)
        return ans

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class Solution {
    public long maximumSum(List<Integer> nums) {
        long ans = 0;
        int n = nums.size();
        boolean[] used = new boolean[n + 1];
        int bound = (int) Math.floor(Math.sqrt(n));
        int[] squares = new int[bound + 1];
        for (int i = 1; i <= bound + 1; i++) {
            squares[i - 1] = i * i;
        }
        for (int i = 1; i <= n; i++) {
            long res = 0;
            int idx = 0;
            int curr = i * squares[idx];
            while (curr <= n) {
                res += nums.get(curr - 1);
                curr = i * squares[++idx];
            }
            ans = Math.max(ans, res);
        }
        return ans;
    }
}

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class Solution {
    public long maximumSum(List<Integer> nums) {
        long ans = 0;
        int n = nums.size();
        for (int k = 1; k <= n; ++k) {
            long t = 0;
            for (int j = 1; k * j * j <= n; ++j) {
                t += nums.get(k * j * j - 1);
            }
            ans = Math.max(ans, t);
        }
        return ans;
    }
}

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class Solution {
public:
    long long maximumSum(vector<int>& nums) {
        long long ans = 0;
        int n = nums.size();
        for (int k = 1; k <= n; ++k) {
            long long t = 0;
            for (int j = 1; k * j * j <= n; ++j) {
                t += nums[k * j * j - 1];
            }
            ans = max(ans, t);
        }
        return ans;
    }
};

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func maximumSum(nums []int) (ans int64) {
    n := len(nums)
    for k := 1; k <= n; k++ {
        var t int64
        for j := 1; k*j*j <= n; j++ {
            t += int64(nums[k*j*j-1])
        }
        ans = max(ans, t)
    }
    return
}

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function maximumSum(nums: number[]): number {
    let ans = 0;
    const n = nums.length;
    for (let k = 1; k <= n; ++k) {
        let t = 0;
        for (let j = 1; k * j * j <= n; ++j) {
            t += nums[k * j * j - 1];
        }
        ans = Math.max(ans, t);
    }
    return ans;
}

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