LeetCode 3411: Maximum Subarray With Equal Products Solution

Master LeetCode problem 3411 (Maximum Subarray With Equal Products), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3411. Maximum Subarray With Equal Products

Easy

Array Math Sliding Window Enumeration Number Theory

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible subarrays and check if each subarray is product equivalent. We can calculate the product, GCD, and LCM of each subarray and compare them. If they are equal, then the subarray is product equivalent. We can keep track of the maximum length of such subarrays and return the result.

Algorithm:

Initialize a variable maxLen to store the maximum length of product equivalent subarrays.
Iterate through all possible subarrays using two nested loops.
For each subarray, calculate the product, GCD, and LCM.
Compare if the product equals the product of GCD and LCM. If they are equal, update maxLen if the current subarray length is greater.
Return maxLen as the result.

Time Complexity:

The time complexity of this algorithm is O(n^3), where n is the length of the input array nums.

Space Complexity:

The space complexity of this algorithm is O(1) as we are not using any extra space apart from a few variables.

Solution Code

class Solution {
    public int maxSubarrayWithEqualProducts(int[] nums) {
        int n = nums.length;
        int maxLen = 0;
        
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                int product = 1, gcd = 0, lcm = 1;
                
                for (int k = i; k <= j; k++) {
                    product *= nums[k];
                    gcd = gcd(gcd, nums[k]);
                    lcm = lcm(lcm, nums[k]);
                }
                
                if (product == gcd * lcm) {
                    maxLen = Math.max(maxLen, j - i + 1);
                }
            }
        }
        
        return maxLen;
    }
    
    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
    
    private int lcm(int a, int b) {
        return a * b / gcd(a, b);
    }
}

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