LeetCode 1868: Product of Two Run-Length Encoded Arrays Solution

Master LeetCode problem 1868 (Product of Two Run-Length Encoded Arrays), a medium 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.

1868. Product of Two Run-Length Encoded Arrays

Medium

Array Two Pointers

Problem Explanation

Explanation:

In this problem, we are given two run-length encoded arrays nums1 and nums2. Each element of the array is a pair (val, freq) where val represents the value and freq represents the frequency of that value. The task is to find the product of each pair of elements from nums1 and nums2 and return the result in a new run-length encoded array.

To solve this problem, we can first decode the run-length encoded arrays into regular arrays. Then, iterate over each pair of elements from the decoded arrays, multiply the values, and update the frequency accordingly. Finally, encode the product array back into a run-length encoded array.

Algorithm:

Decode both run-length encoded arrays nums1 and nums2 into regular arrays.
Initialize an empty product array to store the results.
Iterate over each pair of elements from the decoded arrays, multiply the values, and update the frequency.
Encode the product array into a run-length encoded array.

Time Complexity: O(max(n, m)) where n and m are the lengths of the two input arrays. Space Complexity: O(max(n, m)) for the product array.

Solution Code

class Solution {
    public int[][] findRLEArray(int[][] nums1, int[][] nums2) {
        List<int[]> product = new ArrayList<>();
        
        int i = 0, j = 0;
        while (i < nums1.length && j < nums2.length) {
            int val = nums1[i][0] * nums2[j][0];
            int freq = Math.min(nums1[i][1], nums2[j][1]);
            product.add(new int[]{val, freq});
            
            if (nums1[i][1] > nums2[j][1]) {
                nums1[i][1] -= nums2[j][1];
                j++;
            } else if (nums1[i][1] < nums2[j][1]) {
                nums2[j][1] -= nums1[i][1];
                i++;
            } else {
                i++;
                j++;
            }
        }
        
        return product.toArray(new int[0][]);
    }
}

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