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1868. Product of Two Run-Length Encoded Arrays πŸ”’

Description

Run-length encoding is a compression algorithm that allows for an integer array nums with many segments of consecutive repeated numbers to be represented by a (generally smaller) 2D array encoded. Each encoded[i] = [vali, freqi] describes the ith segment of repeated numbers in nums where vali is the value that is repeated freqi times.

  • For example, nums = [1,1,1,2,2,2,2,2] is represented by the run-length encoded array encoded = [[1,3],[2,5]]. Another way to read this is "three 1's followed by five 2's".

The product of two run-length encoded arrays encoded1 and encoded2 can be calculated using the following steps:

  1. Expand both encoded1 and encoded2 into the full arrays nums1 and nums2 respectively.
  2. Create a new array prodNums of length nums1.length and set prodNums[i] = nums1[i] * nums2[i].
  3. Compress prodNums into a run-length encoded array and return it.

You are given two run-length encoded arrays encoded1 and encoded2 representing full arrays nums1 and nums2 respectively. Both nums1 and nums2 have the same length. Each encoded1[i] = [vali, freqi] describes the ith segment of nums1, and each encoded2[j] = [valj, freqj] describes the jth segment of nums2.

Return the product of encoded1 and encoded2.

Note: Compression should be done such that the run-length encoded array has the minimum possible length.

 

Example 1:

Input: encoded1 = [[1,3],[2,3]], encoded2 = [[6,3],[3,3]]
Output: [[6,6]]
Explanation: encoded1 expands to [1,1,1,2,2,2] and encoded2 expands to [6,6,6,3,3,3].
prodNums = [6,6,6,6,6,6], which is compressed into the run-length encoded array [[6,6]].

Example 2:

Input: encoded1 = [[1,3],[2,1],[3,2]], encoded2 = [[2,3],[3,3]]
Output: [[2,3],[6,1],[9,2]]
Explanation: encoded1 expands to [1,1,1,2,3,3] and encoded2 expands to [2,2,2,3,3,3].
prodNums = [2,2,2,6,9,9], which is compressed into the run-length encoded array [[2,3],[6,1],[9,2]].

 

Constraints:

  • 1 <= encoded1.length, encoded2.length <= 105
  • encoded1[i].length == 2
  • encoded2[j].length == 2
  • 1 <= vali, freqi <= 104 for each encoded1[i].
  • 1 <= valj, freqj <= 104 for each encoded2[j].
  • The full arrays that encoded1 and encoded2 represent are the same length.

Solutions

Solution 1

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class Solution:
    def findRLEArray(
        self, encoded1: List[List[int]], encoded2: List[List[int]]
    ) -> List[List[int]]:
        ans = []
        j = 0
        for vi, fi in encoded1:
            while fi:
                f = min(fi, encoded2[j][1])
                v = vi * encoded2[j][0]
                if ans and ans[-1][0] == v:
                    ans[-1][1] += f
                else:
                    ans.append([v, f])
                fi -= f
                encoded2[j][1] -= f
                if encoded2[j][1] == 0:
                    j += 1
        return ans

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class Solution {
    public List<List<Integer>> findRLEArray(int[][] encoded1, int[][] encoded2) {
        List<List<Integer>> ans = new ArrayList<>();
        int j = 0;
        for (var e : encoded1) {
            int vi = e[0], fi = e[1];
            while (fi > 0) {
                int f = Math.min(fi, encoded2[j][1]);
                int v = vi * encoded2[j][0];
                int m = ans.size();
                if (m > 0 && ans.get(m - 1).get(0) == v) {
                    ans.get(m - 1).set(1, ans.get(m - 1).get(1) + f);
                } else {
                    ans.add(new ArrayList<>(List.of(v, f)));
                }
                fi -= f;
                encoded2[j][1] -= f;
                if (encoded2[j][1] == 0) {
                    ++j;
                }
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<vector<int>> findRLEArray(vector<vector<int>>& encoded1, vector<vector<int>>& encoded2) {
        vector<vector<int>> ans;
        int j = 0;
        for (auto& e : encoded1) {
            int vi = e[0], fi = e[1];
            while (fi) {
                int f = min(fi, encoded2[j][1]);
                int v = vi * encoded2[j][0];
                if (!ans.empty() && ans.back()[0] == v) {
                    ans.back()[1] += f;
                } else {
                    ans.push_back({v, f});
                }
                fi -= f;
                encoded2[j][1] -= f;
                if (encoded2[j][1] == 0) {
                    ++j;
                }
            }
        }
        return ans;
    }
};

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func findRLEArray(encoded1 [][]int, encoded2 [][]int) (ans [][]int) {
    j := 0
    for _, e := range encoded1 {
        vi, fi := e[0], e[1]
        for fi > 0 {
            f := min(fi, encoded2[j][1])
            v := vi * encoded2[j][0]
            if len(ans) > 0 && ans[len(ans)-1][0] == v {
                ans[len(ans)-1][1] += f
            } else {
                ans = append(ans, []int{v, f})
            }
            fi -= f
            encoded2[j][1] -= f
            if encoded2[j][1] == 0 {
                j++
            }
        }
    }
    return
}

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