Q2. Maximum Total Subarray Value I
Status: Medium 🌴
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Problem: Q3. Split and Merge Array Transformation©leetcode🖖
Status: Medium 🧟‍♂️

Problem: Q3. ⚡️
Status: Medium 🏝
Code: Kotlin 🌚

class Solution {
    fun minSplitMerge(nums1: IntArray, nums2: IntArray): Int {
        val start = nums1.toList()
        val target = nums2.toList()
        if (start == target) return 0

        val visited = mutableSetOf<List<Int>>()
        val queue: ArrayDeque<Pair<List<Int>, Int>> = ArrayDeque()
        queue.add(start to 0)
        visited.add(start)

        while (queue.isNotEmpty()) {
            val (curr, steps) = queue.removeFirst()
            val n = curr.size

            for (l in 0 until n) {
                for (r in l until n) {
                    val sub = curr.subList(l, r + 1)
                    val remain = curr.subList(0, l) + curr.subList(r + 1, n)

                    for (pos in 0..remain.size) {
                        val next = remain.subList(0, pos) + sub + remain.subList(pos, remain.size)
                        if (next == target) return steps + 1
                        if (visited.add(next)) {
                            queue.add(next to steps + 1)
                        }
                    }
                }
            }
        }

        return -1
    }
}

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🫵 Problem: Q4
✅ Mode: Hard (6pt)
💡 Code: Cpp

#include <vector>
#include <queue>
#include <set>
#include <tuple>
#include <algorithm>

using namespace std;

class Solution {
public:
    long long maxTotalValue(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> log_table(n+1, 0);
        if (n >= 1) log_table[1] = 0;
        for (int i = 2; i <= n; i++) {
            log_table[i] = log_table[i/2] + 1;
        }
        int LOG = log_table[n] + 1;
        
        vector<vector<int>> st_max(n, vector<int>(LOG, 0));
        vector<vector<int>> st_min(n, vector<int>(LOG, 0));
        for (int i = 0; i < n; i++) {
            st_max[i][0] = nums[i];
            st_min[i][0] = nums[i];
        }
        for (int j = 1; j < LOG; j++) {
            int step = (1 << (j-1));
            for (int i = 0; i + (1 << j) <= n; i++) {
                st_max[i][j] = max(st_max[i][j-1], st_max[i+step][j-1]);
                st_min[i][j] = min(st_min[i][j-1], st_min[i+step][j-1]);
            }
        }
        
        auto query = [&](int l, int r) {
            int len = r - l + 1;
            int j = log_table[len];
            int max_val = max(st_max[l][j], st_max[r - (1 << j) + 1][j]);
            int min_val = min(st_min[l][j], st_min[r - (1 << j) + 1][j]);
            return (long long) max_val - min_val;
        };
        
        priority_queue<tuple<long long, int, int>> pq;
        set<pair<int, int>> visited;
        
        long long initial_range = query(0, n-1);
        pq.push({initial_range, 0, n-1});
        visited.insert({0, n-1});
        
        long long ans = 0;
        for (int i = 0; i < k; i++) {
            if (pq.empty()) break;
            auto [range, l, r] = pq.top();
            pq.pop();
            ans += range;
            if (l + 1 <= r) {
                pair<int, int> next1 = {l+1, r};
                if (visited.find(next1) == visited.end()) {
                    visited.insert(next1);
                    long long new_range = query(l+1, r);
                    pq.push({new_range, l+1, r});
                }
            }
            if (l <= r - 1) {
                pair<int, int> next2 = {l, r-1};
                if (visited.find(next2) == visited.end()) {
                    visited.insert(next2);
                    long long new_range = query(l, r-1);
                    pq.push({new_range, l, r-1});
                }
            }
        }
        
        return ans;
    } };

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📚 Weekly Contest 468 q1-q4 solutions

1⃣ Q1. 🗯
Language: Cpp | Difficulty: Easy🗽
🔗 Link

2⃣ Q2. Maximum Total Subarray Value I 🖖
Language: Kotlin | Difficulty: Medium) 👀🌴
🔗 Link

3⃣ Q3. Split and Merge Array Transformation ⚡️
Language: Kotlin | Difficulty: Medium) 🐇
🔗 Link

4️⃣ Q4 Maximum Total Subarray Value II 🎨
Language: Cpp | Difficulty: Hard 🧑‍🚀
🔗 Link
@leetcode @leetcode7 @leetcode8 #contest #leetcode #weekly468 #contest #leetcodecontest

Q1 😁
Mode: Easy @leetcode7 @weekly @biweekly @contest #biweekly