Add Solution to LeetCode Problem 1547 (Hard)

dsa-solutions/lc-solutions/1500-1599/1547-Minimum-cost-to-cut-a-stick.md

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+---
+id: minimum-cost-to-cut-a-stick
+title: Minimum Cost to Cut a Stick
+sidebar_label: 1547 - Minimum Cost to Cut a Stick
+tags: [Dynamic Programming, Array, C++]
+description: Solve the problem of finding the minimum cost to cut a stick into smaller pieces at specified positions, using dynamic programming.
+---
+
+## Problem Statement
+
+### Problem Description
+
+Given a wooden stick of length `n` units. The stick is labeled from 0 to `n`. For example, a stick of length 6 is labeled as follows:
+```
+0 - 1 - 2 - 3 - 4 - 5 - 6
+```
+
+- Given an integer array `cuts` where `cuts[i]` denotes a position you should perform a cut at.
+
+- You should perform the cuts in order, but you can change the order of the cuts as you wish.
+
+- The cost of one cut is the length of the stick to be cut. The total cost is the sum of costs of all cuts. When you cut a stick, it will be split into two smaller sticks (i.e., the sum of their lengths is the length of the stick before the cut).
+
+- Return the minimum total cost of the cuts.
+
+### Example
+
+**Example 1:**
+```
+Input:
+n = 7, cuts = [1, 3, 4, 5]
+
+Output:
+16
+```
+
+**Explanation:**
+
+- Using cuts order = [1, 3, 4, 5] as in the input leads to the following scenario:
+
+- The first cut is done to a rod of length 7 so the cost is 7.
+
+- The second cut is done to a rod of length 6 (i.e., the second part of the first cut), the third is done to a rod of length 4 and the last cut is to a rod of length 3.
+
+- The total cost is 7 + 6 + 4 + 3 = 20.
+
+- Rearranging the cuts to be [3, 5, 1, 4] for example will lead to a scenario with total cost = 16 (as shown in the example photo 7 + 4 + 3 + 2 = 16).
+
+
+### Constraints
+
+- 2 <= `n` <= 10^6
+- 1 <= `cuts.length` <= min(`n` - 1, 100)
+- 1 <= `cuts[i]` <= `n` - 1
+- All the integers in the `cuts` array are distinct.
+
+## Solution
+
+### Intuition 
+
+- Sort the cuts in ascending order.
+- Use dynamic programming to minimize the cost of cuts.
+- Define `dp[i][j]` as the minimum cost to cut the stick between cuts[i-1] and cuts[j-1].
+- Use a bottom-up approach to solve the subproblems and combine them to get the final result.
+
+### Time Complexity and Space Complexity Analysis
+
+- **Initialization**:
+  - Sorting the cuts array takes $O(m log m)$ time, where m is the number of cuts.
+  - Initializing the `dp` array takes $O(m^3)$ time.
+  - Overall initialization time complexity is $O(m log m + m^2)$.
+
+- **DP Table Calculation**:
+  - Filling the `dp` table involves iterating over all possible subintervals and calculating the minimum cost for each subinterval using nested loops.
+  - This takes $O(m^3)$ time in the worst case.
+
+- **Overall Time Complexity**:
+  - The overall time complexity is $O(m^3)$.
+
+- **Space Complexity**:
+  - The `dp` table requires $O(m^2)$ space.
+  - The space complexity is $O(m^2)$.
+
+### Code
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int minCost(int n, std::vector<int>& cuts) {
+        std::sort(cuts.begin(), cuts.end());
+        int m = cuts.size();
+        std::vector<std::vector<int>> dp(m + 2, std::vector<int>(m + 2, 0));
+
+        for (int l = 2; l <= m + 1; l++) {
+            for (int i = 0; i + l <= m + 1; i++) {
+                int j = i + l;
+                dp[i][j] = INT_MAX;
+                for (int k = i + 1; k < j; k++) {
+                    dp[i][j] = std::min(dp[i][j], dp[i][k] + dp[k][j]);
+                }
+                int left = (i == 0) ? 0 : cuts[i - 1];
+                int right = (j == m + 1) ? n : cuts[j - 1];
+                dp[i][j] += right - left;
+            }
+        }
+
+        return dp[0][m + 1];
+    }
+};
+```