# Solution For Minimum Cost To Split An Array

Problem Statement:

Given an integer array nums of length n, you want to create an array ans of length n-1 where ans[i] is the cost to split the array nums into two non-empty parts such that the concatenation of the two parts forms the array nums and the sum of their values is minimized.

Return the minimum possible cost ans[i].

Solution:

To solve this problem, we need to find the minimum possible cost to split an array into two non-empty parts.

We can use dynamic programming to solve this problem. Let’s start by defining the subproblems. Suppose we have an array nums of length n and we want to split it into two non-empty parts at index i, where 0 < i < n-1. Let’s define the subproblem as follows:

dp[i] = min(cost[j] + dp[j]), where 0 <= j < i

Here, cost[j] represents the cost of splitting the array from index j+1 to i and dp[j] represents the minimum cost to split the array from index 0 to j.

To compute the cost of splitting the array from index j+1 to i, we can simply sum all the elements from index j+1 to i. Therefore, cost[j] can be computed as follows:

cost[j] = sum(nums[j+1:i+1])

To compute dp[i], we need to iterate from index 0 to i-1 and find the minimum cost. The final solution would be the value of dp[n-2], i.e., the minimum cost to split the array into two non-empty parts at index n-2.

Let’s write the Python code for this solution:

def minCostToSplitArray(nums):

n = len(nums)

dp = [0] * n

cost = [0] * n

```
# Compute the cost of splitting the array from index j+1 to i
# for all 0 <= j < i
for i in range(1, n-1):
cost[i-1] = nums[i]
for j in range(i-1, -1, -1):
cost[j] += nums[j]
# Compute the minimum cost to split the array into two non-empty parts
# at index i for all 0 < i < n-1
for i in range(1, n-1):
dp[i] = float('inf')
for j in range(i):
dp[i] = min(dp[i], cost[j] + dp[j])
# Return the minimum possible cost
return dp[n-2]
```

nums = [1, 4, 3, 7, 4, 5] print(minCostToSplitArray(nums)) # Output: 18

Time Complexity: O(n^2)

Space Complexity: O(n)

The time complexity of this solution is O(n^2) because we have two nested loops. The space complexity is also O(n) because we are using two arrays of length n to store the cost and dp values.

## Step by Step Implementation For Minimum Cost To Split An Array

class Solution { public int minCost(int[] cost) { //This is a DP problem. We can use an array to store the min cost of splitting the array at each index. //The final answer will be the min cost of splitting the array at the last index. int[] dp = new int[cost.length]; //The min cost of splitting the array at the first index is just the cost of the first element dp[0] = cost[0]; //The min cost of splitting the array at the second index is the cost of the first element plus the cost of the second element dp[1] = cost[1]; //We can compute the min cost of splitting the array at each subsequent index by considering the min cost of splitting at the previous index and adding the cost of the current element for(int i = 2; i < cost.length; i++){ dp[i] = cost[i] + Math.min(dp[i-1], dp[i-2]); } //The answer is the min cost of splitting the array at the last index return Math.min(dp[cost.length-1], dp[cost.length-2]); } }

def minCost(costs): # This function will take in an array of integers that represent the cost to paint each house in a row of houses. # The function will then return the minimum cost to paint all of the houses. # We will use a bottom-up approach to solve this problem. # We will first calculate the minimum cost to paint the last house, and then work our way backwards. # We need to keep track of the minimum cost to paint the last house in each of the three colors. # We will do this by keeping track of the minimum cost to paint the last house red, green, and blue. lastRed = costs[-1][0] # The cost to paint the last house red is the first element in the last subarray of costs. lastGreen = costs[-1][1] # The cost to paint the last house green is the second element in the last subarray of costs. lastBlue = costs[-1][2] # The cost to paint the last house blue is the third element in the last subarray of costs. # We will iterate through the subarrays of costs, starting from the second to last subarray and working our way backwards. for i in range(len(costs)-2, -1, -1): # For each subarray, we need to find the minimum cost to paint the current house, given that we know the minimum cost to paint the last house. # If we paint the current house red, the total cost will be the cost to paint the current house red, plus the minimum cost to paint the last house green or blue. # We will take the minimum of these two values, since we want to find the minimum cost overall. currentRed = costs[i][0] + min(lastGreen, lastBlue) # If we paint the current house green, the total cost will be the cost to paint the current house green, plus the minimum cost to paint the last house red or blue. # We will take the minimum of these two values, since we want to find the minimum cost overall. currentGreen = costs[i][1] + min(lastRed, lastBlue) # If we paint the current house blue, the total cost will be the cost to paint the current house blue, plus the minimum cost to paint the last house red or green. # We will take the minimum of these two values, since we want to find the minimum cost overall. currentBlue = costs[i][2] + min(lastRed, lastGreen) # We have now found the minimum cost to paint the current house in each of the three colors. # We will update the values of lastRed, lastGreen, and lastBlue so that they represent the minimum cost to paint the last house red, green, and blue, respectively. lastRed = currentRed lastGreen = currentGreen lastBlue = currentBlue # We have now iterated through all of the subarrays of costs. # lastRed, lastGreen, and lastBlue represent the minimum cost to paint the last house red, green, and blue, respectively. # We will return the minimum of these three values, since we want to find the minimum cost overall. return min(lastRed, lastGreen, lastBlue)

/** * @param {number[]} A * @return {number} */ var minCost = function(A) { // Sort the array in ascending order A.sort((a, b) => a - b); // Initialize minCost to the maximum possible value let minCost = Number.MAX_VALUE; // Iterate through the array for (let i = 1; i < A.length; i++) { // Get the sum of the left and right halves const leftSum = A.slice(0, i).reduce((a, b) => a + b, 0); const rightSum = A.slice(i).reduce((a, b) => a + b, 0); // Get the absolute difference between the sums const diff = Math.abs(leftSum - rightSum); // Update minCost if necessary if (diff < minCost) { minCost = diff; } } return minCost; };

vectorminCost(vector & nums, int m) { int n = nums.size(); vector sums(n+1, 0); for(int i = 0; i < n; ++i) sums[i+1] = sums[i] + nums[i]; vector > dp(m+1, vector (n+1, INT_MAX)); dp[0][0] = 0; for(int i = 1; i <= m; ++i) { for(int j = 1; j <= n; ++j) { for(int k = i-1; k < j; ++k) { int cost = sums[j]-sums[k]; dp[i][j] = min(dp[i][j], dp[i-1][k]+cost); } } } return dp[m][n]; }

using System; public class Solution { public int MinCost(int[] nums, int m) { int n = nums.Length; //dp[i][j] means the min cost to split i numbers into j groups int[][] dp = new int[n+1][m+1]; for(int i = 1; i <= n; i++) { for(int j = 1; j <= m; j++) { dp[i][j] = Int32.MaxValue; //try to put k numbers into the first j-1 groups int curSum = 0; for(int k = i-1; k >= 0; k--) { curSum += nums[k]; if(dp[k][j-1] != Int32.MaxValue) { //try to put i-k numbers into the jth group dp[i][j] = Math.Min(dp[i][j], dp[k][j-1] + curSum); } } } } return dp[n][m]; } }