最小化对的总和，在删除时将数组分成 3 个子数组

给定一个大小为N的数组arr ，任务是找到具有最小和的元素对，在删除时将数组分解为原始数组的 3 个非空子数组。打印这对中元素的总和。

Input: arr[]: {4, 2, 1, 2, 4}
Output: 4
Explanation:
Minimum sum pairs are values (2, 1) and (1, 2) but selecting a pair with adjacent indices won’t break the array into 3 parts. Next minimum value pair is (2, 2), which divides the array into {4}, {1}, {4}. Hence the answer is 2 + 2 = 4.

Input: arr[] = {5, 2, 4, 6, 3, 7}
Output: 5
Explanation:
Among all the pairs which break the array into 3 subparts, pair with values (2, 3) gives minimum sum.

天真的方法：将数组分成三个子数组。需要删除的两个元素都应遵循以下条件：

• 无法选择任何端点（索引0和N-1 ）。
• 不能选择相邻的索引。

因此，找到所有可能的配对组合，并选择符合上述条件且总和最小的配对。

时间复杂度： O(N ² )
辅助空间： O(N)

有效方法：按照以下步骤有效解决此问题：

1. 创建一个数组 prefixMin，其中第 i 个元素表示数组 arr 中从第 0 到第 i 个索引的最小元素。
2. 创建一个变量 ans，用于存储最终答案并使用 INT_MAX 对其进行初始化。
3. 现在，运行 i=3 到 i=N-1 的循环（从 3 开始，因为不能选择第 0 个，这个循环是选择对的第二个元素，这意味着它甚至不能从 2 开始因为那么第一个元素的唯一剩余选项是 1 和 2，它们都不遵循给定条件）并且在每次迭代中将 ans 更改为 ans 和 arr[i]+prefixMin[i-2] 中的最小值.
4. 执行上述步骤后，打印 ans。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
 
#include 
using namespace std;
 
// Function to find minimum possible sum
// of pair which breaks the array into 3
// non-empty subarrays
int minSumPair(int arr[], int N)
{
 
    if (N < 5) {
        return -1;
    }
 
    int prefixMin[N];
    prefixMin[0] = arr[0];
 
    // prefixMin[i] contains minimum element till i
    for (int i = 1; i < N - 1; i++) {
        prefixMin[i] = min(arr[i], prefixMin[i - 1]);
    }
 
    int ans = INT_MAX;
 
    for (int i = 3; i < N - 1; i++) {
        ans = min(ans, arr[i] + prefixMin[i - 2]);
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 5, 2, 4, 6, 3, 7 };
    int N = sizeof(arr) / sizeof(int);
 
    cout << minSumPair(arr, N) << endl;
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG{
 
// Function to find minimum possible sum
// of pair which breaks the array into 3
// non-empty subarrays
static int minSumPair(int arr[], int N)
{
 
    if (N < 5) {
        return -1;
    }
 
    int []prefixMin = new int[N];
    prefixMin[0] = arr[0];
 
    // prefixMin[i] contains minimum element till i
    for (int i = 1; i < N - 1; i++) {
        prefixMin[i] = Math.min(arr[i], prefixMin[i - 1]);
    }
 
    int ans = Integer.MAX_VALUE;
 
    for (int i = 3; i < N - 1; i++) {
        ans = Math.min(ans, arr[i] + prefixMin[i - 2]);
    }
 
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    // Given array
    int arr[] = { 5, 2, 4, 6, 3, 7 };
    int N = arr.length;
 
    System.out.print(minSumPair(arr, N) +"\n");
 
}
}
 
// This code is contributed by Princi Singh

Python3

# Python 3 implementation of the above approach
import sys
 
# Function to find minimum possible sum
# of pair which breaks the array into 3
# non-empty subarrays
def minSumPair(arr, N):
    if(N < 5):
        return -1
 
    prefixMin = [0 for i in range(N)]
    prefixMin[0] = arr[0]
 
    # prefixMin[i] contains minimum element till i
    for i in range(1,N - 1,1):
        prefixMin[i] = min(arr[i], prefixMin[i - 1])
 
    ans = sys.maxsize
 
    for i in range(3,N - 1,1):
        ans = min(ans, arr[i] + prefixMin[i - 2])
 
    return ans
 
# Driver Code
if __name__ == '__main__':
   
    # Given array
    arr = [5, 2, 4, 6, 3, 7]
    N = len(arr)
    print(minSumPair(arr, N))
     
    # This code is contributed by ipg201107.

C#

// C# implementation of the above approach
using System;
 
public class GFG{
 
// Function to find minimum possible sum
// of pair which breaks the array into 3
// non-empty subarrays
static int minSumPair(int []arr, int N)
{
 
    if (N < 5) {
        return -1;
    }
 
    int []prefixMin = new int[N];
    prefixMin[0] = arr[0];
 
    // prefixMin[i] contains minimum element till i
    for (int i = 1; i < N - 1; i++) {
        prefixMin[i] = Math.Min(arr[i], prefixMin[i - 1]);
    }
 
    int ans = int.MaxValue;
 
    for (int i = 3; i < N - 1; i++) {
        ans = Math.Min(ans, arr[i] + prefixMin[i - 2]);
    }
 
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
   
    // Given array
    int []arr = { 5, 2, 4, 6, 3, 7 };
    int N = arr.Length;
 
    Console.Write(minSumPair(arr, N) +"\n");
 
}
}
 
// This code is contributed by shikhasingrajput

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to find minimum possible sum
        // of pair which breaks the array into 3
        // non-empty subarrays
        function minSumPair(arr, N) {
 
            if (N < 5) {
                return -1;
            }
 
            let prefixMin = new Array(N).fill(0);
            prefixMin[0] = arr[0];
 
            // prefixMin[i] contains minimum element till i
            for (let i = 1; i < N - 1; i++) {
                prefixMin[i] = Math.min(arr[i], prefixMin[i - 1]);
            }
 
            let ans = Number.MAX_VALUE;
 
            for (let i = 3; i < N - 1; i++) {
                ans = Math.min(ans, arr[i] + prefixMin[i - 2]);
            }
 
            return ans;
        }
 
        // Driver Code
 
        // Given array
        let arr = [5, 2, 4, 6, 3, 7];
        let N = arr.length;
 
        document.write(minSumPair(arr, N));
 
     // This code is contributed by Potta Lokesh
 
    

输出

5

时间复杂度： O(n)

空间复杂度： O(n)