最小总和子序列，以使每四个连续元素中的至少一个被选择

给定正整数的数组arr []。任务是从数组中找到最小总和子序列，以便在四个连续元素的所有组中选取至少一个值。

例子 ：

Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8}
Output: 6
6 is sum of output subsequence {1, 5}
Note that we have following subarrays of four
consecutive elements
{(1, 2, 3, 4), 
 (2, 3, 4, 5),
 (3, 4, 5, 6)
 (4, 5, 6, 7)
 (5, 6, 7, 8)}
Our subsequence {1, 5} has an element from
all above groups of four consecutive elements.
And this subsequence is minimum sum such
subsequence.

Input : arr[] = {1, 2, 3, 3, 4, 5, 6, 1}
Output : 4
The subsequence is {3, 1}. Here we consider
second three.

Input: arr[] = {1, 2, 3, 2, 1}
Output: 2
The subsequence can be {1, 1} or {2}

Input: arr[] = {6, 7, 8}
Output: 6

Input: arr[] = {6, 7}
Output: 6

这个想法类似于LIS问题。我们存储以arr []的每个元素结尾的最小和子序列。最后，我们返回最后四个值中的最小值。

dp[i] stores minimum sum subsequence (with at least
one of every four consecutive elements) of arr[0..i] 
such that arr[i] is part of the solution. Note that 
this may not be the best solution for subarray
arr[0..i].

We can recursively compute dp[i] using below formula
dp[i] = arr[i] + min(dp[i-1], dp[i-2], dp[i-3], dp[i-4])

Finally we return minimum of dp[n-1], dp[n-2], 
dp[n-4] and dp[n-3]

以下是上述想法的实现。

C++

// C++ program to find minimum sum subsequence
// of an array such that one of every four
// consecutive elements is picked.
#include 
using namespace std;
 
// Returns sum of minimum sum subsequence
// such that one of every four consecutive
// elements is picked from arr[].
int minSum(int arr[], int n)
{
    // dp[i] is going to store minimum sum
    // subsequence of arr[0..i] such that arr[i]
    // is part of the solution. Note that this
    // may not be the best solution for subarray
    // arr[0..i]
    int dp[n];
 
    // If there is single value, we get the
    // minimum sum equal to arr[0]
    if (n == 1)
        return arr[0];
 
    // If there are two values, we get the
    // minimum sum equal to the minimum of
    // two values
    if (n == 2)
        return min(arr[0], arr[1]);
 
    // If there are three values, return
    // minimum of the three elements of
    // array
    if (n == 3)
        return min(arr[0], min(arr[1], arr[2]));
 
    // If there are four values, return minimum
    // of the four elements of array
    if (n == 4)
        return min(min(arr[0], arr[1]),
                   min(arr[2], arr[3]));
 
    dp[0] = arr[0];
    dp[1] = arr[1];
    dp[2] = arr[2];
    dp[3] = arr[3];
 
    for (int i = 4; i < n; i++)
        dp[i] = arr[i] + min(min(dp[i - 1], dp[i - 2]),
                             min(dp[i - 3], dp[i - 4]));
 
    // Return the minimum of last 4 index
    return min(min(dp[n - 1], dp[n - 2]),
               min(dp[n - 4], dp[n - 3]));
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 3, 4, 5, 6, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << minSum(arr, n);
    return 0;
}

Java

// Java program to find minimum sum subsequence
// of an array such that one of every four
// consecutive elements is picked.
import java.io.*;
 
class GFG {
 
    // Returns sum of minimum sum subsequence
    // such that one of every four consecutive
    // elements is picked from arr[].
    static int minSum(int[] arr, int n)
    {
        // dp[i] is going to store minimum sum
        // subsequence of arr[0..i] such that arr[i]
        // is part of the solution. Note that this
        // may not be the best solution for subarray
        // arr[0..i]
        int[] dp = new int[n];
 
        // If there is single value, we get the
        // minimum sum equal to arr[0]
        if (n == 1)
            return arr[0];
 
        // If there are two values, we get the
        // minimum sum equal to the minimum of
        // two values
        if (n == 2)
            return Math.min(arr[0], arr[1]);
 
        // If there are three values, return
        // minimum of the three elements of
        // array
        if (n == 3)
        return Math.min(arr[0], Math.min(arr[1], arr[2]));
 
        // If there are four values, return minimum
        // of the four elements of array
        if (n == 4)
            return Math.min(Math.min(arr[0], arr[1]),
                            Math.min(arr[2], arr[3]));
 
        dp[0] = arr[0];
        dp[1] = arr[1];
        dp[2] = arr[2];
        dp[3] = arr[3];
 
        for (int i = 4; i < n; i++)
        dp[i] = arr[i] + Math.min(Math.min(dp[i - 1], dp[i - 2]),
                         Math.min(dp[i - 3], dp[i - 4]));
 
        // Return the minimum of last 4 index
        return Math.min(Math.min(dp[n - 1], dp[n - 2]),
                        Math.min(dp[n - 4], dp[n - 3]));
    }
 
    // Driver code
    static public void main(String[] args)
    {
        int[] arr = { 1, 2, 3, 3, 4, 5, 6, 1 };
        int n = arr.length;
        System.out.println(minSum(arr, n));
    }
}
 
// This Code is contributed by vt_m.

Python3

# Python 3 program to find minimum sum
# subsequence of an array such that one
# of every four consecutive elements is picked.
 
# Returns sum of minimum sum subsequence
# such that one of every four consecutive
# elements is picked from arr[].
def minSum(arr, n):
 
    # dp[i] is going to store minimum sum
    # subsequence of arr[0..i] such that
    # arr[i] is part of the solution. Note
    # that this may not be the best solution
    # for subarray arr[0..i]
    dp = [0] * n
 
    # If there is single value, we get
    # the minimum sum equal to arr[0]
    if (n == 1):
        return arr[0]
 
    # If there are two values, we get the
    # minimum sum equal to the minimum of
    # two values
    if (n == 2):
        return min(arr[0], arr[1])
 
    # If there are three values, return
    # minimum of the three elements of
    # array
    if (n == 3):
        return min(arr[0],
               min(arr[1], arr[2]))
 
    # If there are four values,
    # return minimum of the four
    # elements of array
    if (n == 4):
        return min(min(arr[0], arr[1]),
                   min(arr[2], arr[3]))
 
    dp[0] = arr[0]
    dp[1] = arr[1]
    dp[2] = arr[2]
    dp[3] = arr[3]
 
    for i in range( 4, n):
        dp[i] = arr[i] + min(min(dp[i - 1], dp[i - 2]),
                             min(dp[i - 3], dp[i - 4]))
 
    # Return the minimum of last 4 index
    return min(min(dp[n - 1], dp[n - 2]),
                min(dp[n - 4], dp[n - 3]))
 
# Driver code
if __name__ == "__main__":
     
    arr = [ 1, 2, 3, 3, 4, 5, 6, 1 ]
    n = len(arr)
    print(minSum(arr, n))
 
# This code is contributed by ita_c

C#

// C# program to find minimum sum subsequence
// of an array such that one of every four
// consecutive elements is picked.
using System;
 
class GFG {
 
    // Returns sum of minimum sum subsequence
    // such that one of every four consecutive
    // elements is picked from arr[].
    static int minSum(int[] arr, int n)
    {
        // dp[i] is going to store minimum sum
        // subsequence of arr[0..i] such that arr[i]
        // is part of the solution. Note that this
        // may not be the best solution for subarray
        // arr[0..i]
        int[] dp = new int[n];
 
        // If there is single value, we get the
        // minimum sum equal to arr[0]
        if (n == 1)
            return arr[0];
 
        // If there are two values, we get the
        // minimum sum equal to the minimum of
        // two values
        if (n == 2)
            return Math.Min(arr[0], arr[1]);
 
        // If there are three values, return
        // minimum of the three elements of
        // array
        if (n == 3)
            return Math.Min(arr[0], Math.Min(arr[1], arr[2]));
 
        // If there are four values, return minimum
        // of the four elements of array
        if (n == 4)
            return Math.Min(Math.Min(arr[0], arr[1]),
                            Math.Min(arr[2], arr[3]));
 
        dp[0] = arr[0];
        dp[1] = arr[1];
        dp[2] = arr[2];
        dp[3] = arr[3];
 
        for (int i = 4; i < n; i++)
            dp[i] = arr[i] + Math.Min(Math.Min(dp[i - 1], dp[i - 2]),
                             Math.Min(dp[i - 3], dp[i - 4]));
 
        // Return the minimum of last 4 index
        return Math.Min(Math.Min(dp[n - 1], dp[n - 2]),
                        Math.Min(dp[n - 4], dp[n - 3]));
    }
 
    // Driver code
    static public void Main()
    {
        int[] arr = { 1, 2, 3, 3, 4, 5, 6, 1 };
        int n = arr.Length;
        Console.WriteLine(minSum(arr, n));
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// JavaScript program to find minimum sum subsequence
// of an array such that one of every four
// consecutive elements is picked.
 
// Returns sum of minimum sum subsequence
// such that one of every four consecutive
// elements is picked from arr[].
function minSum(arr, n)
{
     
    // dp[i] is going to store minimum sum
    // subsequence of arr[0..i] such that arr[i]
    // is part of the solution. Note that this
    // may not be the best solution for subarray
    // arr[0..i]
    let dp = [];
 
    // If there is single value, we get the
    // minimum sum equal to arr[0]
    if (n == 1)
        return arr[0];
 
    // If there are two values, we get the
    // minimum sum equal to the minimum of
    // two values
    if (n == 2)
        return Math.min(arr[0], arr[1]);
 
    // If there are three values, return
    // minimum of the three elements of
    // array
    if (n == 3)
    return Math.min(arr[0],
           Math.min(arr[1], arr[2]));
 
    // If there are four values, return minimum
    // of the four elements of array
    if (n == 4)
        return Math.min(Math.min(arr[0], arr[1]),
                        Math.min(arr[2], arr[3]));
 
    dp[0] = arr[0];
    dp[1] = arr[1];
    dp[2] = arr[2];
    dp[3] = arr[3];
 
    for(let i = 4; i < n; i++)
        dp[i] = arr[i] +
        Math.min(Math.min(dp[i - 1], dp[i - 2]),
                 Math.min(dp[i - 3], dp[i - 4]));
 
    // Return the minimum of last 4 index
    return Math.min(Math.min(dp[n - 1], dp[n - 2]),
                    Math.min(dp[n - 4], dp[n - 3]));
}
 
// Driver Code
let arr = [ 1, 2, 3, 3, 4, 5, 6, 1 ];
let n = arr.length;
 
document.write(minSum(arr, n));
 
// This code is contributed by code_hunt
 

输出：

4

替代解决方案：
首先，认为我们只有四个要素，那么我们的结果至少是四个给定的要素。现在，如果我们有四个以上的元素，则必须维护一个数组sum []，其中sum [i]包括直到第i个元素的可能的最小和，并且第i个元素也必须是解决方案的一部分。
在计算sum [i]时，我们的基本条件是arr [i]必须是sum [i]的一部分，然后我们必须有一个来自最后四个元素的元素。因此，我们可以递归计算sum [i]作为arr [i]和最小值(sum [i-1]，sum [i-2]，sum [i-3]，sum [i-4])之和。由于我们的问题的递归结构中存在重叠的子问题，因此可以使用动态编程来解决此问题。对于最终结果，我们必须计算sum []数组的后四个值中的最小值，因为结果必须包含后四个元素中的一个。

C++

// CPP program to calculate
// minimum possible sum  for given constraint
#include 
typedef long long ll;
using namespace std;
 
// function to calculate min sum using dp
int minSum(int ar[], int n)
{
    // if elements are less than or equal to 4
    if (n <= 4)
        return *min_element(ar, ar + n);
 
    // save start four element as it is
    int sum[n];
    sum[0] = ar[0];
    sum[1] = ar[1];
    sum[2] = ar[2];
    sum[3] = ar[3];
 
    // compute sum[] for all rest elements
    for (int i = 4; i < n; i++)
        sum[i] = ar[i] + (*min_element(sum + i - 4, sum + i));
 
    // Since one of the last 4 elements must be
    // present
    return *min_element(sum + n - 4, sum + n);
}
 
// driver program
int main()
{
    int ar[] = { 2, 4, 1, 5, 2, 3, 6, 1, 2, 4 };
    int n = sizeof(ar) / sizeof(ar[0]);
    cout << "Minimum sum = " << minSum(ar, n);
    return 0;
}

Java

// Java program to calculate
// minimum possible sum for given constraint
import java.util.Arrays;
 
class GFG
{
 
// function to calculate min sum using dp
static int minSum(int ar[], int n)
{
    // if elements are less than or equal to 4
    if (n <= 4)
        return Arrays.stream(ar).min().getAsInt();
 
    // save start four element as it is
    int []sum = new int[n];
    sum[0] = ar[0];
    sum[1] = ar[1];
    sum[2] = ar[2];
    sum[3] = ar[3];
 
    // compute sum[] for all rest elements
    for (int i = 4; i < n; i++)
        //sum[i] = ar[i] + (*min_element(sum + i - 4, sum + i));
        sum[i] = ar[i] + Arrays.stream(Arrays.copyOfRange(
                           sum, i - 4, i)).min().getAsInt();
 
    // Since one of the last 4 elements must be
    // present
    return Arrays.stream(Arrays.copyOfRange(
            sum, n - 4, n)).min().getAsInt();
}
 
// Driver Code
public static void main(String[] args)
{
    int ar[] = { 2, 4, 1, 5, 2, 3, 6, 1, 2, 4 };
    int n = ar.length;
    System.out.println("Minimum sum = " + minSum(ar, n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to calculate
# minimum possible sum for given constraint
 
# function to calculate min sum using dp
def minSum(ar, n):
     
    # if elements are less than or equal to 4
    if (n <= 4):
        return min(ar)
 
    # save start four element as it is
    sum = [0 for i in range(n)]
    sum[0] = ar[0]
    sum[1] = ar[1]
    sum[2] = ar[2]
    sum[3] = ar[3]
 
    # compute sum[] for all rest elements
    for i in range(4, n):
        sum[i] = ar[i] + min(sum[i - 4:i])
 
    # Since one of the last 4 elements must be
    # present
    return min(sum[n - 4:n])
 
# Driver Code
ar = [2, 4, 1, 5, 2, 3, 6, 1, 2, 4]
n = len(ar)
print("Minimum sum = ", minSum(ar, n))
 
# This code is contributed by Mohit Kumar

C＃

// C# program to calculate
// minimum possible sum for given constraint
using System;
using System.Linq;
 
class GFG
{
 
// function to calculate min sum using dp
static int minSum(int []ar, int n)
{
    // if elements are less than or equal to 4
    if (n <= 4)
        return ar.Min();
 
    // save start four element as it is
    int []sum = new int[n];
    sum[0] = ar[0];
    sum[1] = ar[1];
    sum[2] = ar[2];
    sum[3] = ar[3];
    int []tempArr;
     
    // compute sum[] for all rest elements
    for (int i = 4; i < n; i++)
    {
        //sum[i] = ar[i] + (*min_element(sum + i - 4, sum + i));
        tempArr = new int[4];
        Array.Copy(sum, i - 4, tempArr, 0, 4);
        sum[i] = ar[i] + tempArr.Min();
    }
     
    // Since one of the last 4 elements must be
    // present
    tempArr = new int[4];
    Array.Copy(sum,n-4,tempArr,0,4);
    return tempArr.Min();
}
 
// Driver Code
public static void Main(String[] args)
{
    int []ar = { 2, 4, 1, 5, 2, 3, 6, 1, 2, 4 };
    int n = ar.Length;
    Console.WriteLine("Minimum sum = " +
                         minSum(ar, n));
}
}
 
// This code is contributed by Rajput-Ji

输出：

Minimum sum = 4

时间复杂度： O(n)。