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📜  找到最小和,使得每三个连续元素中的一个被采用

📅  最后修改于: 2021-09-17 07:18:30             🧑  作者: Mango

给定一个由n 个非负数组成的数组,任务是找到元素的最小总和(从数组中选取),使得至少从数组中的每 3 个连续元素中选取一个元素。
例子 :

Input : arr[] = {1, 2, 3}
Output : 1

Input : arr[] = {1, 2, 3, 6, 7, 1}
Output : 4
We pick 3 and 1  (3 + 1 = 4)
Note that there are following subarrays
of three consecutive elements
{1, 2, 3}, {2, 3, 6}, {3, 6, 7} and {6, 7, 1}
We have picked one element from every subarray.

Input : arr[] = {1, 2, 3, 6, 7, 1, 8, 6, 2,
                 7, 7, 1}
Output : 7
The result is obtained as sum of 3 + 1 + 2 + 1

当 arr[i] 是解和的一部分(不一定是结果)并且是最后选择的元素时,让 sum(i) 是最小可能的和。那么我们的结果是 sum(n-1)、sum(n-2) 和 sum(n-3) 的最小值 [我们必须至少选择最后三个元素中的一个]。
我们可以递归地计算 sum(i) 作为 arr[i] 和 minimum(sum(i-1), sum(i-2), sum(i-3)) 的和。由于问题的递归结构中存在重叠的子问题,我们可以使用动态规划来解决这个问题。
下面是上述想法的实现。

C++
// A Dynamic Programming based C++ program to
// find minimum possible sum of elements of array
// such that an element out of every three
// consecutive is picked.
#include 
using namespace std;
 
// A utility function to find minimum of
// 3 elements
int minimum(int a, int b, int c)
{
    return min(min(a, b), c);
}
 
// Returns minimum possible sum of elements such
// that an element out of every three consecutive
// elements is picked.
int findMinSum(int arr[], int n)
{
    // Create a DP table to store results of
    // subproblems. sum[i] is going to store
    // minimum possible sum when arr[i] is
    // part of the solution.
    int sum[n];
 
    // When there are less than or equal to
    // 3 elements
    sum[0] = arr[0];
    sum[1] = arr[1];
    sum[2] = arr[2];
 
    // Iterate through all other elements
    for (int i=3; i


Java
// A Dynamic Programming based java program to
// find minimum possible sum of elements of array
// such that an element out of every three
// consecutive is picked.
import java.io.*;
 
class GFG
{
    // A utility function to find minimum of
    // 3 elements
    static int minimum(int a, int b, int c)
    {
        return Math. min(Math.min(a, b), c);
    }
     
    // Returns minimum possible sum of elements such
    // that an element out of every three consecutive
    // elements is picked.
    static int findMinSum(int arr[], int n)
    {
        // Create a DP table to store results of
        // subproblems. sum[i] is going to store
        // minimum possible sum when arr[i] is
        // part of the solution.
        int sum[] =new int[n];
     
        // When there are less than or equal to
        // 3 elements
        sum[0] = arr[0];
        sum[1] = arr[1];
        sum[2] = arr[2];
     
        // Iterate through all other elements
        for (int i = 3; i < n; i++)
        sum[i] = arr[i] + minimum(sum[i - 3],
                         sum[i - 2], sum[i - 1]);
     
        return minimum(sum[n - 1], sum[n - 2], sum[n - 3]);
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int arr[] = {1, 2, 3, 20, 2, 10, 1};
        int n = arr.length;
        System.out.println("Min Sum is " + findMinSum(arr, n));
             
    }
}
 
// This code is contributed by vt_m


Python3
# A Dynamic Programming based python 3 program to
# find minimum possible sum of elements of array
# such that an element out of every three
# consecutive is picked.
 
# A utility function to find minimum of
# 3 elements
def minimum(a, b, c):
    return min(min(a, b), c);
 
# Returns minimum possible sum of elements such
# that an element out of every three consecutive
# elements is picked.
def findMinSum(arr,n):
    # Create a DP table to store results of
    # subproblems. sum[i] is going to store
    # minimum possible sum when arr[i] is
    # part of the solution.
    sum = []
 
    # When there are less than or equal to
    # 3 elements
    sum.append(arr[0])
    sum.append(arr[1])
    sum.append(arr[2])
     
    # Iterate through all other elements
    for i in range(3, n):
        sum.append( arr[i] + minimum(sum[i-3],
                           sum[i-2], sum[i-1]))
 
    return minimum(sum[n-1], sum[n-2], sum[n-3])
 
# Driver code
arr = [1, 2, 3, 20, 2, 10, 1]
n = len(arr)
print( "Min Sum is ",findMinSum(arr, n))
 
# This code is contributed by Sam007


C#
// A Dynamic Programming based C# program to
// find minimum possible sum of elements of array
// such that an element out of every three
// consecutive is picked.
using System;
 
class GFG
{
    // A utility function to find minimum of
    // 3 elements
    static int minimum(int a, int b, int c)
    {
        return Math. Min(Math.Min(a, b), c);
    }
     
    // Returns minimum possible sum of elements such
    // that an element out of every three consecutive
    // elements is picked.
    static int findMinSum(int []arr, int n)
    {
        // Create a DP table to store results of
        // subproblems. sum[i] is going to store
        // minimum possible sum when arr[i] is
        // part of the solution.
        int []sum =new int[n];
     
        // When there are less than or equal to
        // 3 elements
        sum[0] = arr[0];
        sum[1] = arr[1];
        sum[2] = arr[2];
     
        // Iterate through all other elements
        for (int i = 3; i < n; i++)
        sum[i] = arr[i] + minimum(sum[i - 3],
                     sum[i - 2], sum[i - 1]);
     
        return minimum(sum[n - 1], sum[n - 2], sum[n - 3]);
    }
     
    // Driver code
    public static void Main ()
    {
        int []arr = {1, 2, 3, 20, 2, 10, 1};
        int n = arr.Length;
        Console.WriteLine("Min Sum is " + findMinSum(arr, n));
             
    }
}
 
//This code is contributed by Sam007


PHP


Javascript


输出:

Min Sum is 4

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