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_mPython3
# 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 Sam007C#
// 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 Sam007PHP
Javascript
输出：
Min Sum is 4如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。