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Minimum sum subsequence such that at least one of every four consecutive elements is picked

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  • Difficulty Level : Medium
  • Last Updated : 29 Oct, 2022
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Given an array arr[] of positive integers. The task is to find minimum sum subsequence from the array such that at least one value among all groups of four consecutive elements is picked. 

Examples :  

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

The idea is similar to LIS problem. We store minimum sum subsequence ending with every element of arr[]. We finally return minimum of last four values.

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]
OR 
///Idea is simple store values of first 4 numbers in four variables 
let arr[] = {5,6,7,8,9}
a=5;
b=6;
c=7;
d=8;
now Imagine if size is only five then what are  the possible sets we can have?
{5,6,7,8} and {6,7,8,9}
if u can simply analyse these both the first three numbers in set two are common and 
the first number in set 1 is  different and last number in set 2 is also different
now what is the minimum sum we can have 
 (min(6,7,8) ,(5+9) ) why minimum of (6,7,8) because they are common 
 why (5+9) because we have to choose one element each from both sets ;
 
 This is code implementation 
class Solution{
    public:
    int minSum(int arr[], int n){
        //Write your code here
        if(n<=4)
            return *min_element(arr, arr+n);
        int a, b, c, d, e, ans;
        a = arr[0];
        b = arr[1];
        c = arr[2];
        d = arr[3];
        for(int i=4; i<n; i++){
            e = arr[i] + min({a, b, c, d});
            a = b;
            b = c;
            c = d;
            d = e;
        }
        return min({a, b, c, d, e});

    }
};
copied from :chessnoobdj;

///

Below is the implementation of above idea.

C++




// C++ program to find minimum sum subsequence
// of an array such that one of every four
// consecutive elements is picked.
#include <iostream>
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




<script>
 
// 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
 
</script>

Output

4

Time Complexity: O(n)
Auxiliary Space: O(n)

Alternate Solution : 

First of all, think that we have only four elements then our result is at least four given elements. Now, in the case, if we have more than four elements then we must maintain an array sum[] where sum[i] includes the possible minimal sum up to i-th element and also i-th element must be a part of the solution. 

While computing sum[i], our base condition is that arr[i] must be a part of sum[i] and then we must have an element from last four elements. So, we can recursively compute sum[i] as sum of arr[i] and minimum (sum[i-1], sum[i-2], sum[i-3], sum[i-4]). Since there are overlapping subproblems in the recursive structure of our problem, we can use Dynamic Programming to solve this problem. And for the final result we must compute minimum of last four values of sum[] array as result must contain an element from last four elements. 

C++




// CPP program to calculate
// minimum possible sum  for given constraint
#include <bits/stdc++.h>
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

Javascript




<script>
 
// Javascript program to calculate
// minimum possible sum  for given constraint
 
// function to calculate min sum using dp
function minSum(ar, n)
{
    // if elements are less than or equal to 4
    if (n <= 4)
        return Math.min.apply(Math,ar);
    var i;
    // save start four element as it is
    var sum = Array(n).fill(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 = 4; i < n; i++){
        var temp = [];
        var it;
        for(it= i-4;it<i;it++)
          temp.push(sum[it]);
      //(*min_element(sum + i - 4, sum + i));
        sum[i] = ar[i] + Math.min.apply(Math,temp);
    }
 
    // Since one of the last 4 elements must be
    // present
    var temp1 = [];
    for(i=n-4;i<n;i++)
        temp1.push(sum[i]);
 
    return Math.min.apply(Math,temp1);
}
 
// driver program
 
    var ar = [2, 4, 1, 5, 2, 3, 6, 1, 2, 4];
    var n = ar.length;
    document.write("Minimum sum = "+minSum(ar, n));
 
</script>

Output

Minimum sum = 4

Time Complexity: O(n)
Auxiliary Space: O(n)

Thanks to Shivam Pradhan (anuj_charm) for providing this alternate solution.
This article is contributed by Roshni Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. 


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