Given an array of n non-negative numbers, the task is to find the minimum sum of elements (picked from the array) such that at least one element is picked out of every 3 consecutive elements in the array.
Examples :
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
Let sum(i) be the minimum possible sum when arr[i] is part of a solution sum (not necessarily result) and is last picked element. Then our result is minimum of sum(n-1), sum(n-2) and sum(n-3) [We must pick at least one of the last three elements].
We can recursively compute sum(i) as sum of arr[i] and minimum(sum(i-1), sum(i-2), sum(i-3)). Since there are overlapping subproblems in recursive structure of problem, we can use Dynamic Programming to solve this problem.
Below is the implementation of above idea.
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 <iostream>using namespace std;// A utility function to find minimum of 3 elementsint 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 < 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 codeint main(){ int arr[] = { 1, 2, 3, 20, 2, 10, 1 }; int n = sizeof(arr) / sizeof(arr[0]); printf("Min Sum is %d", findMinSum(arr, n)); return 0;} |
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 <stdio.h>// Find minimum between two numbers.int min(int num1, int num2){ return (num1 > num2) ? num2 : num1;}// A utility function to find minimum of 3 elementsint 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 < 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 codeint main(){ int arr[] = { 1, 2, 3, 20, 2, 10, 1 }; int n = sizeof(arr) / sizeof(arr[0]); printf("Min Sum is %d", findMinSum(arr, n)); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
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 elementsdef 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 codearr = [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
<?php// A Dynamic Programming based // PHP program to find minimum // possible sum of elements of// array such that an element // out of every three consecutive// is picked.// function to find minimum of// 3 elementsfunction minimum($a, $b, $c){ return min(min($a, $b), $c);}// Returns minimum possible sum// of elements such that an// element out of every three// consecutive elements is picked.function findMinSum($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. $sum[$n] = 0; // When there are less // than or equal to // 3 elements $sum[0] = $arr[0]; $sum[1] = $arr[1]; $sum[2] = $arr[2]; // Iterate through all other elements for ($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 $arr = array(1, 2, 3, 20, 2, 10, 1); $n = sizeof($arr); echo "Min Sum is " , findMinSum($arr, $n); // This code is contributed by nitin mittal. ?> |
Javascript
<script>// A Dynamic Programming based Javascript 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 elementsfunction minimum(a, b, 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.function 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. var sum= Array(n).fill(0); // 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 (var 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 codevar arr = [1, 2, 3, 20, 2, 10, 1];var n = arr.length;document.write( "Min Sum is " + findMinSum(arr, n));</script> |
Output
Min Sum is 4
Time Complexity : O(n)
Auxiliary Space : O(n), since n extra space has been taken.
This problem and solution are contributed by Ayush Saluja. If you like neveropen 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 neveropen main page and help other Geeks.
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