Dada una secuencia de números positivos, encuentre la suma máxima que se puede formar que no tenga tres elementos consecutivos presentes.
Ejemplos:
Input: arr[] = {1, 2, 3}
Output: 5
We can't take three of them, so answer is
2 + 3 = 5
Input: arr[] = {3000, 2000, 1000, 3, 10}
Output: 5013
3000 + 2000 + 3 + 10 = 5013
Input: arr[] = {100, 1000, 100, 1000, 1}
Output: 2101
100 + 1000 + 1000 + 1 = 2101
Input: arr[] = {1, 1, 1, 1, 1}
Output: 4
Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8}
Output: 27
Este problema es principalmente una extensión del siguiente problema.
Suma máxima tal que no hay dos elementos adyacentes
Mantenemos una array auxiliar sum[] (del mismo tamaño que la array de entrada) para encontrar el resultado.
sum[i] : Stores result for subarray arr[0..i], i.e.,
maximum possible sum in subarray arr[0..i]
such that no three elements are consecutive.
sum[0] = arr[0]
// Note : All elements are positive
sum[1] = arr[0] + arr[1]
// We have three cases
// 1) Exclude arr[2], i.e., sum[2] = sum[1]
// 2) Exclude arr[1], i.e., sum[2] = sum[0] + arr[2]
// 3) Exclude arr[0], i.e., sum[2] = arr[1] + arr[2]
sum[2] = max(sum[1], arr[0] + arr[2], arr[1] + arr[2])
In general,
// We have three cases
// 1) Exclude arr[i], i.e., sum[i] = sum[i-1]
// 2) Exclude arr[i-1], i.e., sum[i] = sum[i-2] + arr[i]
// 3) Exclude arr[i-2], i.e., sum[i-3] + arr[i] + arr[i-1]
sum[i] = max(sum[i-1], sum[i-2] + arr[i],
sum[i-3] + arr[i] + arr[i-1])
A continuación se muestra la implementación de la idea anterior.
C++
// C++ program to find the maximum sum such that
// no three are consecutive
#include <bits/stdc++.h>
using namespace std;
// Returns maximum subsequence sum such that no three
// elements are consecutive
int maxSumWO3Consec(int arr[], int n)
{
// Stores result for subarray arr[0..i], i.e.,
// maximum possible sum in subarray arr[0..i]
// such that no three elements are consecutive.
int sum[n];
// Base cases (process first three elements)
if (n >= 1)
sum[0] = arr[0];
if (n >= 2)
sum[1] = arr[0] + arr[1];
if (n > 2)
sum[2] = max(sum[1], max(arr[1] +
arr[2], arr[0] + arr[2]));
// Process rest of the elements
// We have three cases
// 1) Exclude arr[i], i.e., sum[i] = sum[i-1]
// 2) Exclude arr[i-1], i.e., sum[i] = sum[i-2] + arr[i]
// 3) Exclude arr[i-2], i.e., sum[i-3] + arr[i] + arr[i-1]
for (int i = 3; i < n; i++)
sum[i] = max(max(sum[i - 1], sum[i - 2] + arr[i]),
arr[i] + arr[i - 1] + sum[i - 3]);
return sum[n - 1];
}
// Driver code
int main()
{
int arr[] = { 100, 1000 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxSumWO3Consec(arr, n);
return 0;
}
C
// C program to find the maximum sum such that
// no three are consecutive
#include <stdio.h>
// Find maximum between two numbers.
int max(int num1, int num2)
{
return (num1 > num2) ? num1 : num2;
}
// Returns maximum subsequence sum such that no three
// elements are consecutive
int maxSumWO3Consec(int arr[], int n)
{
// Stores result for subarray arr[0..i], i.e.,
// maximum possible sum in subarray arr[0..i]
// such that no three elements are consecutive.
int sum[n];
// Base cases (process first three elements)
if (n >= 1)
sum[0] = arr[0];
if (n >= 2)
sum[1] = arr[0] + arr[1];
if (n > 2)
sum[2] = max(sum[1],
max(arr[1] + arr[2], arr[0] + arr[2]));
// Process rest of the elements
// We have three cases
// 1) Exclude arr[i], i.e., sum[i] = sum[i-1]
// 2) Exclude arr[i-1], i.e., sum[i] = sum[i-2] + arr[i]
// 3) Exclude arr[i-2], i.e., sum[i-3] + arr[i] +
// arr[i-1]
for (int i = 3; i < n; i++)
sum[i] = max(max(sum[i - 1], sum[i - 2] + arr[i]),
arr[i] + arr[i - 1] + sum[i - 3]);
return sum[n - 1];
}
// Driver code
int main()
{
int arr[] = { 100, 1000 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d \n", maxSumWO3Consec(arr, n));
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
Java
// java program to find the maximum sum
// such that no three are consecutive
import java.io.*;
class GFG {
// Returns maximum subsequence sum such that no three
// elements are consecutive
static int maxSumWO3Consec(int arr[], int n)
{
// Stores result for subarray arr[0..i], i.e.,
// maximum possible sum in subarray arr[0..i]
// such that no three elements are consecutive.
int sum[] = new int[n];
// Base cases (process first three elements)
if (n >= 1)
sum[0] = arr[0];
if (n >= 2)
sum[1] = arr[0] + arr[1];
if (n > 2)
sum[2] = Math.max(sum[1], Math.max(arr[1] + arr[2], arr[0] + arr[2]));
// Process rest of the elements
// We have three cases
// 1) Exclude arr[i], i.e., sum[i] = sum[i-1]
// 2) Exclude arr[i-1], i.e., sum[i] = sum[i-2] + arr[i]
// 3) Exclude arr[i-2], i.e., sum[i-3] + arr[i] + arr[i-1]
for (int i = 3; i < n; i++)
sum[i] = Math.max(Math.max(sum[i - 1], sum[i - 2] + arr[i]),
arr[i] + arr[i - 1] + sum[i - 3]);
return sum[n - 1];
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 100, 1000, 100, 1000, 1 };
int n = arr.length;
System.out.println(maxSumWO3Consec(arr, n));
}
}
// This code is contributed by vt_m
Python3
# Python program to find the maximum sum such that # no three are consecutive # Returns maximum subsequence sum such that no three # elements are consecutive def maxSumWO3Consec(arr, n): # Stores result for subarray arr[0..i], i.e., # maximum possible sum in subarray arr[0..i] # such that no three elements are consecutive. sum = [0 for k in range(n)] # Base cases (process first three elements) if n >= 1 : sum[0] = arr[0] if n >= 2 : sum[1] = arr[0] + arr[1] if n > 2 : sum[2] = max(sum[1], max(arr[1] + arr[2], arr[0] + arr[2])) # Process rest of the elements # We have three cases # 1) Exclude arr[i], i.e., sum[i] = sum[i-1] # 2) Exclude arr[i-1], i.e., sum[i] = sum[i-2] + arr[i] # 3) Exclude arr[i-2], i.e., sum[i-3] + arr[i] + arr[i-1] for i in range(3, n): sum[i] = max(max(sum[i-1], sum[i-2] + arr[i]), arr[i] + arr[i-1] + sum[i-3]) return sum[n-1] # Driver code arr = [100, 1000, 100, 1000, 1] n = len(arr) print (maxSumWO3Consec(arr, n)) # This code is contributed by Afzal Ansari
C#
// C# program to find the maximum sum
// such that no three are consecutive
using System;
class GFG {
// Returns maximum subsequence
// sum such that no three
// elements are consecutive
static int maxSumWO3Consec(int[] arr,
int n)
{
// Stores result for subarray
// arr[0..i], i.e., maximum
// possible sum in subarray
// arr[0..i] such that no
// three elements are consecutive.
int[] sum = new int[n];
// Base cases (process
// first three elements)
if (n >= 1)
sum[0] = arr[0];
if (n >= 2)
sum[1] = arr[0] + arr[1];
if (n > 2)
sum[2] = Math.Max(sum[1], Math.Max(arr[1] + arr[2], arr[0] + arr[2]));
// Process rest of the elements
// We have three cases
// 1) Exclude arr[i], i.e.,
// sum[i] = sum[i-1]
// 2) Exclude arr[i-1], i.e.,
// sum[i] = sum[i-2] + arr[i]
// 3) Exclude arr[i-2], i.e.,
// sum[i-3] + arr[i] + arr[i-1]
for (int i = 3; i < n; i++)
sum[i] = Math.Max(Math.Max(sum[i - 1],
sum[i - 2] + arr[i]),
arr[i] + arr[i - 1] + sum[i - 3]);
return sum[n - 1];
}
// Driver code
public static void Main()
{
int[] arr = { 100, 1000, 100, 1000, 1 };
int n = arr.Length;
Console.Write(maxSumWO3Consec(arr, n));
}
}
// This code is contributed by nitin mittal.
PHP
<?php
// PHP program to find the maximum
// sum such that no three are consecutive
// Returns maximum subsequence
// sum such that no three
// elements are consecutive
function maxSumWO3Consec($arr, $n)
{
// Stores result for subarray
// arr[0..i], i.e., maximum
// possible sum in subarray
// arr[0..i] such that no three
// elements are consecutive$.
$sum = array();
// Base cases (process
// first three elements)
if ( $n >= 1)
$sum[0] = $arr[0];
if ($n >= 2)
$sum[1] = $arr[0] + $arr[1];
if ( $n > 2)
$sum[2] = max($sum[1], max($arr[1] + $arr[2],
$arr[0] + $arr[2]));
// Process rest of the elements
// We have three cases
// 1) Exclude arr[i], i.e.,
// sum[i] = sum[i-1]
// 2) Exclude arr[i-1], i.e.,
// sum[i] = sum[i-2] + arr[i]
// 3) Exclude arr[i-2], i.e.,
// sum[i-3] + arr[i] + arr[i-1]
for ($i = 3; $i < $n; $i++)
$sum[$i] = max(max($sum[$i - 1],
$sum[$i - 2] + $arr[$i]),
$arr[$i] + $arr[$i - 1] +
$sum[$i - 3]);
return $sum[$n-1];
}
// Driver code
$arr = array(100, 1000, 100, 1000, 1);
$n =count($arr);
echo maxSumWO3Consec($arr, $n);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// JavaScript program to find the maximum sum
// such that no three are consecutive
// Returns maximum subsequence sum such that no three
// elements are consecutive
function maxSumWO3Consec(arr, n)
{
// Stores result for subarray arr[0..i], i.e.,
// maximum possible sum in subarray arr[0..i]
// such that no three elements are consecutive.
let sum = [];
// Base cases (process first three elements)
if (n >= 1)
sum[0] = arr[0];
if (n >= 2)
sum[1] = arr[0] + arr[1];
if (n > 2)
sum[2] = Math.max(sum[1], Math.max(arr[1] + arr[2], arr[0] + arr[2]));
// Process rest of the elements
// We have three cases
// 1) Exclude arr[i], i.e., sum[i] = sum[i-1]
// 2) Exclude arr[i-1], i.e., sum[i] = sum[i-2] + arr[i]
// 3) Exclude arr[i-2], i.e., sum[i-3] + arr[i] + arr[i-1]
for (let i = 3; i < n; i++)
sum[i] = Math.max(Math.max(sum[i - 1], sum[i - 2] + arr[i]),
arr[i] + arr[i - 1] + sum[i - 3]);
return sum[n - 1];
}
// Driver Code
let arr = [ 100, 1000, 100, 1000, 1 ];
let n = arr.length;
document.write(maxSumWO3Consec(arr, n));
// This code is contributed by chinmoy1997pal.
</script>
Producción:
1100
Complejidad de tiempo: O(n)
Espacio auxiliar: O(n)
Otro enfoque: (usando la recursividad)
C++
// C++ program to find the maximum sum such that
// no three are consecutive using recursion.
#include<bits/stdc++.h>
using namespace std;
// Returns maximum subsequence sum such that no three
// elements are consecutive
int maxSum(int arr[], int i, vector<int> &dp)
{
// base case
if (i < 0)
return 0;
// this condition check is necessary to avoid segmentation fault at line 21
if (i == 0)
return arr[i];
// returning maxSum for already processed indexes of array
if (dp[i] != -1)
return dp[i];
// including current element and the next consecutive element in subsequence
int a = arr[i] + arr[i - 1] + maxSum(arr, i - 3, dp);
// not including the current element in subsequence
int b = maxSum(arr, i - 1, dp);
// including current element but skipping next consecutive element
int c = arr[i] + maxSum(arr, i - 2, dp);
// returning the max of above 3 cases
return dp[i] = max(a, max(b, c));
}
// Driver code
int main()
{
int arr[] = {100, 1000, 100, 1000, 1};
int n = sizeof(arr) / sizeof(arr[0]);
vector<int> dp(n, -1); // declaring and initializing dp vector
cout << maxSum(arr, n - 1, dp) << endl;
return 0;
}
// This code is contributed by Ashish Kumar Yadav
Java
// Java program to find the maximum
// sum such that no three are
// consecutive using recursion.
import java.util.Arrays;
class GFG
{
static int arr[] = {100, 1000, 100, 1000, 1};
static int sum[] = new int[10000];
// Returns maximum subsequence
// sum such that no three
// elements are consecutive
static int maxSumWO3Consec(int n)
{
if(sum[n] != -1)
return sum[n];
//Base cases (process first three elements)
if(n == 0)
return sum[n] = 0;
if(n == 1)
return sum[n] = arr[0];
if(n == 2)
return sum[n] = arr[1] + arr[0];
// Process rest of the elements
// We have three cases
return sum[n] = Math.max(Math.max(maxSumWO3Consec(n - 1),
maxSumWO3Consec(n - 2) + arr[n]),
arr[n] + arr[n - 1] + maxSumWO3Consec(n - 3));
}
// Driver code
public static void main(String[] args)
{
int n = arr.length;
Arrays.fill(sum, -1);
System.out.println(maxSumWO3Consec(n));
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 program to find the maximum # sum such that no three are consecutive # using recursion. arr = [100, 1000, 100, 1000, 1] sum = [-1] * 10000 # Returns maximum subsequence sum such # that no three elements are consecutive def maxSumWO3Consec(n) : if(sum[n] != -1): return sum[n] # 3 Base cases (process first # three elements) if(n == 0) : sum[n] = 0 return sum[n] if(n == 1) : sum[n] = arr[0] return sum[n] if(n == 2) : sum[n] = arr[1] + arr[0] return sum[n] # Process rest of the elements # We have three cases sum[n] = max(max(maxSumWO3Consec(n - 1), maxSumWO3Consec(n - 2) + arr[n-1]), arr[n-1] + arr[n - 2] + maxSumWO3Consec(n - 3)) return sum[n] # Driver code if __name__ == "__main__" : n = len(arr) print(maxSumWO3Consec(n)) # This code is contributed by Ryuga
C#
// C# program to find the maximum
// sum such that no three are
// consecutive using recursion.
using System;
class GFG
{
static int []arr = {100, 1000,
100, 1000, 1};
static int []sum = new int[10000];
// Returns maximum subsequence
// sum such that no three
// elements are consecutive
static int maxSumWO3Consec(int n)
{
if(sum[n] != -1)
return sum[n];
//Base cases (process first
// three elements)
if(n == 0)
return sum[n] = 0;
if(n == 1)
return sum[n] = arr[0];
if(n == 2)
return sum[n] = arr[1] + arr[0];
// Process rest of the elements
// We have three cases
return sum[n] = Math.Max(Math.Max(maxSumWO3Consec(n - 1),
maxSumWO3Consec(n - 2) + arr[n]),
arr[n] + arr[n - 1] + maxSumWO3Consec(n - 3));
}
// Driver code
public static void Main(String[] args)
{
int n = arr.Length;
for(int i = 0; i < sum.Length; i++)
sum[i] = -1;
Console.WriteLine(maxSumWO3Consec(n));
}
}
// This code is contributed by 29AjayKumar
PHP
<?php
// PHP program to find the maximum sum such that
// no three are consecutive using recursion.
$arr = array(100, 1000, 100, 1000, 1);
$sum = array_fill(0, count($arr) + 1, -1);
// Returns maximum subsequence sum such that
// no three elements are consecutive
function maxSumWO3Consec($n)
{
global $sum,$arr;
if($sum[$n] != -1)
return $sum[$n];
// Base cases (process first three elements)
if($n == 0)
return $sum[$n] = 0;
if($n == 1)
return $sum[$n] = $arr[0];
if($n == 2)
return $sum[$n] = $arr[1] + $arr[0];
// Process rest of the elements
// We have three cases
return $sum[$n] = max(max(maxSumWO3Consec($n - 1),
maxSumWO3Consec($n - 2) + $arr[$n]),
$arr[$n] + $arr[$n - 1] +
maxSumWO3Consec($n - 3));
}
// Driver code
$n = count($arr);
echo maxSumWO3Consec($n);
// This code is contributed by mits
?>
Javascript
<script>
// JavaScript program to find the maximum
// sum such that no three are
// consecutive using recursion.
let arr = [100, 1000, 100, 1000, 1];
let sum = new Array(10000);
for(let i = 0; i < 10000; i++)
{
sum[i] = -1;
}
// Returns maximum subsequence
// sum such that no three
// elements are consecutive
function maxSumWO3Consec(n)
{
if(sum[n] != -1)
{
return sum[n];
}
//Base cases (process first three elements)
if(n == 0)
{
return sum[n] = 0;
}
if(n == 1)
{
return sum[n] = arr[0];
}
if(n == 2)
{
return sum[n] = arr[1] + arr[0];
}
// Process rest of the elements
// We have three cases
return sum[n] =
500+Math.max(Math.max(maxSumWO3Consec(n - 1),
maxSumWO3Consec(n - 2) + arr[n]),
arr[n] + arr[n - 1] + maxSumWO3Consec(n - 3));
}
let n = arr.length - 1;
document.write(maxSumWO3Consec(n) + 1);
</script>
Producción:
2101
Complejidad de tiempo: O(n)
Espacio auxiliar: O(n)
Este artículo es una contribución de Roshni Agarwal . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA