Maximum subsequence sum such that no three are consecutive
Given a sequence of positive numbers, find the maximum sum that can be formed which has no three consecutive elements present.
Examples :
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
This problem is mainly an extension of below problem.
Maximum sum such that no two elements are adjacent
We maintain an auxiliary array sum[] (of same size as input array) to find the result.
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])
Below is implementation of above idea.
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. |
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> |
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. ?> |
Output:
1100
Time Complexity: O(n)
Auxiliary Space: O(n)
Another approach: (Using recursion)
Java
// 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(arr[n]+maxSumWO3Consec(n - 1 ), maxSumWO3Consec(n - 3 ) + arr[n]+arr[n- 1 ]); } // Driver code public static void main(String[] args) { int n = arr.length; Arrays.fill(sum, - 1 ); System.out.println(maxSumWO3Consec(n- 1 )); } } // 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 |
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> |
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 ?> |
Output:
2101
Time Complexity: O(n)
Auxiliary Space: O(n)