Given an array arr[], the task is to fnd the sum of a subsequence which forms a perfect square. If there are multiple subsequences having sum equal to a perfect square, print the maximum sum.
Explanation:
Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Output: 36
Explanation:
Maximum possible sum which is a perfect square that can be obtained from the array is 36 obtained from the subsequence {1, 5, 6, 7, 8, 9}.
Input: arr[] = {9, 10}
Output: 9
Naive Approach: Generate all the possible subsequences of a given array and for each subsequence, check if its sum is a Perfect Square. If such a sum is found, update the maximum sum. Finally, print the maximum sum obtained.
Time Complexity: O(N * 2N)
Auxiliary Space: O(N)
Efficient Approach:
The above approach can be optimized by using Dynamic Programming approach.
Follow the steps given below:
- Calculate the sum of the array.
- Iterate k in the range [√sum, 0] and check if any subsequence exists with that sum k2. For every k, follow the steps below:
- Initialize boolean matrix subset[][] of dimensions N * sum, where sum denotes the current value of k2.
- subset[i][j] denotes whether a subsequence of size i with sum j exists or not.
- Initialize first column and first row as true and false respectively of subset[][].
- Run two nested loops, outer from i in the range [1, N] and inner j in the range [1, sum]to fill the subset[][] matrix in bottom up manner based on the following conditions:
- if (j < arr[i – 1]), then subset[i][j] = subset[i – 1][j]
- if (j >= arr[i – 1]), then subset[i][j] = subset[i – 1][j] || subset[i – 1][j – arr[i – 1]]
- Finally, return the subset[n][sum].
- The first k for which subset[n][k] is true, gives the required maximum possible subsequence sum k2.
Below is the implementation of the above approach:
C++
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std; bool isSubsetSum( int arr[], int n, int sum) { bool subset[n + 1][sum + 1]; // If sum is 0, then answer is true for ( int i = 0; i <= n; i++) subset[i][0] = true ; // If sum is not 0 and arr[] is empty, // then answer is false for ( int i = 1; i <= sum; i++) subset[0][i] = false ; // Fill the subset table in bottom up manner for ( int i = 1; i <= n; i++) { for ( int j = 1; j <= sum; j++) { if (j < arr[i - 1]) subset[i][j] = subset[i - 1][j]; if (j >= arr[i - 1]) subset[i][j] = subset[i - 1][j] || subset[i - 1][j - arr[i - 1]]; } } return subset[n][sum]; } // Function to find the sum int findSum( int * arr, int n) { int sum = 0; // Find sum of all values for ( int i = 0; i < n; i++) { sum += arr[i]; } int val = sqrt (sum); for ( int i = val; i >= 0; i--) { if (isSubsetSum(arr, n, i * i)) { // return the value; return i * i; } } return 0; } // Driver Code int main() { int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; int n = sizeof (arr) / sizeof (arr[0]); cout << findSum(arr, n) << endl; return 0; } |
Java
// Java program to implement // the above approach import java.util.*; class GFG{ static boolean isSubsetSum( int arr[], int n, int sum) { boolean [][] subset = new boolean [n + 1 ][sum + 1 ]; // If sum is 0, then answer is true for ( int i = 0 ; i <= n; i++) subset[i][ 0 ] = true ; // If sum is not 0 and arr[] is empty, // then answer is false for ( int i = 1 ; i <= sum; i++) subset[ 0 ][i] = false ; // Fill the subset table in bottom up manner for ( int i = 1 ; i <= n; i++) { for ( int j = 1 ; j <= sum; j++) { if (j < arr[i - 1 ]) subset[i][j] = subset[i - 1 ][j]; if (j >= arr[i - 1 ]) subset[i][j] = subset[i - 1 ][j] || subset[i - 1 ][j - arr[i - 1 ]]; } } return subset[n][sum]; } // Function to find the sum static int findSum( int [] arr, int n) { int sum = 0 ; // Find sum of all values for ( int i = 0 ; i < n; i++) { sum += arr[i]; } int val = ( int )Math.sqrt(sum); for ( int i = val; i >= 0 ; i--) { if (isSubsetSum(arr, n, i * i)) { // return the value; return i * i; } } return 0 ; } // Driver code public static void main(String[] args) { int arr[] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 }; int n = arr.length; System.out.println(findSum(arr, n)); } } // This code is contributed by offbeat |
Python3
# Python3 program to implement # the above approach import math def isSubsetSum(arr, n, sum ): subset = [ [ True for x in range ( sum + 1 ) ] for y in range (n + 1 ) ] # If sum is 0, then answer is true for i in range (n + 1 ): subset[i][ 0 ] = True # If sum is not 0 and arr[] is empty, # then answer is false for i in range ( 1 , sum + 1 ): subset[ 0 ][i] = False # Fill the subset table in bottom up manner for i in range ( 1 , n + 1 ): for j in range ( 1 , sum + 1 ): if (j < arr[i - 1 ]): subset[i][j] = subset[i - 1 ][j] if (j > = arr[i - 1 ]): subset[i][j] = (subset[i - 1 ][j] or subset[i - 1 ][j - arr[i - 1 ]]) return subset[n][ sum ] # Function to find the sum def findSum(arr, n): sum = 0 # Find sum of all values for i in range (n): sum + = arr[i] val = int (math.sqrt( sum )) for i in range (val, - 1 , - 1 ): if (isSubsetSum(arr, n, i * i)): # Return the value; return i * i return 0 # Driver Code if __name__ = = "__main__" : arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 ] n = len (arr) print (findSum(arr, n)) # This code is contributed by chitranayal |
C#
// C# program to implement // the above approach using System; class GFG{ static bool isSubsetSum( int []arr, int n, int sum) { bool [,] subset = new bool [n + 1, sum + 1]; // If sum is 0, then answer is true for ( int i = 0; i <= n; i++) subset[i, 0] = true ; // If sum is not 0 and []arr is empty, // then answer is false for ( int i = 1; i <= sum; i++) subset[0, i] = false ; // Fill the subset table in bottom up manner for ( int i = 1; i <= n; i++) { for ( int j = 1; j <= sum; j++) { if (j < arr[i - 1]) subset[i, j] = subset[i - 1, j]; if (j >= arr[i - 1]) subset[i, j] = subset[i - 1, j] || subset[i - 1, j - arr[i - 1]]; } } return subset[n, sum]; } // Function to find the sum static int findSum( int [] arr, int n) { int sum = 0; // Find sum of all values for ( int i = 0; i < n; i++) { sum += arr[i]; } int val = ( int )Math.Sqrt(sum); for ( int i = val; i >= 0; i--) { if (isSubsetSum(arr, n, i * i)) { // return the value; return i * i; } } return 0; } // Driver code public static void Main(String[] args) { int []arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; int n = arr.Length; Console.WriteLine(findSum(arr, n)); } } // This code is contributed by Rohit_ranjan |
36
Time Complexity: O(N * SUM)
Auxiliary Space: O(N * SUM)
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