Maximum subsequence sum from a given array which is a perfect square

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 * 2^N)
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 

Output:

Time Complexity: O(N * SUM)
Auxiliary Space: O(N * SUM)

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C++

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