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 * 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; } |
chevron_right
filter_none
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 |
chevron_right
filter_none
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 |
chevron_right
filter_none
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 |
chevron_right
filter_none
Output:
36
Time Complexity: O(N * SUM)
Auxiliary Space: O(N * SUM)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
