Given a positive integer N, the task is to print all possible sum of perfect squares such that the total sum of all perfect squares is equal to N.
Example:
Input: N=8
Output: 4 4
1 1 1 1 4
1 1 1 1 1 1 1 1
Explanation: There are three possible ways to make sum equal to NInput: N=3
Output: 1 1 1
Approach: The given problem can be solved using recursion and backtracking. The idea is to generate a list of perfect squares less than N, then calculate the possible combinations of perfect squares sum which are equal to N. In the recursive function, at every index either the element can be chosen for the list or not be chosen. Follow the steps below to solve the problem:
- Initialize an ArrayList to store all perfect squares less than N
- Iterate the integer N from 1 to square root of N using variable i
- If the square of i is less than or equal to N, then add i * i into the list
- Use recursion and backtracking to calculate sum of perfect squares that are equal to N
- At every index ind do the following:
- Do not include element at the current index and make a recursive call to the next index
- Include the element at current index and make a recursive call on the same index
- Following two base cases will be needed for the recursive function:
- If N becomes less than zero, or if ind reached the end of list then return
- If N becomes equal to zero then print the list
Below is the implementation of the above approach:
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Function to print all combinations of// sum of perfect squares equal to Nvoid combinationSum( vector<int> list, int N, int ind, vector<int> perfSquares){ // Sum of perfect squares exceeded N if (N < 0 || ind == list.size()) return; // Sum of perfect squares is equal to N // therefore a combination is found if (N == 0) { for (int i : perfSquares) { cout << i << " "; } cout << endl; return; } // Do not include the current element combinationSum(list, N, ind + 1, perfSquares); // Include the element at current index perfSquares.push_back(list[ind]); combinationSum(list, N - list[ind], ind, perfSquares); // Remove the current element perfSquares.pop_back();}// Function to check whether the// number is a perfect square or notvoid sumOfPerfectSquares(int N){ // Initialize an arraylist to store // all perfect squares less than N vector<int> list; int sqrtN = (int)(sqrt(N)); // Iterate till square root of N for (int i = 1; i <= sqrtN; i++) { // Add all perfect squares // to the list list.push_back(i * i); } vector<int> perfSquares; combinationSum(list, N, 0, perfSquares);}// Driver codeint main(){ int N = 8; // Call the function sumOfPerfectSquares(N); return 0;}// This code is contributed by Potta Lokesh |
Java
// Java implementation for the above approachimport java.io.*;import java.util.*;import java.lang.Math;class GFG { // Function to check whether the // number is a perfect square or not public static void sumOfPerfectSquares(int N) { // Initialize an arraylist to store // all perfect squares less than N ArrayList<Integer> list = new ArrayList<>(); int sqrtN = (int)(Math.sqrt(N)); // Iterate till square root of N for (int i = 1; i <= sqrtN; i++) { // Add all perfect squares // to the list list.add(i * i); } combinationSum(list, N, 0, new ArrayList<>()); } // Function to print all combinations of // sum of perfect squares equal to N static void combinationSum( List<Integer> list, int N, int ind, List<Integer> perfSquares) { // Sum of perfect squares exceeded N if (N < 0 || ind == list.size()) return; // Sum of perfect squares is equal to N // therefore a combination is found if (N == 0) { for (int i : perfSquares) { System.out.print(i + " "); } System.out.println(); return; } // Do not include the current element combinationSum(list, N, ind + 1, perfSquares); // Include the element at current index perfSquares.add(list.get(ind)); combinationSum(list, N - list.get(ind), ind, perfSquares); // Remove the current element perfSquares.remove(perfSquares.size() - 1); } // Driver code public static void main(String[] args) { int N = 8; // Call the function sumOfPerfectSquares(N); }} |
Python3
# Python code for the above approach # Function to print all combinations of# sum of perfect squares equal to Nimport mathdef combinationSum(lst, N, ind, perfSquares): # Sum of perfect squares exceeded N if N < 0 or ind == len(lst): return # Sum of perfect squares is equal to N # therefore a combination is found if N == 0: for i in perfSquares: print(i, end = ' ') print('') return; # Do not include the current element combinationSum(lst,N,ind+1,perfSquares) # Include the element at current index perfSquares.append(lst[ind]) combinationSum(lst, N - lst[ind], ind, perfSquares) # Remove the current element perfSquares.pop() # Function to check whether the# number is a perfect square or not def sumOfPerfectSquares(N): # Initialize an arraylist to store # all perfect squares less than N lst = [] sqrtN = int(math.sqrt(N)) # Iterate till square root of N for i in range(1, sqrtN + 1): # Add all perfect squares # to the list lst.append(i*i) perfSquares = [] combinationSum(lst, N, 0, perfSquares)# Driver code N = 8# Call the functionsumOfPerfectSquares(N)# This code is contributed by rdtank. |
C#
// C# implementation for the above approachusing System;using System.Collections.Generic;class GFG { // Function to check whether the // number is a perfect square or not public static void sumOfPerfectSquares(int N) { // Initialize an arraylist to store // all perfect squares less than N List<int> list = new List<int>(); int sqrtN = (int)(Math.Sqrt(N)); // Iterate till square root of N for (int i = 1; i <= sqrtN; i++) { // Add all perfect squares // to the list list.Add(i * i); } combinationSum(list, N, 0, new List<int>()); } // Function to print all combinations of // sum of perfect squares equal to N static void combinationSum(List<int> list, int N, int ind, List<int> perfSquares) { // Sum of perfect squares exceeded N if (N < 0 || ind == list.Count) return; // Sum of perfect squares is equal to N // therefore a combination is found if (N == 0) { foreach(int i in perfSquares) { Console.Write(i + " "); } Console.WriteLine(); return; } // Do not include the current element combinationSum(list, N, ind + 1, perfSquares); // Include the element at current index perfSquares.Add(list[ind]); combinationSum(list, N - list[ind], ind, perfSquares); // Remove the current element perfSquares.RemoveAt(perfSquares.Count - 1); } // Driver code public static void Main(string[] args) { int N = 8; // Call the function sumOfPerfectSquares(N); }}// This code is contributed by ukasp. |
Javascript
<script>// Javascript code for the above approach// Function to print all combinations of// sum of perfect squares equal to Nfunction combinationSum(list, N, ind, perfSquares){ // Sum of perfect squares exceeded N if (N < 0 || ind == list.length) return; // Sum of perfect squares is equal to N // therefore a combination is found if (N == 0) { for (let i = 0; i < perfSquares.length; i++) { document.write(perfSquares[i] + " "); } document.write("\n"); return; } // Do not include the current element combinationSum(list, N, ind + 1, perfSquares); // Include the element at current index perfSquares.push(list[ind]); combinationSum(list, N - list[ind], ind, perfSquares); // Remove the current element perfSquares.pop();}// Function to check whether the// number is a perfect square or notfunction sumOfPerfectSquares(N){ // Initialize an arraylist to store // all perfect squares less than N let list = []; let sqrtN = Math.sqrt(N); // Iterate till square root of N for (let i = 1; i <= sqrtN; i++) { // Add all perfect squares // to the list list.push(i * i); } let perfSquares = []; combinationSum(list, N, 0, perfSquares);}// Driver code let N = 8; // Call the function sumOfPerfectSquares(N);// This code is contributed by Samim Hossain Mondal.</script> |
Output
4 4 1 1 1 1 4 1 1 1 1 1 1 1 1
Time Complexity: O(N * 2^N), Where N is the number given
Auxiliary Space: O(N)
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