Count ways to represent a number as sum of perfect squares
Last Updated :
26 Jul, 2025
Given an integer N, the task is to find the number of ways to represent the number N as sum of perfect squares.
Examples:
Input: N = 9
Output: 4
Explanation:
There are four ways to represent 9 as the sum of perfect squares:
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9
1 + 1 + 1 + 1 + 1 + 4 = 9
1 + 4 + 4 = 9
9 = 9
Input: N = 8
Output: 3
Naive Approach: The idea is to store all the perfect squares less than or equal to N in an array. The problem now reduces to finding the ways to sum to N using array elements with repetition allowed which can be solved using recursion. Follow the steps below to solve the problem:
- Store all the perfect squares less than or equal to N and in an array psquare[].
- Create a recursion function countWays(index, target) that takes two parameters index, (initially N-1) and target (initially N):
- Handle the base cases:
- If the target is 0, return 1.
- If either index or target is less than 0, return 0.
- Otherwise, include the element, psquare[index] in the sum by subtracting it from the target and recursively calling for the remaining value of target.
- Exclude the element, psquare[index] from the sum by moving to the next index and recursively calling for the same value of target.
- Return the sum obtained by including and excluding the element.
- Print the value of countWays(N-1, N) as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Store perfect squares
// less than or equal to N
vector<int> psquare;
// Utility function to calculate perfect
// squares less than or equal to N
void calcPsquare(int N)
{
for (int i = 1; i * i <= N; i++)
psquare.push_back(i * i);
}
// Function to find the number
// of ways to represent a number
// as sum of perfect squares
int countWays(int index, int target)
{
// Handle the base cases
if (target == 0)
return 1;
if (index < 0 || target < 0)
return 0;
// Include the i-th index element
int inc = countWays(
index, target - psquare[index]);
// Exclude the i-th index element
int exc = countWays(index - 1, target);
// Return the result
return inc + exc;
}
// Driver Code
int main()
{
// Given Input
int N = 9;
// Precalculate perfect
// squares <= N
calcPsquare(N);
// Function Call
cout << countWays(psquare.size() - 1, N);
return 0;
}
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Store perfect squares
// less than or equal to N
static ArrayList<Integer> psquare = new ArrayList<>();
// Utility function to calculate perfect
// squares less than or equal to N
static void calcPsquare(int N)
{
for (int i = 1; i * i <= N; i++)
psquare.add(i * i);
}
// Function to find the number
// of ways to represent a number
// as sum of perfect squares
static int countWays(int index, int target)
{
// Handle the base cases
if (target == 0)
return 1;
if (index < 0 || target < 0)
return 0;
// Include the i-th index element
int inc
= countWays(index, target - psquare.get(index));
// Exclude the i-th index element
int exc = countWays(index - 1, target);
// Return the result
return inc + exc;
}
// Driver Code
public static void main(String[] args)
{
// Given Input
int N = 9;
// Precalculate perfect
// squares <= N
calcPsquare(N);
// Function Call
System.out.print(countWays(psquare.size() - 1, N));
}
}
// This code is contributed by Kingash.
# Python3 program for the above approach
# Store perfect squares
# less than or equal to N
psquare = []
# Utility function to calculate perfect
# squares less than or equal to N
def calcPsquare(N):
for i in range(1, N):
if i * i > N:
break
psquare.append(i * i)
# Function to find the number
# of ways to represent a number
# as sum of perfect squares
def countWays(index, target):
# Handle the base cases
if (target == 0):
return 1
if (index < 0 or target < 0):
return 0
# Include the i-th index element
inc = countWays(index, target - psquare[index])
# Exclude the i-th index element
exc = countWays(index - 1, target)
# Return the result
return inc + exc
# Driver Code
if __name__ == '__main__':
# Given Input
N = 9
# Precalculate perfect
# squares <= N
calcPsquare(N)
# Function Call
print (countWays(len(psquare) - 1, N))
# This code is contributed by mohit kumar 29
using System.IO;
using System;
using System.Collections;
class GFG {
// Store perfect squares
// less than or equal to N
static ArrayList psquare = new ArrayList();
// Utility function to calculate perfect
// squares less than or equal to N
static void calcPsquare(int N)
{
for (int i = 1; i * i <= N; i++)
psquare.Add(i * i);
}
// Function to find the number
// of ways to represent a number
// as sum of perfect squares
static int countWays(int index, int target)
{
// Handle the base cases
if (target == 0)
return 1;
if (index < 0 || target < 0)
return 0;
// Include the i-th index element
int inc = countWays(index,
target - (int)psquare[index]);
// Exclude the i-th index element
int exc = countWays(index - 1, target);
// Return the result
return inc + exc;
}
static void Main()
{
// Given Input
int N = 9;
// Precalculate perfect
// squares <= N
calcPsquare(N);
// Function Call
Console.WriteLine(countWays(psquare.Count - 1, N));
}
}
// This code is contributed by abhinavjain194.
<script>
// JavaScript program for the above approach
// Store perfect squares
// less than or equal to N
var psquare = []
// Utility function to calculate perfect
// squares less than or equal to N
function calcPsquare(N)
{
var i;
for (i = 1; i * i <= N; i++)
psquare.push(i * i);
}
// Function to find the number
// of ways to represent a number
// as sum of perfect squares
function countWays(index, target)
{
// Handle the base cases
if (target == 0)
return 1;
if (index < 0 || target < 0)
return 0;
// Include the i-th index element
var inc = countWays(
index, target - psquare[index]);
// Exclude the i-th index element
var exc = countWays(index - 1, target);
// Return the result
return inc + exc;
}
// Driver Code
// Given Input
var N = 9;
// Precalculate perfect
// squares <= N
calcPsquare(N);
// Function Call
document.write(countWays(psquare.length - 1, N));
</script>
Time Complexity: O(2K), where K is the number of perfect squares less than or equal to N
Auxiliary Space: O(1)
Efficient approach: This problem has overlapping subproblems and optimal substructure property. To optimize the above approach, the idea is to use dynamic programming by memoizing the above recursive calls using a 2D array of size K*N, where K is the number of perfect squares less than or equal to N.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Store perfect squares
// less than or equal to N
vector<int> psquare;
// Utility function to calculate
// perfect squares <= N
void calcPsquare(int N)
{
for (int i = 1; i * i <= N; i++)
psquare.push_back(i * i);
}
// DP array for memoization
vector<vector<int> > dp;
// Recursive function to count
// number of ways to represent
// a number as a sum of perfect squares
int countWaysUtil(int index, int target)
{
// Handle base cases
if (target == 0)
return 1;
if (index < 0 || target < 0)
return 0;
// If already computed, return the result
if (dp[index][target] != -1)
return dp[index][target];
// Else, compute the result
return dp[index][target]
= countWaysUtil(
index, target - psquare[index])
+ countWaysUtil(
index - 1, target);
}
// Function to find the number of ways to
// represent a number as a sum of perfect squares
int countWays(int N)
{
// Precalculate perfect squares less
// than or equal to N
calcPsquare(N);
// Create dp array to memoize
dp.resize(psquare.size() + 1,
vector<int>(N + 1, -1));
// Function call to fill dp array
return countWaysUtil(psquare.size() - 1, N);
}
// Driver Code
int main()
{
// Given Input
int N = 9;
// Function Call
cout << countWays(N);
return 0;
}
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Store perfect squares
// less than or equal to N
private static ArrayList<Integer> psquare;
// Utility function to calculate
// perfect squares <= N
private static void calcPsquare(int n) {
for (int i = 1; i * i <= n; i++)
psquare.add(i * i);
}
// DP array for memoization
private static int[][] dp;
// Recursive function to count
// number of ways to represent
// a number as a sum of perfect squares
private static int countWaysUtil(int index, int target) {
// Handle base cases
if (target == 0)
return 1;
if (index < 0 || target < 0)
return 0;
// If already computed, return the result
if (dp[index][target] != -1)
return dp[index][target];
// Else, compute the result
return dp[index][target]
= countWaysUtil(
index, target - psquare.get(index))
+ countWaysUtil(
index - 1, target);
}
// Function to find the number of ways to
// represent a number as a sum of perfect squares
private static int countWays(int n) {
// Precalculate perfect squares less
// than or equal to N
psquare = new ArrayList<Integer>();
calcPsquare(n);
// Create dp array to memoize
dp = new int[psquare.size()+1][n+1];
for(int i = 0; i<=psquare.size(); i++)Arrays.fill(dp[i], -1);
// Function call to fill dp array
return countWaysUtil(psquare.size() - 1, n);
}
// Driver Code
public static void main(String[] args)
{
// Given Input
int N = 9;
// Function Call
System.out.print(countWays(N));
}
}
// This code is contributed by Dheeraj Bhagchandani.
# Python3 program for the above approach
from math import sqrt
# Store perfect squares
# less than or equal to N
psquare = []
# DP array for memoization
dp = []
# Utility function to calculate
# perfect squares <= N
def calcPsquare(N):
global psquare
for i in range(1, int(sqrt(N)) + 1, 1):
psquare.append(i * i)
# Recursive function to count
# number of ways to represent
# a number as a sum of perfect squares
def countWaysUtil(index, target):
global dp
# Handle base cases
if (target == 0):
return 1
if (index < 0 or target < 0):
return 0
# If already computed, return the result
if (dp[index][target] != -1):
return dp[index][target]
dp[index][target] = (countWaysUtil(
index, target - psquare[index]) +
countWaysUtil(index - 1, target))
# Else, compute the result
return dp[index][target]
# Function to find the number of ways to
# represent a number as a sum of perfect squares
def countWays(N):
global dp
global psquare
# Precalculate perfect squares less
# than or equal to N
calcPsquare(N)
temp = [-1 for i in range(N + 1)]
# Create dp array to memoize
dp = [temp for i in range(len(psquare) + 1)]
# Function call to fill dp array
return countWaysUtil(len(psquare) - 1, N) - 1
# Driver Code
if __name__ == '__main__':
# Given Input
N = 9
# Function Call
print(countWays(N))
# This code is contributed by ipg2016107
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Store perfect squares
// less than or equal to N
static List<int> psquare;
// Utility function to calculate
// perfect squares <= N
private static void calcPsquare(int n)
{
for(int i = 1; i * i <= n; i++)
psquare.Add(i * i);
}
// DP array for memoization
private static int[,]dp;
// Recursive function to count
// number of ways to represent
// a number as a sum of perfect squares
private static int countWaysUtil(int index,
int target)
{
// Handle base cases
if (target == 0)
return 1;
if (index < 0 || target < 0)
return 0;
// If already computed, return the result
if (dp[index, target] != -1)
return dp[index, target];
// Else, compute the result
return dp[index, target] = countWaysUtil(index,
target - psquare[index]) +
countWaysUtil(index - 1, target);
}
// Function to find the number of ways to
// represent a number as a sum of perfect squares
private static int countWays(int n)
{
// Precalculate perfect squares less
// than or equal to N
psquare = new List<int>();
calcPsquare(n);
// Create dp array to memoize
dp = new int[psquare.Count + 1, n + 1];
for(int i = 0; i <= psquare.Count; i++)
{
for(int j = 0; j <= n; j++)
{
dp[i, j] = -1;
}
//Array.Fill(dp[i], -1);
}
// Function call to fill dp array
return countWaysUtil(psquare.Count - 1, n);
}
// Driver Code
static void Main()
{
// Given Input
int N = 9;
// Function Call
Console.Write(countWays(N));
}
}
// This code is contributed by SoumikMondal
<script>
// JavaScript program for the above approach
let psquare;
function calcPsquare(n)
{
for (let i = 1; i * i <= n; i++)
psquare.push(i * i);
}
// DP array for memoization
let dp;
// Recursive function to count
// number of ways to represent
// a number as a sum of perfect squares
function countWaysUtil(index,target)
{
// Handle base cases
if (target == 0)
return 1;
if (index < 0 || target < 0)
return 0;
// If already computed, return the result
if (dp[index][target] != -1)
return dp[index][target];
// Else, compute the result
return dp[index][target]
= countWaysUtil(
index, target - psquare[index])
+ countWaysUtil(
index - 1, target);
}
// Function to find the number of ways to
// represent a number as a sum of perfect squares
function countWays(n)
{
// Precalculate perfect squares less
// than or equal to N
psquare = [];
calcPsquare(n);
// Create dp array to memoize
dp = new Array(psquare.length+1);
for(let i=0;i<psquare.length+1;i++)
{
dp[i]=new Array(n+1);
for(let j=0;j<n+1;j++)
{
dp[i][j]=-1;
}
}
// Function call to fill dp array
return countWaysUtil(psquare.length - 1, n);
}
// Driver Code
// Given Input
let N = 9;
// Function Call
document.write(countWays(N));
// This code is contributed by patel2127
</script>
Time Complexity: O(K*N), where K is the number of perfect squares less than or equal to N
Auxiliary Space: O(K*N)
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