Minimum number of squares whose sum equals to given number N | set 2
Last Updated :
05 Nov, 2021
A number can always be represented as a sum of squares of other numbers. Note that 1 is a square, and we can always break a number as (1*1 + 1*1 + 1*1 + …). Given a number N, the task is to represent N as the sum of minimum square numbers.
Examples:
Input : 10
Output : 1 + 9
These are all possible ways
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1 + 1 + 4
1 + 1 + 4 + 4
1 + 9
Choose one with minimum numbers
Input : 25
Output : 25
Prerequisites: Minimum number of squares whose sum equals to given number N
Approach: This is a typical application of dynamic programming. When we start from N = 6, we can reach 2 by subtracting the square of one i.e. one, 4 times, and by subtracting the square of two i.e. four, 1 time. So the subproblem for 2 is called twice.
Since the same subproblems are called again, this problem has the Overlapping Subproblems property. So-min square sum problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputation of the same subproblems can be avoided by constructing a temporary array table[][] in a bottom-up manner.
Below is the implementation of the above approach:
C++
// C++ program to represent N as the
// sum of minimum square numbers.
#include <bits/stdc++.h>
using namespace std;
// Function for finding
// minimum square numbers
vector<int> minSqrNum(int n)
{
// A[i] of array arr store
// minimum count of
// square number to get i
int arr[n + 1], k;
// sqrNum[i] store last
// square number to get i
int sqrNum[n + 1];
vector<int> v;
// Initialize
arr[0] = 0;
sqrNum[0] = 0;
// Find minimum count of
// square number for
// all value 1 to n
for (int i = 1; i <= n; i++)
{
// In worst case it will
// be arr[i-1]+1 we use all
// combination of a[i-1] and add 1
arr[i] = arr[i - 1] + 1;
sqrNum[i] = 1;
k = 1;
// Check for all square
// number less or equal to i
while (k * k <= i)
{
// if it gives less
// count then update it
if (arr[i] > arr[i - k * k] + 1)
{
arr[i] = arr[i - k * k] + 1;
sqrNum[i] = k * k;
}
k++;
}
}
// Vector v stores optimum
// square number whose sum give N
while (n > 0)
{
v.push_back(sqrNum[n]);
n -= sqrNum[n];
}
return v;
}
// Driver code
int main()
{
int n = 10;
vector<int> v;
// Calling function
v = minSqrNum(n);
// Printing vector
for (auto i = v.begin();
i != v.end(); i++)
{
cout << *i;
if (i + 1 != v.end())
cout << " + ";
}
return 0;
}
// Java program to represent
// N as the sum of minimum
// square numbers.
import java.util.*;
class GFG{
// Function for finding
// minimum square numbers
static Vector<Integer> minSqrNum(int n)
{
// A[i] of array arr store
// minimum count of
// square number to get i
int []arr = new int[n + 1];
int k = 0;
// sqrNum[i] store last
// square number to get i
int []sqrNum = new int[n + 1];
Vector<Integer> v = new Vector<>();
// Initialize
arr[0] = 0;
sqrNum[0] = 0;
// Find minimum count of
// square number for
// all value 1 to n
for (int i = 1; i <= n; i++)
{
// In worst case it will
// be arr[i-1]+1 we use all
// combination of a[i-1] and add 1
arr[i] = arr[i - 1] + 1;
sqrNum[i] = 1;
k = 1;
// Check for all square
// number less or equal to i
while (k * k <= i)
{
// if it gives less
// count then update it
if (arr[i] > arr[i - k * k] + 1)
{
arr[i] = arr[i - k * k] + 1;
sqrNum[i] = k * k;
}
k++;
}
}
// Vector v stores optimum
// square number whose sum give N
while (n > 0)
{
v.add(sqrNum[n]);
n -= sqrNum[n];
}
return v;
}
// Driver code
public static void main(String[] args)
{
int n = 10;
Vector<Integer> v;
// Calling function
v = minSqrNum(n);
// Printing vector
for (int i = 0; i <v.size(); i++)
{
System.out.print(v.elementAt(i));
if (i+1 != v.size())
System.out.print(" + ");
}
}
}
// This code is contributed by gauravrajput1
# Python3 program to represent N as the
# sum of minimum square numbers.
# Function for finding
# minimum square numbers
def minSqrNum(n):
# arr[i] of array arr store
# minimum count of
# square number to get i
arr = [0] * (n + 1)
# sqrNum[i] store last
# square number to get i
sqrNum = [0] * (n + 1)
v = []
# Find minimum count of
# square number for
# all value 1 to n
for i in range(n + 1):
# In worst case it will
# be arr[i-1]+1 we use all
# combination of a[i-1] and add 1
arr[i] = arr[i - 1] + 1
sqrNum[i] = 1
k = 1;
# Check for all square
# number less or equal to i
while (k * k <= i):
# If it gives less
# count then update it
if (arr[i] > arr[i - k * k] + 1):
arr[i] = arr[i - k * k] + 1
sqrNum[i] = k * k
k += 1
# v stores optimum
# square number whose sum give N
while (n > 0):
v.append(sqrNum[n])
n -= sqrNum[n];
return v
# Driver code
n = 10
# Calling function
v = minSqrNum(n)
# Printing vector
for i in range(len(v)):
print(v[i], end = "")
if (i < len(v) - 1):
print(" + ", end = "")
# This article is contributed by Apurvaraj
// C# program to represent
// N as the sum of minimum
// square numbers.
using System;
using System.Collections.Generic;
class GFG{
// Function for finding
// minimum square numbers
static List<int> minSqrNum(int n)
{
// A[i] of array arr store
// minimum count of
// square number to get i
int []arr = new int[n + 1];
int k = 0;
// sqrNum[i] store last
// square number to get i
int []sqrNum = new int[n + 1];
List<int> v = new List<int>();
// Initialize
arr[0] = 0;
sqrNum[0] = 0;
// Find minimum count of
// square number for
// all value 1 to n
for (int i = 1; i <= n; i++)
{
// In worst case it will
// be arr[i-1]+1 we use all
// combination of a[i-1] and add 1
arr[i] = arr[i - 1] + 1;
sqrNum[i] = 1;
k = 1;
// Check for all square
// number less or equal to i
while (k * k <= i)
{
// if it gives less
// count then update it
if (arr[i] > arr[i - k * k] + 1)
{
arr[i] = arr[i - k * k] + 1;
sqrNum[i] = k * k;
}
k++;
}
}
// List v stores optimum
// square number whose sum give N
while (n > 0)
{
v.Add(sqrNum[n]);
n -= sqrNum[n];
}
return v;
}
// Driver code
public static void Main(String[] args)
{
int n = 10;
List<int> v;
// Calling function
v = minSqrNum(n);
// Printing vector
for (int i = 0; i <v.Count; i++)
{
Console.Write(v[i]);
if (i+1 != v.Count)
Console.Write(" + ");
}
}
}
// This code is contributed by gauravrajput1
<script>
// Javascript program to represent N as the
// sum of minimum square numbers.
// Function for finding
// minimum square numbers
function minSqrNum(n)
{
// A[i] of array arr store
// minimum count of
// square number to get i
var arr = Array(n+1), k;
// sqrNum[i] store last
// square number to get i
var sqrNum = Array(n+1);
var v = [];
// Initialize
arr[0] = 0;
sqrNum[0] = 0;
// Find minimum count of
// square number for
// all value 1 to n
for (var i = 1; i <= n; i++)
{
// In worst case it will
// be arr[i-1]+1 we use all
// combination of a[i-1] and add 1
arr[i] = arr[i - 1] + 1;
sqrNum[i] = 1;
k = 1;
// Check for all square
// number less or equal to i
while (k * k <= i)
{
// if it gives less
// count then update it
if (arr[i] > arr[i - k * k] + 1)
{
arr[i] = arr[i - k * k] + 1;
sqrNum[i] = k * k;
}
k++;
}
}
// Vector v stores optimum
// square number whose sum give N
while (n > 0)
{
v.push(sqrNum[n]);
n -= sqrNum[n];
}
return v;
}
// Driver code
var n = 10;
var v = [];
// Calling function
v = minSqrNum(n);
// Printing vector
for(var i = 0; i<v.length; i++)
{
document.write(v[i]);
if (i + 1 != v.length)
document.write( " + ");
}
</script>
Time Complexity: O(n3/2)
Auxiliary Space: O(n)