Ways to write n as sum of two or more positive integers

• Difficulty Level : Medium
• Last Updated : 07 Sep, 2021

For a given number n > 0, find the number of different ways in which n can be written as a sum of at two or more positive integers.
Examples:

Input : n = 5
Output : 6
Explanation : All possible six ways are :
4 + 1
3 + 2
3 + 1 + 1
2 + 2 + 1
2 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1

Input : 4
Output : 4
Explanation : All possible four ways are :
3 + 1
2 + 2
2 + 1 + 1
1 + 1 + 1 + 1

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

This problem can be solved in the similar fashion as coin change problem, the difference is only that in this case we should iterate for 1 to n-1 instead of particular values of coin as in coin-change problem.

C++

// Program to find the number of ways, n can be
// written as sum of two or more positive integers.
#include <bits/stdc++.h>
using namespace std;

// Returns number of ways to write n as sum of
// two or more positive integers
int countWays(int n)
{
// table[i] will be storing the number of
// solutions for value i. We need n+1 rows
// as the table is constructed in bottom up
// manner using the base case (n = 0)
int table[n+1];

// Initialize all table values as 0
memset(table, 0, sizeof(table));

// Base case (If given value is 0)
table = 1;

// Pick all integer one by one and update the
// table[] values after the index greater
// than or equal to n
for (int i=1; i<n; i++)
for (int j=i; j<=n; j++)
table[j] += table[j-i];

return table[n];
}

// Driver program
int main()
{
int n = 7;
cout << countWays(n);
return 0;
}

Java

// Program to find the number of ways,
// n can be written as sum of two or
// more positive integers.
import java.util.Arrays;

class GFG {

// Returns number of ways to write
// n as sum of two or more positive
// integers
static int countWays(int n)
{

// table[i] will be storing the
// number of solutions for value
// i. We need n+1 rows as the
// table is constructed in bottom
// up manner using the base case
// (n = 0)
int table[] = new int[n + 1];

// Initialize all table values as 0
Arrays.fill(table, 0);

// Base case (If given value is 0)
table = 1;

// Pick all integer one by one and
// update the table[] values after
// the index greater than or equal
// to n
for (int i = 1; i < n; i++)
for (int j = i; j <= n; j++)
table[j] += table[j - i];

return table[n];
}

//driver code
public static void main (String[] args)
{
int n = 7;

System.out.print(countWays(n));
}
}

// This code is contributed by Anant Agarwal.

Python

# Program to find the number of ways, n can be
# written as sum of two or more positive integers.

# Returns number of ways to write n as sum of
# two or more positive integers
def CountWays(n):

# table[i] will be storing the number of
# solutions for value i. We need n+1 rows
# as the table is constructed in bottom up
# manner using the base case (n = 0)
# Initialize all table values as 0
table = * (n + 1)

# Base case (If given value is 0)
# Only 1 way to get 0 (select no integer)
table = 1

# Pick all integer one by one and update the
# table[] values after the index greater
# than or equal to n
for i in range(1, n ):

for j in range(i , n + 1):

table[j] +=  table[j - i]

return table[n]

# driver program
def main():

n = 7

print CountWays(n)

if __name__ == '__main__':
main()

#This code is contributed by Neelam Yadav

C#

// Program to find the number of ways, n can be
// written as sum of two or more positive integers.
using System;

class GFG {

// Returns number of ways to write n as sum of
// two or more positive integers
static int countWays(int n)
{

// table[i] will be storing the number of
// solutions for value i. We need n+1 rows
// as the table is constructed in bottom up
// manner using the base case (n = 0)
int []table = new int[n+1];

// Initialize all table values as 0
for(int i = 0; i < table.Length; i++)
table[i] = 0;

// Base case (If given value is 0)
table = 1;

// Pick all integer one by one and update the
// table[] values after the index greater
// than or equal to n
for (int i = 1; i < n; i++)
for (int j = i; j <= n; j++)
table[j] += table[j-i];

return table[n];
}

//driver code
public static void Main()
{
int n = 7;

Console.Write(countWays(n));
}
}

// This code is contributed by Anant Agarwal.

PHP

<?php
// Program to find the number of ways, n can be
// written as sum of two or more positive integers.

// Returns number of ways to write n as sum
// of two or more positive integers
function countWays(\$n)
{
// table[i] will be storing the number of
// solutions for value i. We need n+1 rows
// as the table is constructed in bottom up
// manner using the base case (n = 0)
\$table = array_fill(0, \$n + 1, NULL);

// Base case (If given value is 0)
\$table = 1;

// Pick all integer one by one and update
// the table[] values after the index
// greater than or equal to n
for (\$i = 1; \$i < \$n; \$i++)
for (\$j = \$i; \$j <= \$n; \$j++)
\$table[\$j] += \$table[\$j - \$i];

return \$table[\$n];
}

// Driver Code
\$n = 7;
echo countWays(\$n);

// This code is contributed by ita_c
?>

Javascript

<script>

function countWays(n)
{
// table[i] will be storing the
// number of solutions for value
// i. We need n+1 rows as the
// table is constructed in bottom
// up manner using the base case
// (n = 0)
let table = new Array(n + 1);

// Initialize all table values as 0
for(let i = 0; i < n + 1; i++)
{
table[i]=0;
}

// Base case (If given value is 0)
table = 1;

// Pick all integer one by one and
// update the table[] values after
// the index greater than or equal
// to n
for (let i = 1; i < n; i++)
for (let j = i; j <= n; j++)
table[j] += table[j - i];

return table[n];
}

let n = 7;
document.write(countWays(n));

// This code is contributed by avanitrachhadiya2155
</script>

Output:

14

Time complexity O(n2)