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

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/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 consturcted 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[0] = 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; ` `} ` |

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## 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 consturcted 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[` `0` `] = ` `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. ` |

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## 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 consturcted in bottom up ` ` ` `# manner using the base case (n = 0) ` ` ` `# Initialize all table values as 0 ` ` ` `table ` `=` `[` `0` `] ` `*` `(n ` `+` `1` `) ` ` ` ` ` `# Base case (If given value is 0) ` ` ` `# Only 1 way to get 0 (select no integer) ` ` ` `table[` `0` `] ` `=` `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 ` |

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## 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 consturcted 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[0] = 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. ` |

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## 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(` `$n` `) ` `{ ` ` ` `// table[i] will be storing the number of ` ` ` `// solutions for value i. We need n+1 rows ` ` ` `// as the table is consturcted in bottom up ` ` ` `// manner using the base case (n = 0) ` ` ` `$table` `= ` `array_fill` `(0, ` `$n` `+ 1, NULL); ` ` ` ` ` `// Base case (If given value is 0) ` ` ` `$table` `[0] = 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 ` `?> ` |

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Output:

14

Time complexity O(n^{2})

This article is contributed by **Shivam Pradhan (anuj_charm)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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