# Number of ways to divide a given number as a set of integers in decreasing order

Given two numbers and . The task is to find the number of ways in which **a** can be represented by a set such that and the summation of these numbers is equal to **a**. Also (maximum size of the set cannot exceed **m**).

**Examples**:

Input: a = 4, m = 4

Output: 2 –> ({4}, {3, 1})

Note: {2, 2} is not a valid set as values are not in decreasing order

Input: a = 7, m = 5

Output: 5 –> ({7}, {6, 1}, {5, 2}, {4, 3}, {4, 2, 1})

**Approach:** This problem can be solved by Divide and Conquer using a recursive approach which follows the following conditions:

- If
**a**is equal to zero, one solution has been found. - If a > 0 and m == 0, this set violates the condition as no further values can be added in the set.
- If calculation has already been done for given values of
**a**,**m**and prev (last value included in the current set), return that value. - Start a loop from
**i**=**a**till 0 and if**i**<**prev**, count the number of solutions if we include**i**in the current set and return it.

Below is the implementation of the above approach:

`# Python3 code to calculate the number of ways ` `# in which a given number can be represented ` `# as set of finite numbers ` ` ` `# Import function to initialize the dictionary ` `from` `collections ` `import` `defaultdict ` ` ` `# Initialize dictionary which is used ` `# to check if given solution is already ` `# visited or not to avoid ` `# calculating it again ` `visited ` `=` `defaultdict(` `lambda` `: ` `False` `) ` ` ` `# Initialize dictionary which is used to ` `# store the number of ways in which solution ` `# can be obtained for given values ` `numWays ` `=` `defaultdict(` `lambda` `: ` `0` `) ` ` ` `# This function returns the total number ` `# of sets which satisfy given criteria ` `# a --> number to be divided into sets ` `# m --> maximum possible size of the set ` `# x --> previously selected value ` `def` `countNumOfWays(a, m, prev): ` ` ` ` ` `# number is divided properly and ` ` ` `# hence solution is obtained ` ` ` `if` `a ` `=` `=` `0` `: ` ` ` `return` `1` ` ` ` ` `# Solution can't be obtained ` ` ` `elif` `a > ` `0` `and` `m ` `=` `=` `0` `: ` ` ` `return` `0` ` ` ` ` `# Return the solution if it has ` ` ` `# already been calculated ` ` ` `elif` `visited[(a, m, prev)] ` `=` `=` `True` `: ` ` ` `return` `numWays[(a, m, prev)] ` ` ` ` ` `else` `: ` ` ` `visited[(a, m, prev)] ` `=` `True` ` ` ` ` `for` `i ` `in` `range` `(a, ` `-` `1` `, ` `-` `1` `): ` ` ` `# Continue only if current value is ` ` ` `# smaller compared to previous value ` ` ` `if` `i < prev: ` ` ` `numWays[(a,m,prev)] ` `+` `=` `countNumOfWays(a` `-` `i,m` `-` `1` `,i) ` ` ` ` ` `return` `numWays[(a, m, prev)] ` ` ` `# Values of 'a' and 'm' for which ` `# solution is to be found ` `# MAX_CONST is extremely large value ` `# used for first comparison in the function ` `a, m, MAX_CONST ` `=` `7` `, ` `5` `, ` `10` `*` `*` `5` `print` `(countNumOfWays(a, m, MAX_CONST)) ` |

*chevron_right*

*filter_none*

**Output:**

5

**Time Complexity:** O(a*log(a))

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