Suppose we have an array called nums and another value sum. We have to check whether it is possible to get sum by adding elements present in nums, we may pick a single element multiple times.
So, if the input is like nums = [2, 3, 5] sum = 28, then the output will be True as we can get 26 by using 5 + 5 + 5 + 5 + 3 + 3 + 2
To solve this, we will follow these steps −
- MAX := 1000
- table := an array of size MAX ad fill with 0
- Define a function util() . This will take nums
- table[0] := 1
- sort the list nums
- for i in range 0 to size of nums - 1, do
- val := nums[i]
- if table[val] is non-zero, then
- go for next iteration
- for j in range 0 to MAX - val - 1, do
- if table[j] is non-zero, then
- table[j + val] := 1
- if table[j] is non-zero, then
- From the main method do the following −
- util(nums)
- if table[sum] is non-zero, then
- return True
- return False
Let us see the following implementation to get better understanding −
Example
MAX = 1000 table = [0] * MAX def util(nums): table[0] = 1 nums.sort() for i in range(len(nums)): val = nums[i] if table[val]: continue for j in range(MAX - val): if table[j]: table[j + val] = 1 def solve(nums, sum): util(nums) if table[sum]: return True return False nums = [2, 3, 5] sum = 28 print (solve(nums, sum))
Input
[2, 3, 5], 28
Output
True