Suppose we have a binary list with 0s or 1s. We also have another input called k, we have to find the number of sublists whose sum is same as k.
So, if the input is like nums = [1, 0, 0, 1, 1, 1, 0, 1] k = 3, then the output will be 8 because the sublists are [1,0,0,1,1], [0,0,1,1,1], [0,0,1,1,1,0], [0,1,1,1], [0,1,1,1,0], [1,1,1], [1,1,1,0] [1,1,0,1].
To solve this, we will follow these steps −
- sums := a map initially contains valye 1 for key 0
- r_sum := 0
- ans := 0
- for each x in nums, do
- r_sum := r_sum + x
- ans := ans + (sums[r_sum - k] if (r_sum - k) is present, otherwise 0)
- sums[r_sum] := 1 + (sums[r_sum - k] if (r_sum - k) is present, otherwise 0)
- return ans
Example
Let us see the following implementation to get better understanding −
def solve(nums, k): sums = {0: 1} r_sum = 0 ans = 0 for x in nums: r_sum += x ans += sums.get(r_sum - k, 0) sums[r_sum] = sums.get(r_sum, 0) + 1 return ans nums = [1, 0, 0, 1, 1, 1, 0, 1] k = 3 print(solve(nums, k))
Input
[1, 0, 0, 1, 1, 1, 0, 1], 3
Output
8