# Prefix Sum¶

Done

## What is prefix sum¶

Given a numeric array lst of length n, we can use an array pre_sum of length n+1 to represent the current accumulated sum until now, starting from the empty array, where the sum is initialized as 0:

pre_sum = [0]
for num in lst:
pre_sum.append(pre_sum[-1]+num)


In other word, i-th element in the prefix-sum array pre_sum (i.e., pre_sum[i]) stores the sum of lst[0:i] (the index range of [0,i)), that is, the sum of the first i elements in lst.

Obviously, the remarkable benfit of first traversing the array to get such a prefix-sum array is the less time and space used to evalute the sum of each lst[i:j] later, for which we can simply use the following formula:

sum(lst[i:j]) = pre_sum[j] - pre_sum[i]

• Time Complexity: $$O(n)$$ for obtaining the prefix-sum array and $$O(1)$$ for obtaining any sum within a sub-array.

• Space Complexity : $$O(n)$$

Compare to some naive approaches:

• If you add up all elements in lst[i:j] every time, it costs $$O(j-i)$$ for every query of sum(lst[i:j]).

• If you store all sums between all indexes of i<=j in a 2-D array, it costs $$O(n^2)$$ space.

## When should I use prefix sum to solve problems?¶

• You’re required to evaluate the sums of some elements with successive indexes (i.e., sums of sub-arraies) frequently.

• Always serves as a quick evaluation trick in other methods (e.g., dynamic programming, binary search and etc.) to reduce the cost of calculating the sums of some dynamically sliced sub-arraies.

## Alternatives¶

You may also see some other ways written by coders. For example, some may first assign a (n+1)-zeros list and then update every pre_sum[i] by iterations, which can be regarded as a trivial dynamic programming approach (we refer it as Alt-1 template in the following):

pre_sum = [0]*(n+1)
for i in range(n):
pre_sum[i+1] = pre_sum[i]+lst[i]


while some may utilize itertools.accumulate function directly (we refer it as Alt-2 template in the following) as

from itertools import accumulate
pre_sum = list(accumulate(lst,initial=0))


I also measured the runtime (we use CPU time here) for those three templates using Python 3.8, by running on m lists of n-elements respectively. The source code and raw data are available in this notebook. The results are listed as below:

Scale

Original

Alt 1

Alt 2

n=1e5, m = 100

2.45s

2.78s

516ms

n=5, m=1e4

15.6ms

46.9ms

~ 0ns

n=1e4, m=1e4

20.2s

28.1s

5.25s

To conclude, accumulate (Alt-2 template) outperforms other templates significantly and keeps a very neat code style meanwhile, which is resulted from some optimization in this function implementation.

But I still suggest you stick to any template as long as you’re comfortable with it because LeetCode conducts very loose restrictions on runtime. About 100 test cases with the most extended list as $$10^5$$-elements are usually designed for each LeetCode problem (like row 1 in the table above), in which a template won’t have a great impact on overall runtime, so you can pick any preferred style in a LC weekly contest just for finding the solution fast.

## Usage Examples¶

### Classic¶

LC2256: Minimum Average Difference, a very typical application of prefix sum array: it asks for the average of first i+1 elements and the average of last n-i-1 elements for each 0<=i<n, in which the sum of i+1 elements and the last n-i-1 elements) can be read as pre[i] and pre[-1]-pre[i] respectively.

class Solution:
def minimumAverageDifference(self, nums: List[int]) -> int:
n = len(nums)
pre = [0]
idx,res = -1,float('inf')
for num in nums:
pre.append(pre[-1]+num)
for i in range(n):
diff = abs(pre[i+1]//(i+1) - (pre[-1]-pre[i+1])//(n-i-1)) if i!=n-1 else pre[i+1]//(i+1)
if diff < res:
res = diff
idx = i
return idx


### 2-Direction¶

LC2245 Maximum Trailing Zeros in a Cornered Path: instead of accumulating the number itself, in this problem, we should accumulate the number of factors of 2 and 5 along with all “paths”.

The idea is explained by @votrubac in 1 as the picture below ([a:b]in the table represents there are total a factors of 2 and b factors of 5 so far in this line, upper->lower, left->right):

• First, calculate the prefix sum of all factors of 2 and 5 for each row (stored as rows, with a shape n*(m+1))) and each column (stored as cols, with a shape m*(n+1)) respectively

• Now, for each index (i,j), you can reach it by

1. left -> right: grid[i][:j], the total number of factors so far ((i,j) exclusively) should be rows[i][j]

2. right -> left: grid[i][j+1:], as rows[i][-1]-rows[i][j+1] (yes, suffix sum can be calculated by prefix sum as well, you don’t have to caculate it again reversely)

3. upper -> lower: grid[:i][j], as cols[j][i]

4. lower -> upper: grid[i+1:][j], as cols[j][-1]-cols[j][i+1]

• When selecting any index (i,j) as the “turn point” (you must turn once to maximize the result because at least it won’t reduce the number of factors anyway, why not?), sum up the element itself with two paths that reaches it horizontally and vertically, those are, 1 and 3, 1 and 4, 2 and 3, 2 and 4 above, four combinations in total.

• Find out the max values from all selected “turn points”. The overall time complexity is $$O(nm)$$

class Solution:
def maxTrailingZeros(self, grid: List[List[int]]) -> int:
n,m = len(grid),len(grid[0])
res = 0

def factors(x,f):
res = 0
while x%f == 0:
x//=f
res+=1
return res

def pre_sum(lst):
pre = [(0,0)]
for a,b in lst:
pre.append((pre[-1][0]+a,pre[-1][1]+b))
return pre

for i in range(n):
for j in range(m):
grid[i][j] = factors(grid[i][j],2),factors(grid[i][j],5)

rows = [pre_sum(row) for row in grid]
cols = [pre_sum([grid[i][j] for i in range(n)]) for j in range(m)]

for i in range(n):
for j in range(m):
left_row, upper_col = rows[i][j], cols[j][i]
right_row = rows[i][-1][0] - rows[i][j+1][0],rows[i][-1][1] - rows[i][j+1][1]
lower_col = cols[j][-1][0] -cols[j][i+1][0],cols[j][-1][1] -cols[j][i+1][1]
x0,y0 = grid[i][j]
for x1,y1 in [left_row,right_row]:
for x2,y2 in [lower_col,upper_col]:
res = max(res,min(x0+x1+x2,y0+y1+y2))
return res


### More than sum¶

The idea of prefix sum can be applied to other kinds of operations more than +, as long as they are left-associative, commutative, and invertible. For some special problems, some properities are even not required. For example,

LC238: Product of Array Except Self, which can be solved by calculating product from left to index i exclusively and then reversely calculating from right to index i exclusively as well, multiplying the “prefix product” and “suffix product”, finally getting the product that excludes i-th element:

from itertools import accumulate
from operator import mul
class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
n = len(nums)
prefix = list(accumulate(nums,mul,initial=1))
suffix = list(accumulate(reversed(nums),mul,initial=1))
return [prefix[i]*suffix[n-1-i] for i in range(n)]


In the solution above, I use the accumulate template to show that the operator can be simply replaced by any function of a type as a,b -> a to accumulate for a prefix “result” array.

### 2-D Matrix¶

Besides 1-D array/list, when you’re asked to sum up all numbers in some rectangle areas in a 2-D matrix frequently, you can also consider prefix sum with some subtle modification.

For the 2-D array mat, you can also discretize the matrix as a set of some elements, in which Inclusion-Exclusion Principle help you calculate the union set. In order to get the sum of the sub-matrix mat[:i][:j], elements in the smaller included matrices mat[:i-1][:j] and mat[:i][:j-1] should be added up and then you need to eliminate those elements lying in both two matrices (i.e., mat[:i-1][:j-1]) or they will be added twice. Finally, add the sum of matrices with the new element mat[i][j] to get the sum of mat[:i][:j], that is

sum(mat[:i][:j]) = sum(mat[:i-1][:j]) + sum(mat[:i][:j-1]) - sum(mat[i][j]) + mat[i][j]


In Python, the process of calculating 2-D prefix sum matrix can be described as:

n,m = len(mat),len(mat[0])
pre = [[0]*(m+1) for _ in range(n+1)]
for i in range(n):
for j in range(m):
pre[i+1][j+1] = pre[i+1][j] + pre[i][j+1] + mat[i][j] - pre[i][j]


To get the sum of a sub-matrix mat[a:b][c:d], similary:

sum(mat[a:b][c:d]) = sum(mat[:b][:d]) - sum(mat[:a][:d]) - sum(mat[:b][:c] + sum(mat[:a][:b]))


For example, 1292 Maximum Side Length of a Square with Sum Less than or Equal to Threshold is a classic binary search problem, but in its check(k) functions, sum(mat[i:i+k][j:j+k]) is evaluated frequently, we can use the prefix sum array to avoid repeated calculation:

class Solution:
def maxSideLength(self, mat: List[List[int]], threshold: int) -> int:
n,m = len(mat),len(mat[0])
pre = [[0]*(m+1) for _ in range(n+1)]
min_value = float('inf')

for i in range(n):
for j in range(m):
pre[i+1][j+1] = pre[i+1][j] + pre[i][j+1] + mat[i][j] - pre[i][j]
min_value = min(min_value,mat[i][j])

if min_value>threshold:
return 0

def check(k):
for i in range(k,n+1):
for j in range(k,m+1):
s = pre[i][j] - pre[i-k][j] - pre[i][j-k] + pre[i-k][j-k]
if s<=threshold:
return True
return False

l,r = 1,min(n,m)+1
while l<r:
mid = (l+r)//2
if check(mid):
l = mid + 1
else:
r = mid
return l-1


### Prefix Sum of Prefix Sum¶

LC2281: Sum of Total Strength of Wizards

1. Use monotonic stack to find the nearest smaller number for each A[i]: left[i] is the rightmost position that hosts a number < A[i] on the left side while right[i] is the leftmost position that hosts a number <= A[i]. That is, the longest subarray, in which A[i] plays a role of the min value, is [left[i]+1,right[i]-1] (or (left[i],right[i])).

2. For each A[i]: how to find all subarray that takes A[i] as the min value? Assume a subarray is written as [l,r], then it must satisfy the constraint: left[i]<l<=i<r<=right[i], otherwise it will include some value < A[i] or exclude the min value A[i].

3. How to sum up all numbers in all those subarrays [l,r]? It may be more clear to write it in a math formula:

\begin{split} \begin{aligned} & \sum_{l=left[i]+1}^i{\sum_{r=i}^{right[i]-1}}{sum(\color{green}{A}[l:r+1])} & (1) \\ = & \sum_{l=left[i]+1}^i{\sum_{r=i}^{right[i]-1}}{\color{blue}{presum}[r+1] - \color{blue}{presum}[l]} & \bf{(2)} \\ = & \sum_{l=left[i]+1}^i{\sum_{r=i}^{right[i]-1}}{\color{blue}{presum}[r+1]} - \sum_{l=left[i]+1}^i{\sum_{r=i}^{right[i]-1}}{\color{blue}{presum}[l]} & (3) \\ = & (i-left[i]) {\sum_{r=i}^{right[i]-1}}{\color{blue}{presum}[r+1]} - (right[i]-i)\sum_{l=left[i]+1}^i{\color{blue}{presum}[l]} & (4)\\ = & (i-left[i])(\color{red}{prepresum}[right[i]+1]-\color{red}{prepresum}[i+1]) - (right[i]-i)(\color{red}{prepresum}[i+1]-\color{red}{prepresum}[left[i]+1]) & \bf{(5)} \end{aligned} \end{split}

Note

The formula above is not formal at all. I just use it to demonstrate some points:

1. For the first step, which follows the definition of “sum up all subarrays with [l,r] satisfying the requirement mentioned above”, three layers of sum symbol ($$\sum{\sum{\sum}}$$) should be used here to represent the formula. But I think the inner stuff can be represented more clearly by literally sum notation. I don’t want to introduce more variable symbols here.

2. It’s easy to come up with (2) as you have learned prefix sum and the sum of subarray [l:r+1] is a very obvious sign to use it, and now we call the prefix sum array presum.

3. In (3)-(4), we reorganize the terms and extract something invariant with the loop variable from the summed terms respectively, which can remove one layer of $$\sum$$ immediately.

4. The last step (5) is the most difficult one to handle. But if you view the presum as an array (of course, actually it is), you now come across a problem that requires the sum of subarray again. Just apply the prefix sum of presum array!

In summary:

• A: the raw array

• presum: the first-order prefix sum: prefix sum array of A

• prepresum: the second-order prefix sum: prefix sum array of presum.

By accumulating the origin array twice, along with extracting invariables, we eliminate three $$\sum$$s step by step. For the problem of “sum up all target subarraies for a given index i”, we reduce the complexity from $$O((right[i]-left[i])^3)$$ to $$O(1)$$. Certainly, it first costs $$O(n)$$ time to build prepresum array.

As we have the sum of all [l,r] subarrays for each A[i], just multiply the sum by their min value A[i], then sum up the result for each A[i] we will get the final result. I post my code solution here, which is inspired by @lee2152. Note that in the last step (5) of the formula above, all indexes of prepresum must >= 1, so it’s not necessary to consider initializing prepresum[0]=0, we can build prepresum from prepresum[0]=presum[0].

from itertools import accumulate
class Solution:
def totalStrength(self, A: List[int]) -> int:
n = len(A)
M = 10**9+7

left = [-1]*n
right = [n]*n
stack = []
for i,v in enumerate(A):
while stack and stack[-1][0] > A[i]:
right[stack.pop()[1]] = i
if stack:
left[i] = stack[-1][1]
stack.append((v,i))

res = 0
modplus = lambda x,y:(x+y)%M
acc = list(accumulate(accumulate(A,func=modplus),func=modplus,initial = 0))
for i in range(n):
l,r = left[i],right[i]
lacc = acc[i] - acc[max(l,0)]
racc = acc[r] - acc[i]
ln,rn = i-l,r-i
res =  (res + A[i]*(racc*ln - lacc*rn))%M
return res


### More¶

Some other LC prblems applicable for this template (left as exercises):

1

Prefix Sum of Factors 2 and 5

2

[Java/C++/Python] One Pass Solution