# Find the sums for which an array can be divided into subarrays of equal sum in Python

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Suppose we have an array of integers A; we have to find all the values for sum so that for a value sum[i] the array can be divided into sub-arrays of sum sum[i]. If we cannot divide the array into sub-arrays of equal sum then return -1.

So, if the input is like A = [2, 4, 2, 2, 2, 4, 2, 6], then the output will be [6,8,12] as the array can be divided into sub-arrays of sum 6, 8 and 12. These are as follows: [{2, 4}, {2, 2, 2}, {4, 2}, {6}] [{2, 4, 2}, {2, 2, 4},{2, 6}] [{2, 4, 2, 2, 2},{4, 2, 6

To solve this, we will follow these steps −

• n := size of a

• table := an array of size n and filled with 0

• table[0] := a[0]

• for i in range 1 to n, do

• table[i] := a[i] + table[i - 1]

• S := table[n - 1]

• my_map := a new map

• for i in range 0 to n, do

• my_map[table[i]] := 1

• answer := a new set

• for i in range 1 to integer part of (square root of (S)) + 1, do

• if S mod i is same as 0, then

• is_present := True

• part_1 := i

• part_2 := quotient of S / i

• for j in range part_1 to S + 1, update in each step by part_1, do

• if j not in my_map, then

• is_present := False

• come out from the loop

• if is_present is true and part_1 is not same as S, then

• is_present := True

• for j in range quotient of (S / i) to S + 1, update in each step by S // i, do

• if j not in my_map, then

• is_present := False;

• come out from the loop

• if is_present and part_2 is not same as S, then

• if size of answer is same as 0, then

• return -1

## Example

Let us see the following implementation to get better understanding −

Live Demo

from math import sqrt
def find_sum(a) :
n = len(a)
table = [0] * n
table[0] = a[0]
for i in range(1, n) :
table[i] = a[i] + table[i - 1]
S = table[n - 1]
my_map = {}
for i in range(n) :
my_map[table[i]] = 1
for i in range(1, int(sqrt(S)) + 1) :
if (S % i == 0) :
is_present = True;
part_1 = i
part_2 = S // i
for j in range(part_1 , S + 1, part_1) :
if j not in my_map :
is_present = False
break
if (is_present and part_1 != S) :
is_present = True
for j in range(S // i , S + 1 , S // i) :
if j not in my_map:
is_present = False;
break
if (is_present and part_2 != S) :
return -1
print(find_sum(a))
[2, 4, 2, 2, 2, 4, 2, 6]
{8, 12, 6}