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Find the sums for which an array can be divided into subarrays of equal sum in Python
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
add(part_1) of answer
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
add(part_2) of answer
if size of answer is same as 0, then
return -1
return answer
Example
Let us see the following implementation to get better understanding −
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
answer = set()
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) :
answer.add(part_1)
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) :
answer.add(part_2)
if(len(answer) == 0) :
return -1
return answer
a = [2, 4, 2, 2, 2, 4, 2, 6]
print(find_sum(a))Input
[2, 4, 2, 2, 2, 4, 2, 6]
Output
{8, 12, 6}
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