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

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 −

 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
   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}
Updated on: 2020-08-27T12:24:16+05:30

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