Find an element in array such that sum of left array is equal to sum of right array

Given, an array of size n. Find an element that divides the array into two sub-arrays with equal sums.

Examples: 

Input: 1 4 2 5 0
Output: 2
Explanation: If 2 is the partition, subarrays are : [1, 4] and [5]

Input: 2 3 4 1 4 5
Output: 1
Explanation: If 1 is the partition, Subarrays are : [2, 3, 4] and [4, 5]

Input: 1 2 3
Output: -1
Explanation: No sub-arrays possible. return -1

Method 1 (Simple) 
Consider every element starting from the second element. Compute the sum of elements on its left and the sum of elements on its right. If these two sums are the same, return the element.

Steps to solve the problem:

1. iterate through i=1 to n:

       *declare a leftsum variable to zero.

       *iterate through i-1 till zero and add array element to leftsum.

       *declare a rightsum variable to zero.

       *iterate through i+1 till n and add array element to rightsum.

       *check if leftsum is equal to rightsum then return arr[i].

2. return -1 in case of no point.

2
1

Time Complexity: O(n^2)

Auxiliary Space: O(1)

Method 2 (Using Prefix and Suffix Arrays) : 

We form a prefix and suffix sum arrays

Given array: 1 4 2 5
Prefix Sum:  1  5 7 12
Suffix Sum:  12 11 7 5

Now, we will traverse both prefix arrays.
The index at which they yield equal result,
is the index where the array is partitioned 
with equal sum.

Below is the implementation of the above approach:

2

Time Complexity: O(n)

Auxiliary Space: O(n)

Method 3 (Space efficient) 
We calculate the sum of the whole array except the first element in right_sum, considering it to be the partitioning element. Now, we traverse the array from left to right, subtracting an element from right_sum and adding an element to left_sum. At the point where right_sum equals left_sum, we get the partition.

Below is the implementation :  

1

Time complexity: O(n)
Auxiliary Space: O(1)

Method 4 (Both Time and Space Efficient)

We define two pointers i and j to traverse the array from left and right,
left_sum and right_sum to store sum from right and left respectively

If left_sum is lesser then increment i and if right_sum is lesser then decrement j 
and, find a position where left_sum == right_sum and i and j are next to each other

Note: This solution is only applicable if the array contains only positive elements.

Below is the implementation of the above approach: 

1

Since all loops start traversing from the last updated i and j pointers and do not cross each other, they run n times in the end.

Time Complexity – O(n) 
Auxiliary Space – O(1)

Method 5 (The efficient algorithm):

1. Here we define two pointers to the array -> start = 0, end = n-1
2. Two variables to take care of sum -> left_sum = 0, right_sum = 0

Here our algorithm goes like this:

1. We initialize for loop till the entire size of the array
2. Basically we check if left_sum > right_sum => add the current end element to the right_sum and decrement end
3. If right_sum < left_sum => add the current start element to the left_sum and increment start
4. By these two conditions, we make sure that left_sum and right_sum are nearly balanced, so we can arrive at our solution easily
5. To make this work during all test cases we need to add a couple of conditional statements to our logic:
6. If the equilibrium element is found our start will be equal to the end variable and left_sum will be equal right_sum => return the equilibrium element (here we say start == end because we increment/ decrement the pointer after adding the current start/ end value to respective sum. So if, equilibrium element is found start and end should be at the same location)
7. If start is equal to end variable but left_sum is not equal right_sum => no equilibrium element return -1
8. If left_sum is equal right_sum, but start is not equal to end => we are still in the middle of the algorithm even though we found that left_sum is equal right_sum we haven’t got that one required equilibrium element (So, in this case, add the current end element to the right_sum and decrement end (or) add the current start element to the left_sum and increment start, to make our algorithm continue further).
9. Even here there is one test case that needs to be handled:
10. When there is only one element in the array our algorithm exits without entering for a loop.
11. So we can check if our functions enter the loop if not we can directly return the value as the answer.

Below is the implementation of the above approach:

1

Time complexity: O(n)
Auxiliary Space: O(1)

Method 6: We can use divide and conquer to improve the number of traces to find an equilibrium point, as we know, most of the time a point comes from a mid, which can be considered as an idea to solve this problem

• We can consider that the equilibrium point is mid of the list
• If yes (left sum is equal to right sum) — best case – return the index +1 (as that would be actual count by human)
• If not check where the weight is inclined either the left side or right side

    a) perform the increase and decrease to get the new sum and return the index when equal, Also return -1 if inclination changes to what we had in beginning:

Below is the implementation of the above approach:

1

Time complexity: O(n)
Auxiliary Space: O(1)