Codility - Lesson 3 Time complexity - 3. TapeEquilibrium

Source Link: 
https://codility.com/programmers/lessons/3-time_complexity/tape_equilibrium/

Question: 

A non-empty zero-indexed array A consisting of N integers is given. Array A represents numbers on a tape.
Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].
The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|
In other words, it is the absolute difference between the sum of the first part and the sum of the second part.
For example, consider array A such that:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3 We can split this tape in four places:

• P = 1, difference = |3 − 10| = 7
• P = 2, difference = |4 − 9| = 5
• P = 3, difference = |6 − 7| = 1
• P = 4, difference = |10 − 3| = 7

Write a function:

class Solution { public int solution(int[] A); }

that, given a non-empty zero-indexed array A of N integers, returns the minimal difference that can be achieved.
For example, given:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3 the function should return 1, as explained above.
Assume that:

• N is an integer within the range [2..100,000];
• each element of array A is an integer within the range [−1,000..1,000].

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

My Solution: 
Notes: 
1. Using the concept of Sum
    (sum of the 2nd part) = Sum - (sum of the 1st part)
    importantly, difference = |(sum of the 2nd part) - (sum of the 1st part)|
2. First, compute the sum (will be used for several times)
int sum =0; // initial
for(int i=0; i< A.length; i++){
sum = sum + A[i];
} 
3. then, find the minimum difference
initial setting: Integer.MAX_VALUE
int min_diff = Integer.MAX_VALUE; 
4.  try to compute the above values in "one pass"!
no need to use the second for loop (important)
sum_part_one = sum_part_one + A[p-1]; // the sum of part one
sum_part_two = sum - sum_part_one; // the sum of part two
diff = sum_part_one - sum_part_two; // the difference
if(diff <0) // absolute value
diff = -diff; // all the values can be computed (one pass)
min_diff = Math.min(min_diff, diff); // min difference
5. return the min difference 
return min_diff;