Lession3_3(TapeEquilibrium) - WhyAbout/CodilityAlgorithm GitHub Wiki

Task description
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:

int solution(vector<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.

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)
    unsigned int l = A.size();
    vector<int> sump(l, 0);
    sump[0] = A[0];
    for(unsigned int i = 1; i < l; ++i){
        sump[i] = sump[i - 1] + A[i];
    }
    int sumLeft = sump[0];
    int sumRight = sump[l-1] - sump[0];
    int minDiff = abs(sumLeft - sumRight);
    int pDiff = 0;
    for(unsigned int p = 2; p < l; ++p){
        sumLeft = sump[p - 1];
        sumRight = sump[l - 1] - sump[p - 1];
        pDiff = abs(sumLeft - sumRight);
        if (pDiff < minDiff){
            minDiff = pDiff;
        }
    }
    
    return minDiff;
}