Problem: TapeEquilibrium

P

A non-empty 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(int A[], int N);

that, given a non-empty 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.

Write an efficient algorithm for the following assumptions:

  • 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].
#include <limits.h>  
int solution(int A[], int N) 
{
    int sumL = 0;
    int sumR = 0;
    int minD = INT_MAX;
    
    for(int i=0;i<N;i++)
         sumR+=A[i];      
    
    for(int i=0;i<N-1;i++)     
    {
         sumL+=A[i];
         sumR-=A[i];
         int r = abs(sumL-sumR);
         minD = minD > r ? r :minD;
    }
    return minD;
}

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