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] = 3We 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] = 3the 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].
// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99 (gcc 6.2.0)
int i, sum = 0, result = 10000, min = 0, sum_A = 0;
for(i = 1; i < N; i++)
sum += A[i];
for(i = 0; i < N-1; i++){
min = (sum_A + A[i]) - sum;
sum_A += A[i];
sum -= A[i+1];
if(min < 0)
min *= -1;
if(result > min)
result = min;
}
return result;
}
}// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99 (gcc 6.2.0)
int i, sum = 0, result = 10000, min = 0, sum_A = 0;
for(i = 1; i < N; i++)
sum += A[i];
for(i = 0; i < N-1; i++){
min = (sum_A + A[i]) - sum;
sum_A += A[i];
sum -= A[i+1];
if(min < 0)
min *= -1;
if(result > min)
result = min;
}
return result;
}
}func.c:26:1: error: expected identifier or '(' before '}' token
}
^
// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99 (gcc 6.2.0)
int i, sum = 0, result = 10000, min = 0, sum_A = 0;
for(i = 1; i < N; i++)
sum += A[i];
for(i = 0; i < N-1; i++){
min = (sum_A + A[i]) - sum;
sum_A += A[i];
sum -= A[i+1];
if(min < 0)
min *= -1;
if(result > min)
result = min;
}
return result;
}// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99 (gcc 6.2.0)
int i, sum = 0, result = 10000, min = 0, sum_A = 0;
for(i = 1; i < N; i++)
sum += A[i];
for(i = 0; i < N-1; i++){
min = (sum_A + A[i]) - sum;
sum_A += A[i];
sum -= A[i+1];
if(min < 0)
min *= -1;
if(result > min)
result = min;
}
return result;
}The solution obtained perfect score.