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:
class Solution { public int solution(int[] A); }
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 also use imports, for example:
// import java.math.*;
class Solution {
public int solution(int[] A) {
// write your code in Java SE 7
int N = A.length;
int sum1 = A[0];
int sum2 = 0;
int P = 1;
for (int i = P; i < N; i++) {
sum2 += A[i];
}
int diff = Math.abs(sum1 - sum2);
for (; P < N-1; P++) {
sum1 += A[P];
sum2 -= A[P];
int newDiff = Math.abs(sum1 - sum2);
if (newDiff < diff) {
diff = newDiff;
}
}
return diff;
}
}
None
// you can also use imports, for example:
// import java.math.*;
class Solution {
public int solution(int[] A) {
// write your code in Java SE 7
int N = A.length;
int sum1 = A[0];
int sum2 = 0;
int P = 1;
for (int i = P; i < N; i++) {
sum2 += A[i];
}
int diff = Math.abs(sum1 - sum2);
for (; P < N-1; P++) {
sum1 += A[P];
sum2 -= A[P];
int newDiff = Math.abs(sum1 - sum2);
if (newDiff < diff) {
diff = newDiff;
}
}
return diff;
}
}
None
None
// you can also use imports, for example:
// import java.math.*;
class Solution {
public int solution(int[] A) {
// write your code in Java SE 7
int N = A.length;
int sum1 = A[0];
int sum2 = 0;
int P = 1;
for (int i = P; i < N; i++) {
sum2 += A[i];
}
int diff = Math.abs(sum1 - sum2);
for (; P < N-1; P++) {
sum1 += A[P];
sum2 -= A[P];
int newDiff = Math.abs(sum1 - sum2);
if (newDiff < diff) {
diff = newDiff;
}
}
return diff;
}
}
None
None
// you can also use imports, for example:
// import java.math.*;
class Solution {
public int solution(int[] A) {
// write your code in Java SE 7
int N = A.length;
int sum1 = A[0];
int sum2 = 0;
int P = 1;
for (int i = P; i < N; i++) {
sum2 += A[i];
}
int diff = Math.abs(sum1 - sum2);
for (; P < N-1; P++) {
sum1 += A[P];
sum2 -= A[P];
int newDiff = Math.abs(sum1 - sum2);
if (newDiff < diff) {
diff = newDiff;
}
}
return diff;
}
}
The solution obtained perfect score.