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.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
int difference = 0;
int sumPart1 = 0;
int sumPart2 = 0;
int sum = 0;
for (int i = 0; i < A.length; i++) {
sum += A[i];
}
sumPart1 = A[0];
sumPart2 = sum - sumPart1;
difference = Math.abs(sumPart1 - sumPart2);
for (int p = 2; p < A.length; p++) {
sumPart1 += A[p - 1];
sumPart2 -= A[p - 1];
int temp = Math.abs(sumPart1 - sumPart2);
if (temp < difference) {
difference = temp;
}
}
return difference;
}
}
// you can also use imports, for example:
// import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
int difference = 0;
int sumPart1 = 0;
int sumPart2 = 0;
int sum = 0;
for (int i = 0; i < A.length; i++) {
sum += A[i];
}
sumPart1 = A[0];
sumPart2 = sum - sumPart1;
difference = Math.abs(sumPart1 - sumPart2);
for (int p = 2; p < A.length; p++) {
sumPart1 += A[p - 1];
sumPart2 -= A[p - 1];
int temp = Math.abs(sumPart1 - sumPart2);
if (temp < difference) {
difference = temp;
}
}
return difference;
}
}
// you can also use imports, for example:
// import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
int difference = 0;
int sumPart1 = 0;
int sumPart2 = 0;
int sum = 0;
for (int i = 0; i < A.length; i++) {
sum += A[i];
}
sumPart1 = A[0];
sumPart2 = sum - sumPart1;
difference = Math.abs(sumPart1 - sumPart2);
for (int p = 2; p < A.length; p++) {
sumPart1 += A[p - 1];
sumPart2 -= A[p - 1];
int temp = Math.abs(sumPart1 - sumPart2);
if (temp < difference) {
difference = temp;
}
}
return difference;
}
}
The solution obtained perfect score.