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) {
// write your code in Java SE 8
int minNum = Integer.MAX_VALUE;
int totalSum = 0, leftSum=0;
for(int i : A)
totalSum += i;
for(int i=0; i<A.length-1; i++){
leftSum += A[i];
minNum = Math.min(minNum, Math.abs(leftSum - (totalSum-leftSum)));
}
return minNum;
}
}
// 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) {
// write your code in Java SE 8
int minNum = Integer.MAX_VALUE;
int totalSum = 0, leftSum=0;
for(int i : A)
totalSum += i;
for(int i=0; i<A.length-1; i++){
leftSum += A[i];
minNum = Math.min(minNum, Math.abs(leftSum - (totalSum-leftSum)));
}
return minNum;
}
}
// 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) {
// write your code in Java SE 8
int minNum = Integer.MAX_VALUE;
int totalSum = 0, leftSum=0;
for(int i : A)
totalSum += i;
for(int i=0; i<A.length-1; i++){
leftSum += A[i];
minNum = Math.min(minNum, Math.abs(leftSum - (totalSum-leftSum)));
}
return minNum;
}
}
// 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) {
// write your code in Java SE 8
int minNum = Integer.MAX_VALUE;
int totalSum = 0, leftSum=0;
for(int i : A)
totalSum += i;
for(int i=0; i<A.length-1; i++){
leftSum += A[i];
minNum = Math.min(minNum, Math.abs(leftSum - (totalSum-leftSum)));
}
return minNum;
}
}
The solution obtained perfect score.