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].
import java.util.*;
class Solution {
public int solution(int[] A) {
int answer = -1;
int n = A.length;
for (int i=1; i<n; i++) {
int sumOfFirstArr = 0;
int sumOfSecondArr = 0;
for (int j=0; j<i; j++) {
sumOfFirstArr += A[j];
}
for (int j=i; j<n; j++) {
sumOfSecondArr += A[j];
}
int abs = Math.abs(sumOfFirstArr - sumOfSecondArr);
if (answer<0 || answer > abs) {
answer = abs;
}
}
return answer;
}
}
import java.util.*;
class Solution {
public int solution(int[] A) {
int answer = -1;
int n = A.length;
for (int i=1; i<n; i++) {
int sumOfFirstArr = 0;
int sumOfSecondArr = 0;
for (int j=0; j<i; j++) {
sumOfFirstArr += A[j];
}
for (int j=i; j<n; j++) {
sumOfSecondArr += A[j];
}
int abs = Math.abs(sumOfFirstArr - sumOfSecondArr);
if (answer<0 || answer > abs) {
answer = abs;
}
}
return answer;
}
}
import java.util.*;
class Solution {
public int solution(int[] A) {
int answer = -1;
int n = A.length;
for (int i=1; i<n; i++) {
int sumOfFirstArr = 0;
int sumOfSecondArr = 0;
for (int j=0; j<i; j++) {
sumOfFirstArr += A[j];
}
for (int j=i; j<n; j++) {
sumOfSecondArr += A[j];
}
int abs = Math.abs(sumOfFirstArr - sumOfSecondArr);
if (answer<0 || answer > abs) {
answer = abs;
}
}
return answer;
}
}
The following issues have been detected: timeout errors.
random medium, numbers from 0 to 100, length = ~10,000
running time: 1.304 sec., time limit: 0.100 sec.
random medium, numbers from -1,000 to 50, length = ~10,000
running time: 1.312 sec., time limit: 0.112 sec.
large sequence, numbers from -1 to 1, length = ~100,000
Killed. Hard limit reached: 6.000 sec.
random large, length = ~100,000
Killed. Hard limit reached: 6.000 sec.
large sequence, length = ~100,000
Killed. Hard limit reached: 6.000 sec.
large test with maximal and minimal values, length = ~100,000
Killed. Hard limit reached: 6.000 sec.