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 use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.HashMap;
import java.util.Map;
public class Solution {
public final static int solution(int[] A) {
int min_dif = 0;
int sum = 0;
int[] sums = new int[A.length];
for (int p=0; p<A.length; p++){
sum += A[p];
sums[p] = sum;
}
for (int i=0; i<sums.length; i++){
int val = sums[i];
int dif = val-(sum-val);
if (dif<0)
dif = -dif;
if (min_dif==0 || dif<min_dif){
min_dif = dif;
}
}
return min_dif;
}
}
// you can also use imports, for example:
// import java.util.*;
// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.HashMap;
import java.util.Map;
public class Solution {
public final static int solution(int[] A) {
int min_dif = 0;
int total = 0;
int[] sums = new int[A.length];
for (int p=0; p<A.length; p++){
total += A[p];
sums[p] = sum;
}
for (int i=0; i<sums.length; i++){
int val = sums[i];
int dif = val-(total-val);
if (dif<0)
dif = -dif;
if (min_dif==0 || dif<min_dif){
min_dif = dif;
}
}
return min_dif;
}
}
Solution.java:17: error: cannot find symbol sums[p] = sum; ^ symbol: variable sum location: class Solution 1 error
// you can also use imports, for example:
// import java.util.*;
// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.HashMap;
import java.util.Map;
public class Solution {
public final static int solution(int[] A) {
int min_dif = 0;
int total = 0;
int[] cumuls = new int[A.length];
for (int p=0; p<A.length; p++){
total += A[p];
cumuls[p] = total;
}
for (int i=0; i<cumuls.length; i++){
int val = cumuls[i];
int dif = val-(total-val);
if (dif<0)
dif = -dif;
if (min_dif==0 || dif<min_dif){
min_dif = dif;
}
}
return min_dif;
}
}
// you can also use imports, for example:
// import java.util.*;
// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.HashMap;
import java.util.Map;
public class Solution {
public final static int solution(int[] A) {
int min_dif = 0;
int total = 0;
int[] cumuls = new int[A.length];
for (int p=0; p<A.length; p++){
total += A[p];
cumuls[p] = total;
}
for (int i=0; i<cumuls.length; i++){
int val = cumuls[i];
int dif = val-(total-val);
if (dif<0)
dif = -dif;
if (min_dif==0 || dif<min_dif){
min_dif = dif;
}
}
return min_dif;
}
}
// you can also use imports, for example:
// import java.util.*;
// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.HashMap;
import java.util.Map;
public class Solution {
public final static int solution(int[] A) {
int min_dif = 0;
int total = 0;
int[] cumuls = new int[A.length];
for (int p=0; p<A.length; p++){
total += A[p];
cumuls[p] = total;
}
for (int i=0; i<cumuls.length; i++){
int val = cumuls[i];
int dif = val-(total-val);
if (dif<0)
dif = -dif;
if (min_dif==0 || dif<min_dif){
min_dif = dif;
}
}
return min_dif;
}
}
The following issues have been detected: wrong answers.
large sequence, numbers from -1 to 1, length = ~100,000
got 2 expected 0
large test with maximal and minimal values, length = ~100,000
got 2000 expected 0