Tasks Details
easy
Minimize the value |(A[0] + ... + A[P-1]) - (A[P] + ... + A[N-1])|.
Task Score
92%
Correctness
100%
Performance
83%
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].
Copyright 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Java 8
Time spent on task 1 minutes
Notes
not defined yet
Code: 08:17:47 UTC,
java,
autosave
Code: 08:18:00 UTC,
java,
autosave
// 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 totalSum = IntStream.of(A).sum();
int frontSum = 0;
int backSum = 0;
int minimumDiff = Integer.MAX_VALUE;
for (int i=0; i<A.length-1; i++) {
frontSum += A[i];
backSum = totalSum - frontSum;
minimumDiff = Math.min(minimumDiff, Math.abs(frontSum-backSum));
}
return minimumDiff;
}
}
Code: 08:18:30 UTC,
java,
autosave
// you can also use imports, for example:
// import java.util.*;
import java.util.stream.IntStream;
// 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 totalSum = IntStream.of(A).sum();
int frontSum = 0;
int backSum = 0;
int minimumDiff = Integer.MAX_VALUE;
for (int i=0; i<A.length-1; i++) {
frontSum += A[i];
backSum = totalSum - frontSum;
minimumDiff = Math.min(minimumDiff, Math.abs(frontSum-backSum));
}
return minimumDiff;
}
}
Code: 08:18:33 UTC,
java,
verify,
result: Passed
// you can also use imports, for example:
// import java.util.*;
import java.util.stream.IntStream;
// 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 totalSum = IntStream.of(A).sum();
int frontSum = 0;
int backSum = 0;
int minimumDiff = Integer.MAX_VALUE;
for (int i=0; i<A.length-1; i++) {
frontSum += A[i];
backSum = totalSum - frontSum;
minimumDiff = Math.min(minimumDiff, Math.abs(frontSum-backSum));
}
return minimumDiff;
}
}
User test case 1:
[1, 5, -982, 631, 1, 0, 4, 5]
User test case 2:
[-1000, 1000]
User test case 3:
[-1000, -1000, 8, 1, 3, 1000, 999, 854]
Analysis
expand all
User tests
1.
0.088 s
OK
function result: 337
function result: 337
1.
0.088 s
OK
function result: 2000
function result: 2000
1.
0.088 s
OK
function result: 843
function result: 843
Code: 08:18:43 UTC,
java,
verify,
result: Passed
// you can also use imports, for example:
// import java.util.*;
import java.util.stream.IntStream;
// 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 totalSum = IntStream.of(A).sum();
int frontSum = 0;
int backSum = 0;
int minimumDiff = Integer.MAX_VALUE;
for (int i=0; i<A.length-1; i++) {
frontSum += A[i];
backSum = totalSum - frontSum;
minimumDiff = Math.min(minimumDiff, Math.abs(frontSum-backSum));
}
return minimumDiff;
}
}
User test case 1:
[1, 5, -982, 631, 1, 0, 4, 5]
User test case 2:
[-1000, 1000]
User test case 3:
[-1000, -1000, 8, 1, 3, 1000, 999, 854]
Analysis
expand all
User tests
1.
0.088 s
OK
function result: 337
function result: 337
1.
0.088 s
OK
function result: 2000
function result: 2000
1.
0.088 s
OK
function result: 843
function result: 843
Code: 08:18:47 UTC,
java,
final,
score:
92
// you can also use imports, for example:
// import java.util.*;
import java.util.stream.IntStream;
// 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 totalSum = IntStream.of(A).sum();
int frontSum = 0;
int backSum = 0;
int minimumDiff = Integer.MAX_VALUE;
for (int i=0; i<A.length-1; i++) {
frontSum += A[i];
backSum = totalSum - frontSum;
minimumDiff = Math.min(minimumDiff, Math.abs(frontSum-backSum));
}
return minimumDiff;
}
}
Analysis summary
The following issues have been detected: timeout errors.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.088 s
OK
2.
0.088 s
OK
3.
0.088 s
OK
1.
0.088 s
OK
2.
0.088 s
OK
1.
0.088 s
OK
2.
0.088 s
OK
1.
0.088 s
OK
2.
0.088 s
OK
3.
0.088 s
OK
4.
0.088 s
OK
1.
0.084 s
OK
1.
0.092 s
OK
1.
0.088 s
OK
expand all
Performance tests
medium_random1
random medium, numbers from 0 to 100, length = ~10,000
random medium, numbers from 0 to 100, length = ~10,000
✘
TIMEOUT ERROR
running time: 0.108 sec., time limit: 0.100 sec.
running time: 0.108 sec., time limit: 0.100 sec.
1.
0.108 s
TIMEOUT ERROR,
running time: 0.108 sec., time limit: 0.100 sec.
1.
0.108 s
OK
1.
0.280 s
OK
2.
0.280 s
OK
1.
0.340 s
OK
2.
0.340 s
OK
1.
0.192 s
OK
1.
0.344 s
OK
2.
0.348 s
OK
3.
0.300 s
OK