Tasks Details
medium
Find the maximal sum of any double slice.
Task Score
100%
Correctness
100%
Performance
100%
A non-empty array A consisting of N integers is given.
A triplet (X, Y, Z), such that 0 ≤ X < Y < Z < N, is called a double slice.
The sum of double slice (X, Y, Z) is the total of A[X + 1] + A[X + 2] + ... + A[Y − 1] + A[Y + 1] + A[Y + 2] + ... + A[Z − 1].
For example, array A such that:
A[0] = 3 A[1] = 2 A[2] = 6 A[3] = -1 A[4] = 4 A[5] = 5 A[6] = -1 A[7] = 2contains the following example double slices:
- double slice (0, 3, 6), sum is 2 + 6 + 4 + 5 = 17,
- double slice (0, 3, 7), sum is 2 + 6 + 4 + 5 − 1 = 16,
- double slice (3, 4, 5), sum is 0.
The goal is to find the maximal sum of any double slice.
Write a function:
class Solution { public int solution(int[] A); }
that, given a non-empty array A consisting of N integers, returns the maximal sum of any double slice.
For example, given:
A[0] = 3 A[1] = 2 A[2] = 6 A[3] = -1 A[4] = 4 A[5] = 5 A[6] = -1 A[7] = 2the function should return 17, because no double slice of array A has a sum of greater than 17.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [3..100,000];
- each element of array A is an integer within the range [−10,000..10,000].
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 2 minutes
Notes
not defined yet
Code: 13:18:21 UTC,
java,
autosave
Code: 13:18:26 UTC,
java,
autosave
import java.util.*;
class Solution {
public int solution(int[] A) {
int[] fowardMaxSum = new int[A.length];
int[] backwardMaxSum = new int[A.length];
int maxSum = 0;
for (int i=1; i<A.length-1; i++) {
fowardMaxSum[i] = Math.max(fowardMaxSum[i-1] + A[i], 0);
}
for (int i=A.length-2; i>1; i--) {
backwardMaxSum[i] = Math.max(backwardMaxSum[i+1] + A[i], 0);
}
for (int i=2; i<A.length-1; i++) {
maxSum = Math.max(maxSum, fowardMaxSum[i-1] + backwardMaxSum[i+1]);
}
return maxSum;
}
}
Code: 13:18:30 UTC,
java,
verify,
result: Passed
import java.util.*;
class Solution {
public int solution(int[] A) {
int[] fowardMaxSum = new int[A.length];
int[] backwardMaxSum = new int[A.length];
int maxSum = 0;
for (int i=1; i<A.length-1; i++) {
fowardMaxSum[i] = Math.max(fowardMaxSum[i-1] + A[i], 0);
}
for (int i=A.length-2; i>1; i--) {
backwardMaxSum[i] = Math.max(backwardMaxSum[i+1] + A[i], 0);
}
for (int i=2; i<A.length-1; i++) {
maxSum = Math.max(maxSum, fowardMaxSum[i-1] + backwardMaxSum[i+1]);
}
return maxSum;
}
}
User test case 1:
[6, 1, 5, 6, 4, 2, 9, 4]
Analysis
Code: 13:19:58 UTC,
java,
autosave
import java.util.*;
class Solution {
public int solution(int[] A) {
int[] fowardMaxSum = new int[A.length];
int[] backwardMaxSum = new int[A.length];
int maxSum = 0;
for (int i=1; i<A.length-1; i++) {
fowardMaxSum[i] = Math.max(fowardMaxSum[i-1] + A[i], 0);
}
for (int i=A.length-2; i>1; i--) {
backwardMaxSum[i] = Math.max(backwardMaxSum[i+1] + A[i], 0);
}
for (int i=1; i<A.length-1; i++) {
maxSum = Math.max(maxSum, fowardMaxSum[i-1] + backwardMaxSum[i+1]);
}
return maxSum;
}
}
Code: 13:20:01 UTC,
java,
verify,
result: Passed
import java.util.*;
class Solution {
public int solution(int[] A) {
int[] fowardMaxSum = new int[A.length];
int[] backwardMaxSum = new int[A.length];
int maxSum = 0;
for (int i=1; i<A.length-1; i++) {
fowardMaxSum[i] = Math.max(fowardMaxSum[i-1] + A[i], 0);
}
for (int i=A.length-2; i>1; i--) {
backwardMaxSum[i] = Math.max(backwardMaxSum[i+1] + A[i], 0);
}
for (int i=1; i<A.length-1; i++) {
maxSum = Math.max(maxSum, fowardMaxSum[i-1] + backwardMaxSum[i+1]);
}
return maxSum;
}
}
User test case 1:
[6, 1, 5, 6, 4, 2, 9, 4]
Analysis
Code: 13:20:08 UTC,
java,
verify,
result: Passed
import java.util.*;
class Solution {
public int solution(int[] A) {
int[] fowardMaxSum = new int[A.length];
int[] backwardMaxSum = new int[A.length];
int maxSum = 0;
for (int i=1; i<A.length-1; i++) {
fowardMaxSum[i] = Math.max(fowardMaxSum[i-1] + A[i], 0);
}
for (int i=A.length-2; i>1; i--) {
backwardMaxSum[i] = Math.max(backwardMaxSum[i+1] + A[i], 0);
}
for (int i=1; i<A.length-1; i++) {
maxSum = Math.max(maxSum, fowardMaxSum[i-1] + backwardMaxSum[i+1]);
}
return maxSum;
}
}
User test case 1:
[6, 1, 5, 6, 4, 2, 9, 4]
Analysis
Code: 13:20:13 UTC,
java,
final,
score:
100
import java.util.*;
class Solution {
public int solution(int[] A) {
int[] fowardMaxSum = new int[A.length];
int[] backwardMaxSum = new int[A.length];
int maxSum = 0;
for (int i=1; i<A.length-1; i++) {
fowardMaxSum[i] = Math.max(fowardMaxSum[i-1] + A[i], 0);
}
for (int i=A.length-2; i>1; i--) {
backwardMaxSum[i] = Math.max(backwardMaxSum[i+1] + A[i], 0);
}
for (int i=1; i<A.length-1; i++) {
maxSum = Math.max(maxSum, fowardMaxSum[i-1] + backwardMaxSum[i+1]);
}
return maxSum;
}
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.008 s
OK
2.
0.004 s
OK
3.
0.008 s
OK
1.
0.004 s
OK
2.
0.004 s
OK
1.
0.008 s
OK
1.
0.004 s
OK
1.
0.004 s
OK
1.
0.008 s
OK
expand all
Performance tests
1.
0.004 s
OK
1.
0.008 s
OK
1.
0.012 s
OK
1.
0.176 s
OK
1.
0.244 s
OK
1.
0.268 s
OK
1.
0.168 s
OK
2.
0.188 s
OK