Task description
A non-empty zero-indexed 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:
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.
Assume that:
class Solution {
public int solution(int[] A) { // write your code in Java SE 8 int len=A.length; int sum=0; for(int i=0;i<len;i++){ sum=sum+A[i]; } int diff=0; int retValue=0; int diffS=0; for(int i=0;i<len-1;i++){ diffS=diffS+A[i]; diff=sum-diffS; diff=Math.abs(diff-diffS); if(retValue==0) retValue=diff; if(retValue>diff) retValue=diff; } return retValue; } }
83% 83 out of 100 points
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:
Write a function:
- P = 1, difference = |3 − 10| = 7
- P = 2, difference = |4 − 9| = 5
- P = 3, difference = |6 − 7| = 1
- P = 4, difference = |10 − 3| = 7
that, given a non-empty zero-indexed array A of N integers, returns the minimal difference that can be achieved.class Solution { public int solution(int[] A); }
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.
Assume that:
Complexity:
- 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].
Elements of input arrays can be modified.
- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
class Solution {
public int solution(int[] A) { // write your code in Java SE 8 int len=A.length; int sum=0; for(int i=0;i<len;i++){ sum=sum+A[i]; } int diff=0; int retValue=0; int diffS=0; for(int i=0;i<len-1;i++){ diffS=diffS+A[i]; diff=sum-diffS; diff=Math.abs(diff-diffS); if(retValue==0) retValue=diff; if(retValue>diff) retValue=diff; } return retValue; } }
83% 83 out of 100 points
Detected time complexity:
O(N)
O(N)
expand allExample tests
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large_extreme
large test with maximal and minimal values, length = ~100,000
large test with maximal and minimal values, length = ~100,000
WRONG ANSWER
got 2000 expected 0
got 2000 expected 0
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