This is a demo task.
An array A consisting of N integers is given. An equilibrium index of this array is any integer P such that 0 ≤ P < N and the sum of elements of lower indices is equal to the sum of elements of higher indices, i.e.
A[0] + A[1] + ... + A[P−1] = A[P+1] + ... + A[N−2] + A[N−1].
Sum of zero elements is assumed to be equal to 0. This can happen if P = 0 or if P = N−1.
For example, consider the following array A consisting of N = 8 elements:
A[0] = -1 A[1] = 3 A[2] = -4 A[3] = 5 A[4] = 1 A[5] = -6 A[6] = 2 A[7] = 1P = 1 is an equilibrium index of this array, because:
- A[0] = −1 = A[2] + A[3] + A[4] + A[5] + A[6] + A[7]
P = 3 is an equilibrium index of this array, because:
- A[0] + A[1] + A[2] = −2 = A[4] + A[5] + A[6] + A[7]
P = 7 is also an equilibrium index, because:
- A[0] + A[1] + A[2] + A[3] + A[4] + A[5] + A[6] = 0
and there are no elements with indices greater than 7.
P = 8 is not an equilibrium index, because it does not fulfill the condition 0 ≤ P < N.
Write a function:
class Solution { public int solution(int[] A); }
that, given an array A consisting of N integers, returns any of its equilibrium indices. The function should return −1 if no equilibrium index exists.
For example, given array A shown above, the function may return 1, 3 or 7, as explained above.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
class Solution {
public int solution(int[] A) {
long left, right = 0;
long sum = sum(A);
for(int i = 0; i < A.length; i++) {
right = sum - A[i] - left;
if(left == right) return i;
left += A[i];
}
return -1;
}
private long sum(int[] A) {
long sum = 0;
for(int i = 0; i < A.length; i++) {
sum += A[i];
}
return sum;
}
}
Solution.java:7: error: variable left might not have been initialized right = sum - A[i] - left; ^ 1 error
class Solution {
public int solution(int[] A) {
long left = 0;
long right = 0;
long sum = sum(A);
for (int i = 0; i < A.length; i++) {
right = sum - left - A[i];
if (left == right)
return i;
left += A[i];
}
return -1;
}
private long sum(int[] A) {
long sum = 0;
for (int i = 0; i < A.length; i++) {
sum += A[i];
}
return sum;
}
}
class Solution {
public int solution(int[] A) {
long left = 0;
long right = 0;
long sum = sum(A);
for (int i = 0; i < A.length; i++) {
right = sum - left - A[i];
if (left == right)
return i;
left += A[i];
}
return -1;
}
private long sum(int[] A) {
long sum = 0;
for (int i = 0; i < A.length; i++) {
sum += A[i];
}
return sum;
}
}
class Solution {
public int solution(int[] A) {
long left = 0;
long right = 0;
long sum = sum(A);
for (int i = 0; i < A.length; i++) {
right = sum - left - A[i];
if (left == right)
return i;
left += A[i];
}
return -1;
}
private long sum(int[] A) {
long sum = 0;
for (int i = 0; i < A.length; i++) {
sum += A[i];
}
return sum;
}
}
The solution obtained perfect score.