A zero-indexed 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] = 1
P = 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 a zero-indexed 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.
Assume that:
- 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].
Complexity:
- 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).
- Elements of input arrays can be modified.
Solution:
public static int solution(int[] A) {
// write your code in Java SE 8
BigInteger B[] = new BigInteger[A.length];
BigInteger C[] = new BigInteger[A.length];
int firstEqui = -1;
// START - CALCULATE B
for (int i = 0; i < A.length; i++) {
if (i == 0) {
B[i] = BigInteger.valueOf(A[i]);
} else {
B[i] = B[i - 1].add(BigInteger.valueOf(A[i]));
}
}
// STOP - CALCULATE B
// START - CALCULATE C
for (int i = 0; i < A.length; i++) {
int c = (A.length - 1) - i;
if (i == 0) {
C[c] = BigInteger.valueOf(A[c]);
} else {
C[c] = BigInteger.valueOf(A[c]).add(C[c + 1]);
}
}
// STOP - CALCULATE C
for (int i = 0; i < A.length; i++) {
if (B[i].equals(C[i])) {
firstEqui = i;
i = A.length;
}
}
return firstEqui;
}