An array A consisting of N integers is given. The dominator of array A is the value that occurs in more than half of the elements of A.
For example, consider array A such that
A[0] = 3 A[1] = 4 A[2] = 3
A[3] = 2 A[4] = 3 A[5] = -1
A[6] = 3 A[7] = 3
The dominator of A is 3 because it occurs in 5 out of 8 elements of A (namely in those with indices 0, 2, 4, 6 and 7) and 5 is more than a half of 8.
Write a function
int solution(vector<int> &A);
that, given an array A consisting of N integers, returns index of any element of array A in which the dominator of A occurs. The function should return −1 if array A does not have a dominator.
For example, given array A such that
A[0] = 3 A[1] = 4 A[2] = 3
A[3] = 2 A[4] = 3 A[5] = -1
A[6] = 3 A[7] = 3
the function may return 0, 2, 4, 6 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].
For example, consider array A such that
A[0] = 3 A[1] = 4 A[2] = 3
A[3] = 2 A[4] = 3 A[5] = -1
A[6] = 3 A[7] = 3
The dominator of A is 3 because it occurs in 5 out of 8 elements of A (namely in those with indices 0, 2, 4, 6 and 7) and 5 is more than a half of 8.
Write a function
int solution(vector<int> &A);
that, given an array A consisting of N integers, returns index of any element of array A in which the dominator of A occurs. The function should return −1 if array A does not have a dominator.
For example, given array A such that
A[0] = 3 A[1] = 4 A[2] = 3
A[3] = 2 A[4] = 3 A[5] = -1
A[6] = 3 A[7] = 3
the function may return 0, 2, 4, 6 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].
#include <unordered_map> int solution(vector<int> &A) { // write your code in C++11 (g++ 4.8.2) unordered_map<int, int> um; for (size_t i = 0; i < A.size(); i++) { if (um.find(A[i]) != um.end()) { um[A[i]]++; } else { um[A[i]] = 1; } if (um[A[i]] > A.size() / 2) return i; } return -1; }
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