I was trying to solve this problem.
An integer M and a non-empty zero-indexed array A consisting of N non-negative integers are given. All integers in array A are less than or equal to M.
A pair of integers (P, Q), such that 0 ≤ P ≤ Q < N, is called a slice of array A. The slice consists of the elements A[P], A[P + 1], ..., A[Q]. A distinct slice is a slice consisting of only unique numbers. That is, no individual number occurs more than once in the slice.
For example, consider integer M = 6 and array A such that:
A[0] = 3 A[1] = 4 A[2] = 5 A[3] = 5 A[4] = 2
There are exactly nine distinct slices: (0, 0), (0, 1), (0, 2), (1, 1), (1,2), (2, 2), (3, 3), (3, 4) and (4, 4).
The goal is to calculate the number of distinct slices.
Thanks in advance.
#include <algorithm>
#include <cstring>
#include <cmath>
#define MAX 100002
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
using namespace std;
bool check[MAX];
int solution(int M, vector<int> &A) {
memset(check, false, sizeof(check));
int base = 0;
int fibot = 0;
int sum = 0;
while(fibot < A.size()){
if(check[A[fibot]]){
base = fibot;
}
check[A[fibot]] = true;
sum += fibot - base + 1;
fibot += 1;
}
return min(sum, 1000000000);
}
Using sort function() Calculate the length of an array using the length() function that will return an integer value as per the elements in an array. Call the sort function and pass the array and the size of an array as a parameter. Take a temporary variable that will store the count of distinct elements.
For every index i we can find the largest index j ⩾ i such that (i, j) is a bounded slice. Any slice with a larger value of j will not be bounded. Using so called brute-force approach, we can check increasing indices, remembering the minimum and the maximum values at all times.
The solution is not correct because your algorithm is wrong.
First of all, let me show you a counter example. Let A = {2, 1, 2}
. The first iteration: base = 0
, fibot = 0
, sum += 1.
That's right. The second one: base = 0, fibot = 1
, sum += 2
. That's correct, too. The last step: fibot = 2
, check[A[fibot]] is true
, thus, base = 2
. But it should be 1
. So your code returns1 + 2 + 1 = 4
while the right answer 1 + 2 + 2 = 5
.
The right way to do it could be like this: start with L = 0
. For each R
from 0
to n - 1
, keep moving the L
to the right until the subarray contais only distinct values (you can maintain the number of occurrences of each value in an array and use the fact that A[R]
is the only element that can occur more than once).
There is one more issue with your code: the sum
variable may overflow if int
is 32-bit type on the testing platform (for instance, if all elements of A
are distinct).
As for the question WHY your algorithm is incorrect, I have no idea why it should be correct in the first place. Can you prove it? The base = fibot
assignment looks quite arbitrary to me.
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