Given array of integer, count number of distinct sub array such that it has at most k odd elements and two sub array differ if only when they have at least one member differ.
Example:
{3, 2, 3, 4}, k = 1;
output; 7
[ [3], [2], [4], [3,2], [2,3], [2,3,4],[3,4] ]; Note we did not count [3,2,3] since it has more than k odd elements.
Example 2:
{6, 3, 5, 8}, k = 1;
[ [6], [3], [5], [8] , [6,3], [5,8] ] = 6
We did not count [3,5] as it has > k odd elements
Example 2:
{2, 2, 5, 6, 9, 2, 11, 9, 2, 11, 12}
There are these many arrays ;
[2], [2] , [5], [6], [9] , [2], [11], [9], [2], [11], [12]
[2, 2] , [2, 5] , [5, 6] , [6, 9] , [9, 2] , [2, 11] , [11, 9] , [11, 12] , [2, 2, 5] , [2, 5, 6] , [5, 6, 9] , [6, 9, 2] , [9, 2, 11] , [2, 11, 9] , [11, 9, 2] , [2, 11, 12] , [2, 2, 5, 6] , [2, 5, 6, 9] , [5, 6, 9, 2] , [6, 9, 2, 11] , [9, 2, 11, 9] , [2, 11, 9, 2] , [11, 9, 2, 11] , [9, 2, 11, 12] , [2, 2, 5, 6, 9] , [2, 5, 6, 9, 2] , [5, 6, 9, 2, 11] , [6, 9, 2, 11, 9] , [9, 2, 11, 9, 2] , [2, 11, 9, 2, 11] , [11, 9, 2, 11, 12] , [2, 2, 5, 6, 9, 2] , [2, 5, 6, 9, 2, 11] , [5, 6, 9, 2, 11, 9] , [6, 9, 2, 11, 9, 2] , [9, 2, 11, 9, 2, 11] , [2, 11, 9, 2, 11, 12] , [2, 2, 5, 6, 9, 2, 11] , [2, 5, 6, 9, 2, 11, 9] , [5, 6, 9, 2, 11, 9, 2] , [6, 9, 2, 11, 9, 2, 11] , [9, 2, 11, 9, 2, 11, 12] , [2, 2, 5, 6, 9, 2, 11, 9] , [2, 5, 6, 9, 2, 11, 9, 2] , [5, 6, 9, 2, 11, 9, 2, 11] , [6, 9, 2, 11, 9, 2, 11, 12] , [2, 2, 5, 6, 9, 2, 11, 9, 2] , [2, 5, 6, 9, 2, 11, 9, 2, 11] , [5, 6, 9, 2, 11, 9, 2, 11, 12] , [2, 2, 5, 6, 9, 2, 11, 9, 2, 11] , [2, 5, 6, 9, 2, 11, 9, 2, 11, 12]
But only 18 get qualified as there are duplicates like [9, 2, 11] etc.
Qualified arrays are
[2] , [5], [6], [9] , [11], [12] , [2, 2] , [2, 5] , [5, 6] , [6, 9] , [9, 2] , [2, 11] , [11, 12] , [2, 2, 5] , [2, 5, 6] , [6, 9, 2] , [2, 11, 12] , [2, 2, 5, 6]
MY finding so far;
we can use sliding window technique, such that we start counting all sub array of len = 1 to n such that each sub array are different and have at most k odd elements
Here is the code, that i've written for this approach O(n^2)
static int subArraysBrute(int arr[], int k) {
int count = 0;
Set<Integer> set = new HashSet<>();
//Count single length
for (int i = 0; i < arr.length; i++) {
count += set.contains(arr[i]) ? 0 : 1;
set.add(arr[i]);
}
int len = 2;
int odd;
Set<List<Integer>> setArray = new HashSet<>();
while (len < arr.length) {
setArray.clear();
for (int i = 0; i < arr.length - len + 1; i++) {
int j = i + len - 1;
odd = 0;
List<Integer> ar = new ArrayList<>();
for (int x = i; x <= j; x++) {
if (arr[x] % 2 != 0)
odd++;
ar.add(arr[x]);
}
if (!setArray.contains(ar)) {
if (odd <= k) {
count++;
System.out.print(ar + " , ");
}
}
setArray.add(ar);
}
len++;
}
return count;
}
Other findings;
1. We can't sort the array, as they will ruin the subarray property
2. We can't use simple sliding technique as they will mis so many sub arrays ( moving from left to right window) - I've tried this, this fails like any thing;
Probably, in second idea (sliding window), can be improve further such that once we counted sub arrays, we can run one more time and count those sub array which are left out.
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