HackerRank: Non-Divisible Subset

考慮事項：

• remainder[k/2] = 1
• remainder[0] 裏面只能抽出一個
• 即 remainder[0] = std::min(remainder[0], 1)
• and remainder[k/2] = std::min(remainder[k/2], 1) when k%2 == 0.

計算

for (int i = 0; i<=k/2, i++)
	maxRes += std::max(remainder[i], remainder[k-i]);

static int subsetPairNotDivisibleByK(int arr[], int n, int k) {
    int maximumSubset = 0;
    //creating an array with all possible remainder fo k
    int[] remainders = new int[k];

    //fill the array initially with 0
    Arrays.fill(remainders, 0);

    //loop through the array arr and get the frequency of the remainders
    for(int val: arr){
        int rem = val%k;
        remainders[rem]++;
    }
    //consider a case when k is even. eg. k=6 then remainder 3+3 is 6 so can not be cnsidered.
    if(k%2==0){
        remainders[k/2] = Math.min(remainders[k/2], 1);
    }
    //consider a case where we take number that is divisible by k into consideration,
    //there is only 1 number can be considered e.g. k=3 then we can either consider 3 or 9 both can not be considered together.
    maximumSubset = Math.min(remainders[0], 1);

    //loop through the array of remainders array and get the max of remainder[i] or remainder[i-k]
    for(int i=1;i<=k/2;i++){
        maximumSubset+=Math.max(remainders[i], remainders[k-i]);
    }

    return maximumSubset;

}