Question

I have this problem , where given an array of positive numbers i have to find the maximum sum of elements such that no two adjacent elements are picked. The maximum has to be less than a certain given K. I tried thinking on the lines of the similar problem without the k , but i have failed so far.I have the following dp-ish soln for the latter problem

    int sum1,sum2 = 0;
    int sum = sum1 = a[0];

    for(int i=1; i<n; i++)
    {
        sum = max(sum2 + a[i], sum1);
        sum2 = sum1;
        sum1 = sum;
    }

Could someone give me tips on how to proceed with my present problem??

Answer 1

The best I can think of off the top of my head is an O(n*K) dp:

int sums[n][K+1] = {{0}};
int i, j;
for(j = a[0]; j <= K; ++j) {
    sums[0][j] = a[0];
}
if (a[1] > a[0]) {
    for(j = a[0]; j < a[1]; ++j) {
        sums[1][j] = a[0];
    }
    for(j = a[1]; j <= K; ++j) {
        sums[1][j] = a[1];
    }
} else {
    for(j = a[1]; j < a[0]; ++j) {
        sums[1][j] = a[1];
    }
    for(j = a[0]; j <= K; ++j) {
        sums[1][j] = a[0];
    }
}
for(i = 2; i < n; ++i) {
    for(j = 0; j <= K && j < a[i]; ++j) {
        sums[i][j] = max(sums[i-1][j],sums[i-2][j]);
    }
    for(j = a[i]; j <= K; ++j) {
        sums[i][j] = max(sums[i-1][j],a[i] + sums[i-2][j-a[i]]);
    }
}

sums[i][j] contains the maximal sum of non-adjacent elements of a[0..i] not exceeding j. The solution is then sums[n-1][K] at the end.