CareerCup

Using DP we can solve this.

Code:

public void findMinCostAndQuantity(int q[], int p[], int tot) {
		int c[] = new int[tot + 1];
		int price[] = new int[tot + 1];
		for (int i = 0; i <= tot; i++) {
			c[i] = i;
			price[i] = 10000000;
		}
		for (int i = 1; i <= tot; i++) {
			for (int j = 0; j < q.length && q[j] <= i; j++) {
				if (i == q[j] && price[i] > p[j]) {
					c[i] = 1;
					price[i] = p[j];
				}
				if ((price[i] > price[i - q[j]] + p[j])
						|| (price[i] == price[i - q[j]] + p[j] && (c[i] > c[i
								- q[j]] + 1))) {

					price[i] = price[i - q[j]] + p[j];
					c[i] = c[i - q[j]] + 1;
				}
			}
		}
		System.out.println(c[tot]);
		System.out.println(price[tot]);
	}

Here I have kept the price to be optimum and then the quantity.

How can both be optimum at same time ??

How to we define the optimal combination??? For example say quantity[] = 1,5,10
and corresponding price[] = 10,30,80 . Now if total quantity is 20. We have 3 options
1) 20 bags cost:200. 2)4 bags Cost 120 and 3.) 2bags and cost160 and many more possibilities. In the above defined options which gets priority 2 or 3???

yes India is right here
It should be either on price or on quantity. I think price should matter here as given that it should have Q quantity

Step 1: Get all the possible combinations of quantity which is equal to Q.
Step 2: Get the minimum amount combination from above combinations.
Step 3: If more than one combination have same price then get the one with lower number of bags.