Numeric Interview Question for Backend Developers


Country: United States




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I could not solve it in the interview but solved it later. The idea is to find the unit cost of each shop. Then run a loop and fetch the min most unit cost that is affordable; until all the money is exhausted or no shop is affordable.

Code in Golang:

func budgetShopping(n int32, bundleQuantities []int32,
	bundleCosts []int32) int32 {

	var totalItems int32
	var perUnitCost []float32

	for i := 0; i < len(bundleCosts); i++ {
		perUnitCost = append(perUnitCost,
			float32(bundleCosts[i])/float32(bundleQuantities[i]))
	}

	for {

		var curMinUnitCost float32
		t := -1

		for j := 0; j < len(bundleCosts); j++ {
			if n >= bundleCosts[j] {
				if t == -1 {
					curMinUnitCost = perUnitCost[j]
					t = j
				} else {
					if perUnitCost[j] < curMinUnitCost {
						curMinUnitCost = perUnitCost[j]
						t = j
					}
				}
			}
		}

		if t == -1 {
			return totalItems
		}

		totalItems += bundleQuantities[t]
		n -= bundleCosts[t]
	}
}

- git January 24, 2019 | Flag Reply
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The greedy algorithm you've provided is not always optimal - can you think of a case where it fails?

This is the classic "knapsack" problem, which is solved by dynamic programing.

- Yawn January 24, 2019 | Flag
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Something like this would go well,(Java)

private static int calculate(int amount, int[] q, int[] c) {
    int ret = 0;
    int bal = 0;
    for (int i = 0; i < n; i++) {
        int n = amount / c[i];
        int quantity = q[i] * n;
        if (quantity > ret) {
            ret = quantity;
            bal = amount % c[i];
        }
    }
    
    if (ret!=0&&bal!=0) ret+=calculate(bal,q,c);
    
    return ret;
}

- PeyarTheriyaa January 24, 2019 | Flag Reply
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Is this an optimal solution?

- Les January 26, 2019 | Flag
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of 0 votes

its an (n log(n)) solution assumes 2 things
1 need to use up all the money
2 can come back to the beginning of the shops if there is a balance of money

but to my knowledge this might be the optimal solution

- PeyarTheriyaa January 26, 2019 | Flag
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public int maxNumberChocolates(int amount, int[] quantities, int[] cost, int n) {
        if(amount <= 0 || n < 0) {
            return 0;
        }
        if(amount < cost[n]) {
            return maxNumberChocolates(amount, quantities, cost, n-1);
        } else {
            return Math.max(
                    maxNumberChocolates(amount - cost[n], quantities, cost, n) + quantities[n],
                    maxNumberChocolates(amount, quantities, cost, n-1));
        }
    }

    public int maxNumberChocolatesDpBottomUp(int amount, int[] quantities, int[] cost, int n) {
        int[][] table = new int[amount+1][quantities.length+1];
        for(int i=0; i <= amount; i++) {

            for(int j=0; j <= quantities.length; j++) {
                if(i == 0 || j == 0) {
                    table[i][j] = 0;
                    continue;
                }

                if(i < cost[j-1]) {
                    table[i][j] = table[i][j-1];
                } else {
                    table[i][j] = Math.max(
                                    table[i-cost[j-1]][j] + quantities[j-1],
                                    table[i][j-1]);
                }
            }
        }
        return table[amount][quantities.length];
    }

- PraveenB January 25, 2019 | Flag Reply
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public int maxNumberChocolates(int amount, int[] quantities, int[] cost, int n) {
        if(amount <= 0 || n < 0) {
            return 0;
        }
        if(amount < cost[n]) {
            return maxNumberChocolates(amount, quantities, cost, n-1);
        } else {
            return Math.max(
                    maxNumberChocolates(amount - cost[n], quantities, cost, n) + quantities[n],
                    maxNumberChocolates(amount, quantities, cost, n-1));
        }
    }

    public int maxNumberChocolatesDpBottomUp(int amount, int[] quantities, int[] cost, int n) {
        int[][] table = new int[amount+1][quantities.length+1];
        for(int i=0; i <= amount; i++) {

            for(int j=0; j <= quantities.length; j++) {
                if(i == 0 || j == 0) {
                    table[i][j] = 0;
                    continue;
                }

                if(i < cost[j-1]) {
                    table[i][j] = table[i][j-1];
                } else {
                    table[i][j] = Math.max(
                                    table[i-cost[j-1]][j] + quantities[j-1],
                                    table[i][j-1]);
                }
            }
        }
        return table[amount][quantities.length];

}

- PraveenB January 25, 2019 | Flag Reply


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