Interview Question


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private int findMaxWeapons(int x, int y, int z) throws Exception {
        if (x > y) {
            return x - z * (x / y);
        } else if (x <= y) {
            return x - z;
        } else {
            return -1;
        }
    }

- AEK September 07, 2019 | Flag Reply
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x=3000
y = 1000
z = 1000

output is 833

but with code above it is zero

- anoophky September 07, 2019 | Flag
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If y = 1000 and z = 1000, then even a fully loaded truck would lose all the weapons before reaching its destination, so the output should be 0.

The code above, however, does not account for a possible not fully loaded last trip. Example:
x = 3500
y = 1000
z = 100

Output should be 3100 but the code will output 2700

- Anonymous September 07, 2019 | Flag
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for 3000, 1000, 1000. First fill 1000 in truck move to 500 and drop 500 there repeat 3 times. Now problem reduced to X= 1500 y = 1000, Z = 500 now answer will be 750. But can't satisfy 833

- anoophky September 09, 2019 | Flag
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void f(){
dp[0]=w;
for(int i=0;i<=d;i++){
for ( int j = i+1; j <=d;j++){
int x = (dp[i]/c)*(j-i),y = (dp[i]%c>=j-i)? (j-i) : dp[i]%c ;
dp[j] = max(dp[j],dp[i]-x-y );
}
}
cout<<dp[d];
}

- chandankumar80567 September 08, 2019 | Flag Reply
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explain your code.

- anoophky September 09, 2019 | Flag
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the idea is similar to Longest increasing subsequence problem, you have to check through all possible combinations, and store the most optimal solution.
dp[i] stores the max no of weapons which can be transported from 0 to i.
x, y stores the weapons lost when transporting from i to j.

- chandankumar80567 September 09, 2019 | Flag
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public class MaxWeapons {

  private static int getMaxWeapons(int x, int y, int z) {
    if (x < y) {
      return Math.max(x - z, 0);
    }
    int count = x;
    for (int i = 1; i <= z; i++) {
      count = count - (int) Math.ceil(((double) count) / y);
      if (count <= 0) {
        break;
      }
    }
    return Math.max(0, count);
  }

  public static void main(String[] args) {
    System.out.println(getMaxWeapons(3000, 1000, 1000));
    System.out.println(getMaxWeapons(1000, 1200, 1));
    System.out.println(getMaxWeapons(1000, 1200, 100));
  }

}

To get maximum weapons to the destination we can move all the weapons to one KM and then again full load the truck and carry. So we need to keep doing unloading/loading at 1 KM of distance.

- xyz September 16, 2019 | Flag Reply
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Obvious linear time solution, not sure how to compute it faster:

x = 15
y = 3
z = 2

for i in range(z):
    if x == 0: 
        break
    x = x - (x+y-1)//y
print (x)

- adr September 17, 2019 | Flag Reply


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