Amazon Interview Question for SDE1s


Country: India
Interview Type: In-Person




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0
of 0 vote

The idea is to accommodate "cheapest" tasks first using a cpu with the least free cores.

def count(tasks, cpus):
    tasks = sorted(tasks)
    heapq.heapify(cpus)
    res = 0
    for t in tasks:
        while cpus and cpus[0] < t:
            heapq.heappop(cpus)
        if not cpus:
            return res
        cores = cpus[0]
        res += 1
        heapq.heapreplace(cpus, cores - t)
    return res

- adr July 23, 2018 | Flag Reply
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0
of 0 vote

def count(tasks, cpus):
tasks = sorted(tasks)
heapq.heapify(cpus)
res = 0
for t in tasks:
while cpus and cpus[0] < t:
heapq.heappop(cpus)
if not cpus:
return res
cores = cpus[0]
res += 1
heapq.heapreplace(cpus, cores - t)
return res

- Anonymous August 07, 2018 | Flag Reply
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0
of 0 vote

public class AIQ {
    public static void main(String args[]) {
        int process[] = {3,5,7};
        int core[] = {1,3,5};
        
        System.out.println(findMaxProcessAllocation(process,core));
    }
    
    public static int findMaxProcessAllocation(int[] process,int[] core){
        int maxCore = findMaxCore(core);
        int result = 0;
        for(int i=0;i<process.length;i++){
            if(process[i] <= maxCore){
                result++;
            }
        }
        return result;
    }
    
    public static int findMaxCore(int[] core){
        int max = 0;
        for(int i=0;i<core.length;i++){
            if(core[i] > max){
                max = core[i];
            }
        }
        return max;
    }
}

- Vasanth August 14, 2018 | Flag Reply
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0
of 0 vote

public class AIQ {
    public static void main(String args[]) {
        int process[] = {3,5,7};
        int core[] = {1,3,5};
        
        System.out.println(findMaxProcessAllocation(process,core));
    }
    
    public static int findMaxProcessAllocation(int[] process,int[] core){
        int maxCore = findMaxCore(core);
        int result = 0;
        for(int i=0;i<process.length;i++){
            if(process[i] <= maxCore){
                result++;
            }
        }
        return result;
    }
    
    public static int findMaxCore(int[] core){
        int max = 0;
        for(int i=0;i<core.length;i++){
            if(core[i] > max){
                max = core[i];
            }
        }
        return max;
    }
}

- Vasanth August 14, 2018 | Flag Reply
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0
of 0 vote

public class AIQ {
public static void main(String args[]) {
int process[] = {3,5,7};
int core[] = {1,3,5};

System.out.println(findMaxProcessAllocation(process,core));
}

public static int findMaxProcessAllocation(int[] process,int[] core){
int maxCore = findMaxCore(core);
int result = 0;
for(int i=0;i<process.length;i++){
if(process[i] <= maxCore){
result++;
}
}
return result;
}

public static int findMaxCore(int[] core){
int max = 0;
for(int i=0;i<core.length;i++){
if(core[i] > max){
max = core[i];
}
}
return max;
}
}

- Vasanth August 14, 2018 | Flag Reply
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0
of 0 vote

and

public class AIQ {
public static void main(String args[]) {
int process[] = {3,5,7};
int core[] = {1,3,5};

System.out.println(findMaxProcessAllocation(process,core));
}

public static int findMaxProcessAllocation(int[] process,int[] core){
int maxCore = findMaxCore(core);
int result = 0;
for(int i=0;i<process.length;i++){
if(process[i] <= maxCore){
result++;
}
}
return result;
}

public static int findMaxCore(int[] core){
int max = 0;
for(int i=0;i<core.length;i++){
if(core[i] > max){
max = core[i];
}
}
return max;
}
}

- Vasanth August 14, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

and

public class AIQ {
public static void main(String args[]) {
int process[] = {3,5,7};
int core[] = {1,3,5};

System.out.println(findMaxProcessAllocation(process,core));
}

public static int findMaxProcessAllocation(int[] process,int[] core){
int maxCore = findMaxCore(core);
int result = 0;
for(int i=0;i<process.length;i++){
if(process[i] <= maxCore){
result++;
}
}
return result;
}

public static int findMaxCore(int[] core){
int max = 0;
for(int i=0;i<core.length;i++){
if(core[i] > max){
max = core[i];
}
}
return max;
}
}

and

- Vasanth August 14, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static int findMaxProcessAllocation(int[] process,int[] core){
int maxCore = findMaxCore(core);
int result = 0;
for(int i=0;i<process.length;i++){
if(process[i] <= maxCore){
result++;
}
}
return result;
}

- Vasanth August 14, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

count=0
cpus.sort(reverse=True): # cpu with most cores first
for cores in tasks.sort(reverse=True): # tasks needing most cores first
    if !cpus: break # done if no cpus left 
    cores <= cpus[0]:
        cpus.pop(0) # use this cpu for this task, it has enough cores
        count+=1
return count

- d March 24, 2019 | Flag Reply
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0
of 0 vote

It seems were trying to optimally allocate CPU given a multiple tasks.

public class AllocateCPU{
    
    public static void main(String []args){
        int[] cpu = new int[] {1, 3, 5};
        int[] task = new int[] {3, 5, 7};
        int match = 0;
        
        for(int i = 0; i < cpu.length; i++) {
            for(int j = 0; j < task.length; j++) {
                if(cpu[i] >= task[j]) {
                    match++;
                    break;
                }
            }
        }
        System.out.println(match);
    }
}

- David Deppe September 10, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

This problem cab solved by applying solution for maximum bipartite graph problem.

- Balaraju July 30, 2018 | Flag Reply


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