xyz Interview Question for abcs


Country: United States
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
0
of 0 vote

Depending on meaning, either:

def count_different(input):
    return len(set(input))

or

def count_different(input):
    if not input:
        return 0
    front = back = max_length = this_length = 0
    running_set = set()
    while front < len(input):
        if input[front] in running_set:
            while True:
                back += 1
                max_length = max(max_length, this_length)
                running_set.remove(input[back - 1])
                this_length -= 1
                if input[back - 1] == input[front]:
                    break
        running_set.add(input[front])
        this_length += 1
        front += 1
    return max(this_length, max_length)

Both are O(n)

- pasio April 28, 2018 | Flag Reply
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0
of 0 vote

O(n):

def count_different(input):
    if not input:
        return 0
    front = back = max_length = this_length = 0
    running_set = set()
    while front < len(input):
        if input[front] in running_set:
            while True:
                back += 1
                max_length = max(max_length, this_length)
                running_set.remove(input[back - 1])
                this_length -= 1
                if input[back - 1] == input[front]:
                    break
        running_set.add(input[front])
        this_length += 1
        front += 1
    return max(this_length, max_length)

- yehadut April 28, 2018 | Flag Reply
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0
of 0 vote

C++ solution..

int main(){
	int a[]={1,2,2,3,4,5,6,2,3};
	int n=sizeof(a)/sizeof(a[0]);
	int p=a[0];
	for (int i = 0; i < n; ++i)
	{	if(a[i]>p)
		p=a[i];
		
	}
	
	int d[p];
	
	memset(d,0,p*sizeof(int));

	for (int i = 0; i < n; ++i)
	{
		d[a[i]-1]+=1;
	}

	
	int t=0;
	for (int i = 0; i < p; ++i)
	{
		if(d[i]==1)
			t++;
	}

	cout<<t;

}

- bipinkumar April 29, 2018 | Flag Reply
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0
of 0 vote

private static void usingString(int[] array) {
		String str = "";
		for (int i = 0; i < array.length; i++) {
			String temp = String.valueOf(array[i]);
			if (!str.contains(temp)) {
				str = str + temp;
			}
		}
		System.out.println(str.length());
	}

- Anonymous May 08, 2018 | Flag Reply
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0
of 0 vote

private static void usingString(int[] array) {
		String str = "";
		for (int i = 0; i < array.length; i++) {
			String temp = String.valueOf(array[i]);
			if (!str.contains(temp)) {
				str = str + temp;
			}
		}
		System.out.println(str.length());
	}

- Sujay May 08, 2018 | Flag Reply
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0
of 0 vote

public static int distinctArray(int arr[]){
	HashSet<Integer> set=new HashSet<>();
	for(int x:arr){
	set.add(x);
	}
	set.size();
}

- Code September 13, 2018 | Flag Reply
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0
of 0 vote

import java.util.HashSet;

public class NonRepeatedNum {
	public static void main(String args[])
	{
		HashSet<Integer> realSet = new HashSet<>();
		HashSet<Integer> DupeSet = new HashSet<>();
		int [] arr = {1, 2, 2, 3, 4, 5, 6, 2, 3};
		int [] arrDupe = new int[arr.length-1];
		int count = 0;
		
		
		for(int i =0;i<arr.length;i++)
		{
			arrDupe[arr[i]]++; 
		}
		for(int i = 0;i< arrDupe.length;i++)
		{
			if(arrDupe[i] == 1)
			{
				count++;
			}
		}
		System.out.println(count);
	}

}

- TonyStark September 23, 2018 | Flag Reply
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0
of 0 vote

def countUnique(arr):
	n=len(arr)
	d={}
	for i in range(len(arr)):
		if arr[i] not in d:
			d[arr[i]]=1
		else:
			d[arr[i]]+=1
	count=0
	for x in d:
		if d[x]==1:
			count+=1
	return count

- Tuhin Maji October 24, 2019 | Flag Reply


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