Adobe Interview Question for SDE1s


Country: India




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Recursion:
- Recursion on Shops, Mood Change Count is an extra parameter
- In each recursion of a shop,
-- you do both sell and buy so that you can compute all the possible scenarios
-- In each buy/sell case, check if mood has changed, if so update the mood change count and pass on the updated mood count change to next level recursion
-- In each buy/sell case, push the activity on to stack (also have to pop once the control comes back to this level)

- Recursion ends when all shops are visited
-- When recursion ends, check total mood change count with that of stored count, if the new one is more, store that and also take copy stack of sales/buys

- Laxmi Narsimha Rao Oruganti November 26, 2018 | Flag Reply
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#include<iostream>
using namespace std;
int main()
{
	int arr[]={100,200,100,1,1};
	int k=1900,n=2100;
	int mood=0;
	int m=4;
	for(int i=0;i<m;i++)
	{
		if(k<2100)
		{
			k+=arr[i];
			if(k>=2100)
			mood++;
		}
		else
		{
		k-=arr[i];
		if(k<2100)
		{
			mood++;
			}	
		}
	}
	cout<<mood<<endl;
	return(0);
}

- Tanyaanand0211 April 03, 2019 | Flag Reply
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The question says you must sell all the coconuts that you have.

- Poon May 10, 2019 | Flag
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#include <iostream>
using namespace std;

int main()
{
    int shop[]={100,200,100,1,1};
    int m = sizeof(shop)/sizeof(shop[0]);
    int k=1900;
    int n=2100;
    int count=0;
    bool happy;
    if(k <= n)
        happy = true; 
    else
        happy = false;
    for(int i=0; i<m; i++)
    {
        if(k<=n)
        {
            k += shop[i];
            if(!happy)
            {
                happy = true;
                count++;
            }
        }
        else
        {
            if(k< shop[i])
                k -= k;
            else
                k -= shop[i];
            if(happy)
            {
                happy = false;
                count ++;
            }
        }
    }
    cout << "numof mood swings = "<<count <<endl;
    return 0;
}

- knightriders May 13, 2019 | Flag Reply


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