Walmart Labs Interview Question for Java Developers


Country: India
Interview Type: Written Test




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Typical DP problem, developing a mathematical equation should solve

Assume all the diamonds are in array, arr[] whose size is N

Maximum he can collect with N dragons is
MaxDiam(N)= Math.max{ arr[N-1] + MaxDiam(N-2), MaxDiam(N-1)}

- Anonymous July 28, 2016 | Flag Reply
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Why can't He take "5+2+4" =11?

- Berky July 28, 2016 | Flag Reply
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Why can't he take "5+2+4" = 11?

- Berky July 28, 2016 | Flag Reply
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var input = [1,2,3,4,5];

var output = input.reduce(function(prev,e) {
	return [prev[1],prev[2],Math.max(prev[0],prev[1])+e];
},[0,0,0]);

console.log(output[2]);

- Chaar-Lee July 29, 2016 | Flag Reply
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public class DiamondCollectionPuzzle {
	private static int[] input= {1, 2, 3, 4, 5, 6, 7};
	public static void main(String[] args) {
		PosssibleDiamondCount possibleDiamonds = collectDiamonds(0);
		System.out.println(Math.max(possibleDiamonds.maxExcludingCurrent, possibleDiamonds.maxIncludingCurrent));
	}
	private static PosssibleDiamondCount collectDiamonds(int startIndex) {
		if(startIndex == input.length-1){
			PosssibleDiamondCount possibleDiamondCount = new PosssibleDiamondCount();
			possibleDiamondCount.maxIncludingCurrent = input[startIndex];
			possibleDiamondCount.maxExcludingCurrent = 0;
			return possibleDiamondCount;
		}else{
			PosssibleDiamondCount nextPossibleDiamondCount = collectDiamonds(startIndex + 1);
			PosssibleDiamondCount currPossibleDiamondCount = new PosssibleDiamondCount();
			currPossibleDiamondCount.maxIncludingCurrent = input[startIndex] + nextPossibleDiamondCount.maxExcludingCurrent;
			currPossibleDiamondCount.maxExcludingCurrent = nextPossibleDiamondCount.maxIncludingCurrent;
			return currPossibleDiamondCount;
		}
	}
	
	/**
	 * Data structure to hold possible diamonds including current and excluding current
	 */
	private static class PosssibleDiamondCount{
		int maxIncludingCurrent;
		int maxExcludingCurrent;
	}
}

- Srikanth August 19, 2016 | Flag Reply
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public class DiamondCollectionPuzzle {
	private static int[] input= {1, 2, 3, 4, 5, 6, 7};
	public static void main(String[] args) {
		PosssibleDiamondCount possibleDiamonds = collectDiamonds(0);
		System.out.println(Math.max(possibleDiamonds.maxExcludingCurrent, possibleDiamonds.maxIncludingCurrent));
	}
	private static PosssibleDiamondCount collectDiamonds(int startIndex) {
		if(startIndex == input.length-1){
			PosssibleDiamondCount possibleDiamondCount = new PosssibleDiamondCount();
			possibleDiamondCount.maxIncludingCurrent = input[startIndex];
			possibleDiamondCount.maxExcludingCurrent = 0;
			return possibleDiamondCount;
		}else{
			PosssibleDiamondCount nextPossibleDiamondCount = collectDiamonds(startIndex + 1);
			PosssibleDiamondCount currPossibleDiamondCount = new PosssibleDiamondCount();
			currPossibleDiamondCount.maxIncludingCurrent = input[startIndex] + nextPossibleDiamondCount.maxExcludingCurrent;
			currPossibleDiamondCount.maxExcludingCurrent = nextPossibleDiamondCount.maxIncludingCurrent;
			return currPossibleDiamondCount;
		}
	}
	
	/**
	 * Data structure to hold possible diamonds including current and excluding current
	 */
	private static class PosssibleDiamondCount{
		int maxIncludingCurrent;
		int maxExcludingCurrent;
	}
}

- Srikanth August 19, 2016 | Flag Reply
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basically the problem is to find the maximum sum of the array such that no two elements are consecutive.

- som0890 September 22, 2016 | Flag Reply
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import java.util.Scanner;
class MaxProblem
{
public static void main(String[] args)
{
Scanner scn=new Scanner(System.in);
int n=scn.nextInt();
int arr[] =new int[n];
for(int i=0;i<n;i++)
arr[i]=scn.nextInt();
System.out.println( FindMaxSum(arr, n));
}
public static int FindMaxSum(int arr[], int n)
{
int incl = arr[0];
int excl = 0;
int excl_new;
for (int i = 1; i < n; i++)
{
excl_new = (incl > excl)? incl: excl;
incl = excl + arr[i];
excl = excl_new;
}
return ((incl > excl)? incl : excl);
}
}

- Anonymous December 28, 2016 | Flag Reply
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import java.util.Scanner;
class MaxProblem
{
public static void main(String[] args)
{
Scanner scn=new Scanner(System.in);
int n=scn.nextInt();
int arr[] =new int[n];
for(int i=0;i<n;i++)
arr[i]=scn.nextInt();
System.out.println( FindMaxSum(arr, n));
}
public static int FindMaxSum(int arr[], int n)
{
int incl = arr[0];
int excl = 0;
int excl_new;
for (int i = 1; i < n; i++)
{
excl_new = (incl > excl)? incl: excl;
incl = excl + arr[i];
excl = excl_new;
}
return ((incl > excl)? incl : excl);
}
}

- Ramgopal December 28, 2016 | Flag Reply
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public class MyClass {
    public static void main(String args[]) {
        int n[]=new int[]{5, 5, 10, 100, 10, 5};
      System.out.println(helper(n));
    }
    public static int helper(int[]n)
    {
        
        int dp[]=new int[n.length];
        if(n.length==1)
        {
            return n[n.length-1];
        }
        if(n.length==2)
        {
            return n[n.length-2];
        }
        if(n.length==3)
        {
            return Math.max(n[n.length-3]+n[n.length-1],n[n.length-2]);
        }
        dp[n.length-1]=n[n.length-1];
        dp[n.length-2]=n[n.length-2];
        dp[n.length-3]=n[n.length-3]+n[n.length-1];
        for(int i=n.length-4;i>=0;i--)
        {
            dp[i]=n[i]+Math.max(dp[i+2],dp[i+3]);
            
        }
        int result=0;
        for(int i=0;i<dp.length;i++)
        {
            result=Math.max(dp[i],result);
        }
        return result;
    }
}

- Yash Bansal July 14, 2018 | Flag Reply
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of 0 vote

a

- a July 14, 2018 | Flag Reply


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