CareerCup

Can you clarify - couldn't understand the question.

c++, implementation, O(n)

int countOfMeltDown(vector<int>& arr) {
	vector<int> mins;
	int i, k, count;

	mins.assign(arr.size(), 0);

	count = 0;
	for (i = 0, k = 1; i < arr.size(); i++, k++) {
		mins[i] = min(arr[i], k);
		if (arr[i] <= 0) k = 0;
	}

	for (i = arr.size() - 1, k = 1; i >= 0; i--, k++) {
		count = max(count, min(mins[i], min(arr[i], k)));
		if (arr[i] <= 0) k = 0;
	}

	return count;
}

public int meltHistogram1(int[] a){

        int len = a.length;
        int count = 0;
        boolean flag = true;
        while(flag){
            flag = false;
            int prev = a[0];
            for(int j = 1; j < len-1; ++j){
                int k = a[j];
                if(k == 0)
                    k = 0;
                else if(k == prev && k == a[j+1])
                    k = k - 1;
                else if(k < prev && k < a[j+1])
                    k = k - 1;
                else{
                    if(k > prev)
                        k = prev;
                    if(k > a[j + 1])
                        k = a[j + 1];
                }
                prev = a[j];
                a[j] = k;
            }
            a[0] = 0;
            a[len-1] = 0;
            count++;
            for(int i = 0; i < len; ++i){
                flag = flag || (a[i] != 0);
            }
        }

        return count;

}

public int meltHistogram1(int[] a){
        int len = a.length;
        int count = 0;
        boolean flag = true;
        while(flag){
            flag = false;
            int prev = a[0];
            for(int j = 1; j < len-1; ++j){
                int k = a[j];
                if(k == 0)
                    k = 0;
                else if(k == prev && k == a[j+1])
                    k = k - 1;
                else if(k < prev && k < a[j+1])
                    k = k - 1;
                else{
                    if(k > prev)
                        k = prev;
                    if(k > a[j + 1])
                        k = a[j + 1];
                }
                prev = a[j];
                a[j] = k;
            }
            a[0] = 0;
            a[len-1] = 0;
            count++;
            for(int i = 0; i < len; ++i){
                flag = flag || (a[i] != 0);
            }
        }

        return count;
    }

public int meltHistogram(int[] a){
        int len = a.length;
        int count = 0;
        boolean flag = true;
        while(flag){
            flag = false;
            int prev = a[0];
            for(int j = 1; j < len-1; ++j){
                int k = a[j];
                if(k == 0)
                    k = 0;
                else if(k == prev && k == a[j+1])
                    k = k - 1;
                else if(k < prev && k < a[j+1])
                    k = k - 1;
                else{
                    if(k > prev)
                        k = prev;
                    if(k > a[j + 1])
                        k = a[j + 1];
                }
                prev = a[j];
                a[j] = k;
            }
            a[0] = 0;
            a[len-1] = 0;
            count++;
            for(int i = 0; i < len; ++i){
                flag = flag || (a[i] != 0);
            }
        }

        return count;

}

int countHist(int arr[],int n)
	{
		for(int i=0;i<n;i++)
			if(arr[i]>=5)
				arr[i]=0;
		int maxval=0;
		for(int i=0;i<n;i++)
			maxval=max(max,val);
		return maxval;
	}

What does melt mean in this context?

package goldmansachs;
import java.util.*;
import java.math.*;
public class histogramMelt {
	


	  public static void main(String args[]){
	    //int arr[] = {5,4,3,6,0,1};
	    int arr[] = {0,1,1,1,1,0};
	    int count =histogramMelt(arr);
	    System.out.println("result =" +count);  
	  }
	  
	  
	  public static boolean checkZero(int arr[]){
	   int flag =0;
	    for(int i=0;i<arr.length;i++){
	      if(arr[i]==0){
	        flag=1;
	      }
	      else{
	        flag=0;
	        break;
	      }
	      System.out.print(arr[i] +" ");
	    }
	    if(flag==0) return false;
	    else return true;
	  }
	  
	  public static int  histogramMelt(int arr[]){
	    boolean b ;
	    int count=0;
	    int l = arr.length;
	    int val[] = new int[l];
	    do{
	    for(int i =0 ; i<l;i++){
	      System.out.println(i+" "+arr[i]);
	      if(i==0 || i== l-1){ //corner case first and last elements of the array
	        val[i]=0;
	      System.out.println(i+" "+val[i]);
	      }
	      else{
	          if((arr[i]>arr[i+1])||(arr[i]>arr[i-1])){ // other values are replaced by min(left,right) neighbours
	              val[i] = Math.min(arr[i+1],arr[i-1]);   
	            System.out.println(i+" "+val[i]);
	          }
	          else{
	            
	            val[i] = arr[i]-1; // every column has the top block exposed, so every column gets deducted by 1
	            if(val[i]<0) val[i]=0; // avoid negatives
	            System.out.println(i+" "+val[i]);
	          }
	      }
	    
	    }
	    
	    b =checkZero(val);
	      arr =val;
	      count ++; // number of times the histogram was melted
	    } while(b==false);
	    return count;
	  }
	}

Ruby

def melt_histogram(array)
  count = 0
  while !array.all?(&:zero?)
    prev = 0
    array.each_with_index do |x, i|
      reduce_to = [prev, array[i+1].to_i].min
      array[i] = [0, [reduce_to, x-1].min].max
      prev = x
    end
    count +=1
  end
  count
end

can someone please explain the question and the logic behind the solution?

can you explain the logic behind the solution?