Partition Problem

dynamic-programming/PartitionProblem.java

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+import java.util.*;
+// problem is [10,20,30,40] we have to divide the array  into  k partitions such that max sum of partition is minimum 
+class PartitionProblem
+{
+	// function to sum from i to j both inclusive
+	static int sum(int arr[],int i,int j)
+	{
+		int s=0;
+		for(int k=i;k<=j;k++)
+		{
+			s+=arr[k];
+		}
+		return s;
+	}
+	// simple recursion to calculate above
+	// concept is based on that first calculate min of(k-1 partitions)[calculated due to recursion] and one left one partition is found by sum
+	// ex [10,20,30,40] K=3 then 
+	// in the for loop
+	// initial->0
+	// sum from arr[initial:i] + partition(k-1's sum from i+1 : arr.length-1)
+	static int inefficient(int arr[],int initial,int k,int best)
+	{
+
+		if(k==1){return sum(arr,initial, arr.length-1);}
+		else if(initial==arr.length){return best;}
+		for(int i=initial;i<=arr.length-k;i++)
+		{
+			best=Math.min(best,Math.max(inefficient(arr,i+1,k-1,best),sum(arr,initial,i)));
+		}
+		return best;
+	}
+	// n^3 time complexity
+	static int painterpartitionWithDp(int arr[],int K)
+	{
+
+		int dp[][]=new int[K+1][arr.length+1];
+		int sum=arr[0];
+		// below is the base case ie when the board(or k) is 1 then ans is simply sum of all the eleents upto i(index in array)
+		dp[1][1]=sum;
+		for(int i=2;i<=arr.length;i++)
+		{
+			sum+=arr[i-1];
+			dp[1][i]=sum;
+		}
+		// second base case is that board is k and element is 1 then ans is first element
+		for(int i=1;i<=K;i++)
+		{
+			dp[i][1]=arr[0];
+		}
+		// when k is more than 2 
+		for(int k=2;k<=K;k++)
+		{
+			// when arr length is more than 2
+			for(int j=2;j<=arr.length;j++)
+			{
+				int best=Integer.MAX_VALUE;
+				// partition array[1:j] into k starting from dp[k][p] to arr[p:j];
+				for(int p=1;p<=j;p++)
+				{
+					best=Math.min(best,Math.max(dp[k-1][p],sum(arr,p,arr.length-1)));
+				}
+				dp[k][j]=best;
+
+			}
+
+		}
+		return dp[K][arr.length];	
+	}
+	public static void main(String args[])
+	{
+		int arr[]={40,20,30,40};
+		System.out.println(inefficient(arr,0,2,Integer.MAX_VALUE));
+		System.out.println(dp(arr, 2));
+
+	}
+}