Maximum elements that can be removed from front of two arrays such that their sum is at most K

// C# program for the above approach
using System;

class GFG
{    

// Function to find the maximum number
// of items that can be removed from
// both the arrays
static void maxItems(int n, int m, int[] a,
                     int[] b, int K)
{
    
    // Stores the maximum item count
    int count = 0;

    // Stores the prefix sum of the
    // cost of items
    int[] A = new int[n + 1];
    int[] B= new int[m + 1];

    // Insert the item cost 0 at the
    // front of the arrays
    A[0] = 0;
    B[0] = 0;

    // Build the prefix sum for
    // the array A[]
    for(int i = 1; i <= n; i++) 
    {
        
        // Update the value of A[i]
        A[i] = a[i - 1] + A[i - 1];
    }

    // Build the prefix sum for
    // the array B[]
    for(int i = 1; i <= m; i++)
    {
        
        // Update the value of B[i]
        B[i] = b[i - 1] + B[i - 1];
    }

    // Iterate through each item
    // of the array A[]
    for(int i = 0; i <= n; i++) 
    {
        
        // If A[i] exceeds K
        if (A[i] > K)
            break;

        // Store the remaining amount
        // after taking top i elements
        // from the array A
        int rem = K - A[i];

        // Store the number of items
        // possible to take from the
        // array B[]
        int j = 0;

        // Store low and high bounds
        // for binary search
        int lo = 0, hi = m;

        // Binary search to find
        // number of item that
        // can be taken from stack
        // B in rem amount
        while (lo <= hi)
        {
            
            // Calculate the mid value
            int mid = (lo + hi) / 2;
            if (B[mid] <= rem)
            {
                
                // Update the value
                // of j and lo
                j = mid;
                lo = mid + 1;
            }
            else 
            {
                
                // Update the value
                // of the hi
                hi = mid - 1;
            }
        }

        // Store the maximum of total
        // item count
        count = Math.Max(j + i, count);
    }

    // Print the result
    Console.Write(count);
}


// Driver code
public static void Main(String []args)
{
    int n = 4, m = 5, K = 7;
    int[] A = { 2, 4, 7, 3 };
    int[] B = { 1, 9, 3, 4, 5 };
    
    maxItems(n, m, A, B, K);

}
}

// This code is contributed by code_hunt.

using System;

public class MaxItems
{
    public static void FindMaxItems(int n, int m, int[] a, int[] b, int K)
    {
        Array.Sort(a); // Sort array a in ascending order
        Array.Sort(b); // Sort array b in ascending order

        int i = 0, j = m - 1, count = 0;
        while (i < n && j >= 0)
        {
            if (a[i] + b[j] <= K) // If sum of elements at a[i] and b[j] is less than or equal to K
            {
                count++;
                i++;
                j--;
            }
            else if (a[i] + b[j] > K && j > 0) // If sum is greater than K and there are elements left in array b
            {
                j--;
            }
            else
            {
                i++; // Increment index of array a
            }
        }

        Console.WriteLine(count); // Print the count of valid pairs
    }

    public static void Main(string[] args)
    {
        int n = 4, m = 5, K = 7;
        int[] A = { 2, 4, 7, 3 };
        int[] B = { 1, 9, 3, 4, 5 };
        FindMaxItems(n, m, A, B, K);
    }
}