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

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  • Last Updated : 04 Jun, 2021
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Given an integer K and two arrays A[] and B[] consisting of N and M integers, the task is to maximize the number of elements that can be removed from the front of either array according to the following rules:

  • Remove an element from the front of either array A[] and B[] such that the value of the removed element must be at most K.
  • Decrease the value of K by the value of the element removed.

Examples:

Input: K = 7, A[] = {2, 4, 7, 3}, B[] = {1, 9, 3, 4, 5}
Output: 3
Explanation:
Operation 1: Choose element from the array A[] and decrease K by A[0](=2), then value of K becomes = (7 – 2) = 5.
Operation 2: Choose element from the array B[] and decrease K by B[0](=1), then value of K becomes = (5 – 1) = 4.
Operation 3: Choose element from the array A[] and decrease K by A[1](=4), then value of K becomes = (4 – 4) = 4.
After the above operations, no more element can be removed. Therefore, print 3.

Input: K = 9, A[] = {12, 10, 1, 2, 3}, B[] = {15, 19, 3, 4, 5}
Output: 0

 

Approach: The given problem can be solved by using the Prefix Sum and Binary Search to find the total items possible j to take from stack B after taking i items from stack A and return the result as the maximum value of (i + j). Follow the below steps to solve the given problem:

  • Find prefix sum of the arrays A[] and B[].
  • Initialize a variable, say count as 0, that stores the maximum items that can be taken.
  • Traverse the array, A[] over the range [0, N] using the variable i and perform the following steps:
    • If the value of A[i] is greater than K, then break out of the loop.
    • Store the remaining amount after taking i items from stack A in a variable, rem as K – A[i].
    • Perform a binary search on the array B, to find the maximum items say, j that can be taken in rem amount from stack B (after taking i elements from stack A).
    • Store the maximum value of i + j in the variable count.
  • After completing the above steps, print the value of count as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum number
// of items that can be removed from
// both the arrays
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[n + 1];
    int B[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 = max(j + i, count);
    }
 
    // Print the result
    cout << count;
}
 
// Driver Code
int main()
{
    int n = 4, m = 5, K = 7;
    int A[n] = { 2, 4, 7, 3 };
    int B[m] = { 1, 9, 3, 4, 5 };
    maxItems(n, m, A, B, K);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
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
    System.out.print(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 sanjoy_62

Python3




# Python3 program for the above approach
 
# Function to find the maximum number
# of items that can be removed from
# both the arrays
def maxItems(n, m, a, b, K):
     
    # Stores the maximum item count
    count = 0
 
    # Stores the prefix sum of the
    # cost of items
    A = [0 for i in range(n + 1)]
    B = [0 for i in range(m + 1)]
 
    # Build the prefix sum for
    # the array A[]
    for i in range(1, n + 1, 1):
         
        # Update the value of A[i]
        A[i] = a[i - 1] + A[i - 1]
 
    # Build the prefix sum for
    # the array B[]
    for i in range(1, m + 1, 1):
         
        # Update the value of B[i]
        B[i] = b[i - 1] + B[i - 1]
 
    # Iterate through each item
    # of the array A[]
    for i in range(n + 1):
         
        # If A[i] exceeds K
        if (A[i] > K):
            break
 
        # Store the remaining amount
        # after taking top i elements
        # from the array A
        rem = K - A[i]
 
        # Store the number of items
        # possible to take from the
        # array B[]
        j = 0
 
        # Store low and high bounds
        # for binary search
        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
            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 = max(j + i, count)
 
    # Print the result
    print(count)
 
# Driver Code
if __name__ == '__main__':
     
    n = 4
    m = 5
    K = 7
    A = [ 2, 4, 7, 3 ]
    B = [ 1, 9, 3, 4, 5 ]
     
    maxItems(n, m, A, B, K)
         
# This code is contributed by bgangwar59

C#




// 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.

Javascript




<script>
 
// javascript program for the above approach
 
// Function to find the maximum number
// of items that can be removed from
// both the arrays
function maxItems(n, m, a, b, K)
{
    // Stores the maximum item count
    var count = 0;
 
    // Stores the prefix sum of the
    // cost of items
    var A = new Array(n + 1);
    var B = new Array(m + 1);
 
    // Insert the item cost 0 at the
    // front of the arrays
    A[0] = 0;
    B[0] = 0;
     
    var i;
    // Build the prefix sum for
    // the array A[]
    for (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 (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 (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
        var rem = K - A[i];
 
        // Store the number of items
        // possible to take from the
        // array B[]
        var j = 0;
 
        // Store low and high bounds
        // for binary search
        var 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
            var mid = parseInt((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
    document.write(count);
}
 
// Driver Code
    var n = 4, m = 5, K = 7;
    var A = [2, 4, 7, 3];
    var B = [1, 9, 3, 4, 5];
    maxItems(n, m, A, B, K);
 
// This code is contributed by SURENDRA_GANGWAR.
</script>

Output: 

3

 

Time Complexity: O(N * log(M))
Auxiliary Space: O(N + M)


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