Maximum count of integers to be chosen from given two stacks having sum at most K

Given two stacks stack1[] and stack2[] of size N and M respectively and an integer K, The task is to count the maximum number of integers from two stacks having sum less than or equal to K.

Examples:

Input: stack1[ ] = { 60, 90, 120 }
stack2[ ] = { 100, 10, 10, 250 }, K = 130
Output: 3
Explanation: Take 3 numbers from stack1 which are 100 and 10 and 10.
Total sum 100 + 10 + 10 = 120 

Input: stack1[ ] = { 60, 90, 120 }
stack2[ ] = { 80, 150, 80, 150 }, K = 740
Output: 7
Explanation: Select all the numbers because the value K is enough.

Approach: This problem cannot be solved using the Greedy approach because in greedy at each step the number having the minimum value will be selected but the first example fails according to this. This problem can be solved using prefix sum and binary search. calculate the prefix sum of both the stacks and now iterate for every possible value of 1st stack and take target which is (K – stack1[i]) and apply binary search on the second stack to take lower bound of stack2[].
Follow the steps mentioned below:

• Take two new stacks which are sumA[] and sumB[].
• Calculate the prefix of both the stacks stack1[] and stack2[].
• Iterate on the first stack.
  • Now, take the remValueOfK variable and store (K – stack1[i]).
  • If it is less than 0 so continue the loop.
  • Else take the lower bound of the second stack.
    • If the lower bound is greater than the size of the second stack or the value of the lower bound is greater than the value of remValueOfK just decrement the value of the lower bound variable.
• Store the maximum count of elements selected and return that as the final answer.

Below is the implementation of the above approach.

C++

// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the
// maximum number of elements
int maxNumbers(int stack1[], int N, int stack2[],
               int M, int K)
{

    // Take prefix of both the stack
    vector<int> sumA(N + 1, 0);
    vector<int> sumB(M + 1, 0);
    for (int i = 0; i < N; i++)
        sumA[i + 1] = sumA[i] + stack1[i];

    for (int i = 0; i < M; i++)
        sumB[i + 1] = sumB[i] + stack2[i];

    // Calculate maxNumbers
    int MaxNumbers = 0;
    for (int i = 0; i <= N; i++) {

        // Calculate remaining value of K
        // after selecting numbers
        // from 1st stack
        int remValueOfK = K - sumA[i];

        // If rem value of K is less than 0
        // continue the loop
        if (remValueOfK < 0)
            continue;

        // Calculate lower bound
        int lowerBound
            = lower_bound(sumB.begin(),
                          sumB.end(),
                          remValueOfK)
              - sumB.begin();

        // If size of lower bound is greater
        // than self stack size or
        // value of lower bound element
        // decrement lowerBound
        if (lowerBound > M
            or sumB[lowerBound] > remValueOfK) {
            lowerBound--;
        }

        // Store max possible numbers
        int books = i + lowerBound;
        MaxNumbers = max(MaxNumbers, books);
    }
    return MaxNumbers;
}

// Driver code
int main()
{
    int stack1[] = { 60, 90, 120 };
    int stack2[] = { 100, 10, 10, 200 };
    int K = 130;
    int N = 3;
    int M = 4;
    int ans
        = maxNumbers(stack1, N, stack2, M, K);
    cout << ans;
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG
{

  static int lower_bound(int []a, int val) {
    int lo = 0, hi = a.length - 1;
    while (lo < hi) {
      int mid = (int)Math.floor(lo + (double)(hi - lo) / 2);
      if (a[mid] < val)
        lo = mid + 1;
      else
        hi = mid;
    }
    return lo;
  }

  // Function to find the
  // maximum number of elements
  static int maxNumbers(int []stack1, int N, int []stack2,
                        int M, int K) {

    // Take prefix of both the stack
    int []sumA = new int[N + 1];
    for(int i = 0; i < N + 1; i++) {
      sumA[i] = 0;
    }

    int []sumB = new int[M + 1];
    for(int i = 0; i < M + 1; i++) {
      sumB[i] = 0;
    }

    for (int i = 0; i < N; i++)
      sumA[i + 1] = sumA[i] + stack1[i];

    for (int i = 0; i < M; i++)
      sumB[i + 1] = sumB[i] + stack2[i];

    // Calculate maxNumbers
    int MaxNumbers = 0;
    for (int i = 0; i <= N; i++) {

      // Calculate remaining value of K
      // after selecting numbers
      // from 1st stack
      int remValueOfK = K - sumA[i];

      // If rem value of K is less than 0
      // continue the loop
      if (remValueOfK < 0)
        continue;

      // Calculate lower bound
      int lowerBound
        = lower_bound(sumB,
                      remValueOfK);


      // If size of lower bound is greater
      // than self stack size or
      // value of lower bound element
      // decrement lowerBound
      if (lowerBound > M
          || sumB[lowerBound] > remValueOfK) {
        lowerBound--;
      }

      // Store max possible numbers
      int books = i + lowerBound;
      MaxNumbers = Math.max(MaxNumbers, books);
    }
    return MaxNumbers;
  }

  // Driver Code
  public static void main(String args[])
  {
    int []stack1 = {60, 90, 120};
    int []stack2 = {100, 10, 10, 200};
    int K = 130;
    int N = 3;
    int M = 4;
    int ans = maxNumbers(stack1, N, stack2, M, K);
    System.out.println(ans);

  }
}

// This code is contributed by sanjoy_62.

Python3

# Python code for the above approach
def lower_bound(a, val):
    lo = 0
    hi = len(a) - 1;
    while (lo < hi):
        mid = (lo + (hi - lo) // 2);
        if (a[mid] < val):
            lo = mid + 1;
        else:
            hi = mid;
    return lo;

# Function to find the
# maximum number of elements
def maxNumbers(stack1, N, stack2, M, K):

    # Take prefix of both the stack
    sumA = [0] * (N + 1)
    sumB = [0] * (M + 1)
    for i in range(N):
        sumA[i + 1] = sumA[i] + stack1[i];

    for i in range(M):
        sumB[i + 1] = sumB[i] + stack2[i];

    # Calculate maxNumbers
    MaxNumbers = 0;
    for i in range(N + 1):

        # Calculate remaining value of K
        # after selecting numbers
        # from 1st stack
        remValueOfK = K - sumA[i];

        # If rem value of K is less than 0
        # continue the loop
        if (remValueOfK < 0):
            continue;

        # Calculate lower bound
        lowerBound = lower_bound(sumB, remValueOfK);


        # If size of lower bound is greater
        # than self stack size or
        # value of lower bound element
        # decrement lowerBound
        if (lowerBound > M or sumB[lowerBound] > remValueOfK):
            lowerBound -= 1

        # Store max possible numbers
        books = i + lowerBound;
        MaxNumbers = max(MaxNumbers, books);

    return MaxNumbers;

# Driver code

stack1 = [60, 90, 120];
stack2 = [100, 10, 10, 200];
K = 130;
N = 3;
M = 4;
ans = maxNumbers(stack1, N, stack2, M, K);
print(ans)

# This code is contributed by gfgking

C#

// C# code for the above approach
using System;
class GFG
{
  static int lower_bound(int []a, int val) {
    int lo = 0, hi = a.Length - 1;
    while (lo < hi) {
      int mid = (int)Math.Floor(lo + (double)(hi - lo) / 2);
      if (a[mid] < val)
        lo = mid + 1;
      else
        hi = mid;
    }
    return lo;
  }

  // Function to find the
  // maximum number of elements
  static int maxNumbers(int []stack1, int N, int []stack2,
                        int M, int K) {

    // Take prefix of both the stack
    int []sumA = new int[N + 1];
    for(int i = 0; i < N + 1; i++) {
      sumA[i] = 0;
    }

    int []sumB = new int[M + 1];
    for(int i = 0; i < M + 1; i++) {
      sumB[i] = 0;
    }

    for (int i = 0; i < N; i++)
      sumA[i + 1] = sumA[i] + stack1[i];

    for (int i = 0; i < M; i++)
      sumB[i + 1] = sumB[i] + stack2[i];

    // Calculate maxNumbers
    int MaxNumbers = 0;
    for (int i = 0; i <= N; i++) {

      // Calculate remaining value of K
      // after selecting numbers
      // from 1st stack
      int remValueOfK = K - sumA[i];

      // If rem value of K is less than 0
      // continue the loop
      if (remValueOfK < 0)
        continue;

      // Calculate lower bound
      int lowerBound
        = lower_bound(sumB,
                      remValueOfK);


      // If size of lower bound is greater
      // than self stack size or
      // value of lower bound element
      // decrement lowerBound
      if (lowerBound > M
          || sumB[lowerBound] > remValueOfK) {
        lowerBound--;
      }

      // Store max possible numbers
      int books = i + lowerBound;
      MaxNumbers = Math.Max(MaxNumbers, books);
    }
    return MaxNumbers;
  }

  // Driver code
  public static void Main() {

    int []stack1 = {60, 90, 120};
    int []stack2 = {100, 10, 10, 200};
    int K = 130;
    int N = 3;
    int M = 4;
    int ans = maxNumbers(stack1, N, stack2, M, K);
    Console.Write(ans);

  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
      // JavaScript code for the above approach

      function lower_bound(a, val) {
          let lo = 0, hi = a.length - 1;
          while (lo < hi) {
              let mid = Math.floor(lo + (hi - lo) / 2);
              if (a[mid] < val)
                  lo = mid + 1;
              else
                  hi = mid;
          }
          return lo;
      }

      // Function to find the
      // maximum number of elements
      function maxNumbers(stack1, N, stack2,
          M, K) {

          // Take prefix of both the stack
          let sumA = new Array(N + 1).fill(0);
          let sumB = new Array(M + 1).fill(0);
          for (let i = 0; i < N; i++)
              sumA[i + 1] = sumA[i] + stack1[i];

          for (let i = 0; i < M; i++)
              sumB[i + 1] = sumB[i] + stack2[i];

          // Calculate maxNumbers
          let MaxNumbers = 0;
          for (let i = 0; i <= N; i++) {

              // Calculate remaining value of K
              // after selecting numbers
              // from 1st stack
              let remValueOfK = K - sumA[i];

              // If rem value of K is less than 0
              // continue the loop
              if (remValueOfK < 0)
                  continue;

              // Calculate lower bound
              let lowerBound
                  = lower_bound(sumB,
                      remValueOfK);


              // If size of lower bound is greater
              // than self stack size or
              // value of lower bound element
              // decrement lowerBound
              if (lowerBound > M
                  || sumB[lowerBound] > remValueOfK) {
                  lowerBound--;
              }

              // Store max possible numbers
              let books = i + lowerBound;
              MaxNumbers = Math.max(MaxNumbers, books);
          }
          return MaxNumbers;
      }

      // Driver code

      let stack1 = [60, 90, 120];
      let stack2 = [100, 10, 10, 200];
      let K = 130;
      let N = 3;
      let M = 4;
      let ans
          = maxNumbers(stack1, N, stack2, M, K);
      document.write(ans)

     // This code is contributed by Potta Lokesh
  </script>

Output

3

Time Complexity: O(N * logN)
Auxiliary Space: O(N)