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

Last Updated : 08 Mar, 2024

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>