Given an array **A[]** of N integers and an integer **K**, the task is to select the maximum number of elements from the array whose sum is at most **K**.

**Examples:**

Input:A[] = {1, 12, 5, 111, 200, 1000, 10}, K = 50Output:4Explanation:

Maximum number of selections will be1, 12, 5, 10 that is 1 + 12 + 5 + 10 = 28 < 50.

Input:A[] = {3, 7, 2, 9, 4}, K = 15Output:3Explanation:

Maximum number of selections will be 3, 2, 4 that is 3 + 2 + 4 =9 < 15.

**Naive Approach:** The idea is to generate all possible subsequences of the array and find the sum of elements of all the subsequences generated. Find the subsequence with maximum length and with the sum less than or equal to **K**. **Time Complexity:** O(2^{N}) **Auxiliary Space:** (1)

**Efficient Approach:** The efficient approach can be solved using the Greedy Technique. Below are the steps:

- Sort the given array.
- Iterate in the array and keep the track of the sum of elements until the
**sum is less than or equal to K**. - If the sum while iterating in the above steps exceeds
**K**then break the loop and print the value of count till that index.

Below is the implementation of the above approach:

## C++14

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to select a maximum number of` `// elements in array whose sum is at most K` `int` `maxSelections(` `int` `A[], ` `int` `n, ` `int` `k)` `{` ` ` `// Sort the array` ` ` `sort(A, A + n);` ` ` `// Calculate the sum and count while` ` ` `// iterating the sorted array` ` ` `int` `sum = 0;` ` ` `int` `count = 0;` ` ` `// Iterate for all the` ` ` `// elements in the array` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// Add the current element to sum` ` ` `sum = sum + A[i];` ` ` `if` `(sum > k) {` ` ` `break` `;` ` ` `}` ` ` `// Increment the count` ` ` `count++;` ` ` `}` ` ` `// Return the answer` ` ` `return` `count;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given array` ` ` `int` `A[] = { 3, 7, 2, 9, 4 };` ` ` `// Given sum k` ` ` `int` `k = 15;` ` ` `int` `n = ` `sizeof` `(A) / ` `sizeof` `(A[0]);` ` ` `// Function Call` ` ` `cout << maxSelections(A, n, k);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to select a maximum number of` `// elements in array whose sum is at most K` `static` `int` `maxSelections(` `int` `A[], ` `int` `n, ` `int` `k)` `{` ` ` ` ` `// Sort the array` ` ` `Arrays.sort(A);` ` ` `// Calculate the sum and count while` ` ` `// iterating the sorted array` ` ` `int` `sum = ` `0` `;` ` ` `int` `count = ` `0` `;` ` ` `// Iterate for all the` ` ` `// elements in the array` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `{` ` ` ` ` `// Add the current element to sum` ` ` `sum = sum + A[i];` ` ` `if` `(sum > k)` ` ` `{` ` ` `break` `;` ` ` `}` ` ` `// Increment the count` ` ` `count++;` ` ` `}` ` ` ` ` `// Return the answer` ` ` `return` `count;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ` `// Given array` ` ` `int` `A[] = { ` `3` `, ` `7` `, ` `2` `, ` `9` `, ` `4` `};` ` ` `// Given sum k` ` ` `int` `k = ` `15` `;` ` ` `int` `n = A.length;` ` ` `// Function call` ` ` `System.out.print(maxSelections(A, n, k));` `}` `}` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python3 program for` `# the above approach` `# Function to select a maximum` `# number of elements in array` `# whose sum is at most K` `def` `maxSelections(A, n, k):` ` ` `# Sort the array` ` ` `A.sort();` ` ` ` ` `# Calculate the sum and` ` ` `# count while iterating` ` ` `# the sorted array` ` ` `sum` `=` `0` `;` ` ` `count ` `=` `0` `;` ` ` `# Iterate for all the` ` ` `# elements in the array` ` ` `for` `i ` `in` `range` `(n):` ` ` `# Add the current element to sum` ` ` `sum` `=` `sum` `+` `A[i];` ` ` `if` `(` `sum` `> k):` ` ` `break` `;` ` ` `# Increment the count` ` ` `count ` `+` `=` `1` `;` ` ` `# Return the answer` ` ` `return` `count;` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `# Given array` ` ` `A ` `=` `[` `3` `, ` `7` `, ` `2` `, ` `9` `, ` `4` `];` ` ` `# Given sum k` ` ` `k ` `=` `15` `;` ` ` `n ` `=` `len` `(A);` ` ` `# Function call` ` ` `print` `(maxSelections(A, n, k));` ` ` `# This code is contributed by gauravrajput1` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` `// Function to select a maximum number of` `// elements in array whose sum is at most K` `static` `int` `maxSelections(` `int` `[] A, ` `int` `n, ` `int` `k)` `{` ` ` `// Sort the array` ` ` `Array.Sort(A);` ` ` `// Calculate the sum and count while` ` ` `// iterating the sorted array` ` ` `int` `sum = 0;` ` ` `int` `count = 0;` ` ` `// Iterate for all the` ` ` `// elements in the array` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` ` ` `// Add the current element to sum` ` ` `sum = sum + A[i];` ` ` `if` `(sum > k)` ` ` `{` ` ` `break` `;` ` ` `}` ` ` `// Increment the count` ` ` `count++;` ` ` `}` ` ` `// Return the answer` ` ` `return` `count;` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `// Given array` ` ` `int` `[] A = { 3, 7, 2, 9, 4 };` ` ` `// Given sum k` ` ` `int` `k = 15;` ` ` `int` `n = A.Length;` ` ` `// Function call` ` ` `Console.Write(maxSelections(A, n, k));` `}` `}` `// This code is contributed by gauravrajput1` |

**Output:**

3

**Time Complexity:** O(N*log N) **Auxiliary Space:** O(1)

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