Min steps to empty an Array by removing a pair each time with sum at most K

Given an array arr[] and a target value K. The task is to find the minimum number of steps required to take all elements from the array. In each step, at most two elements can be selected from array such that their sum must not exceed target value K. 
Note: All the elements of the array are less than or equals to K.
 

Input: arr[] = [5, 1, 5, 4], K = 8 
Output: 3 
Explanation: 
We can pick {1, 4}, {5}, {5} in 3 steps: 
Other possible arrangement can be {1, 5}, {4}, {5} in three steps. 
So, the minimum number of steps are required is 3
Input: [1, 2, 1, 1, 3], n = 9 
Output: 3 
Explanation: 
We can pick {1, 1}, {2, 3} and {1} in three steps. 
Other possible choices {1, 3}, {1, 2}, {1} or {1, 1}, {1, 3}, {2} 
So, the minimum number of steps are required is 3 
 

Approach: The above problem can be solved using Greedy Approach along with Two Pointers Technique. The idea is to pick the smallest and the largest element from the array and check if the sum doesn't exceeds N then remove these elements and count this step else remove the largest element and then repeat the above steps until all elements are removed. Below are the steps:
 

1. Sort the given array arr[].
2. Initialize two index i = 0 and j = N - 1.
3. If the sum of elements arr[i] and arr[j] doesn't exceed N then increment i and decrement j.
4. Else decrement j.
5. Repeat the above steps till i <= j and count each step.

Below is the implementation of the above approach:
 

// C++ program for the above approach 
#include <bits/stdc++.h> 
using namespace std; 

// Function to count minimum steps 
int countMinSteps(int arr[], int target, 
                int n) 
{ 

    // Function to sort the array 
    sort(arr, arr + n); 
    int minimumSteps = 0; 
    int i = 0, j = n - 1; 

    // Run while loop 
    while (i <= j) { 

        // Condition to check whether 
        // sum exceed the target or not 
        if (arr[i] + arr[j] <= target) { 
            i++; 
            j--; 
        } 
        else { 
            j--; 
        } 

        // Increment the step 
        // by 1 
        minimumSteps++; 
    } 

    // Return minimum steps 
    return minimumSteps; 
} 

// Driver Code 
int main() 
{ 
    // Given array arr[] 
    int arr[] = { 4, 6, 2, 9, 6, 5, 8, 10 }; 

    // Given target value 
    int target = 11; 

    int size = sizeof(arr) / sizeof(arr[0]); 

    // Function call 
    cout << countMinSteps(arr, target, size); 

    return 0; 
} 

// Java program implementation
// of the above approach
import java.util.*;
import java.io.*;

class GFG{

// Function to count minimum steps     
static int countMinSteps(int arr[],
                         int target, 
                         int n) 
{
    
    // Function to sort the array
    Arrays.sort(arr);

    int minimumSteps = 0;
    int i = 0;
    int j = n - 1;

    // Run while loop
    while (i <= j)
    {
        
        // Condition to check whether 
        // sum exceed the target or not
        if (arr[i] + arr[j] <= target)
        {
            i += 1;
            j -= 1;
        }
        else 
        {
            j -= 1;
        }
    
        // Increment the step by 1 
        minimumSteps += 1;
    }

    // Return minimum steps
    return minimumSteps;
}

// Driver code 
public static void main(String[] args) 
{
    int arr[] = { 4, 6, 2, 9, 6, 5, 8, 10 };

    // Given target value
    int target = 11;

    int size = arr.length;
        
    // Print the minimum flip
    System.out.print(countMinSteps(arr, target,
                                        size));
}
}

// This code is contributed by code_hunt

# Python3 program for the above approach 

# Function to count minimum steps 
def countMinSteps(arr, target, n):
    
    # Function to sort the array 
    arr.sort()

    minimumSteps = 0
    i, j = 0, n - 1
    
    # Run while loop
    while i <= j:
        
        # Condition to check whether 
        # sum exceed the target or not
        if arr[i] + arr[j] <= target:
            i += 1
            j -= 1
        else:
            j -= 1
            
        # Increment the step 
        # by 1 
        minimumSteps += 1
        
    # Return minimum steps
    return minimumSteps

# Driver code
    
# Given array arr[]
arr = [ 4, 6, 2, 9, 6, 5, 8, 10 ]
    
# Given target value
target = 11

size = len(arr)
    
# Function call
print(countMinSteps(arr, target, size))

# This code is contributed by Stuti Pathak

// C# program implementation
// of the above approach
using System;

class GFG{

// Function to count minimum steps     
static int countMinSteps(int[] arr,
                         int target, 
                         int n) 
{
    
    // Function to sort the array
    Array.Sort(arr);

    int minimumSteps = 0;
    int i = 0;
    int j = n - 1;

    // Run while loop
    while (i <= j)
    {
        
        // Condition to check whether 
        // sum exceed the target or not
        if (arr[i] + arr[j] <= target)
        {
            i += 1;
            j -= 1;
        }
        else
        {
            j -= 1;
        }

        // Increment the step by 1 
        minimumSteps += 1;
    }

    // Return minimum steps
    return minimumSteps;
}

// Driver code 
public static void Main() 
{
    int[] arr = new int[]{ 4, 6, 2, 9, 
                           6, 5, 8, 10 };

    // Given target value
    int target = 11;

    int size = arr.Length;
    
    // Print the minimum flip
    Console.Write(countMinSteps(
                  arr, target, size));
}
}

// This code is contributed by sanjoy_62

<script>

// javascript program for the above approach 

// Function to count minimum steps 
function countMinSteps(arr, target, n) 
{ 

    // Function to sort the array 
    arr = arr.sort(function(a, b) {
  return a - b;
});
    var minimumSteps = 0; 
    var i = 0, j = n - 1; 

    // Run while loop 
    while (i <= j) { 

        // Condition to check whether 
        // sum exceed the target or not 
        if (arr[i] + arr[j] <= target) { 
            i++; 
            j--; 
        } 
        else { 
            j--; 
        } 

        // Increment the step 
        // by 1 
        minimumSteps++; 
    } 

    // Return minimum steps 
    return minimumSteps; 
} 

// Driver Code 

    // Given array arr[] 
    var arr = [4, 6, 2, 9, 6, 5, 8, 10]; 

    // Given target value 
    var target = 11; 

    var size = arr.length; 

    // Function call 
    document.write(countMinSteps(arr, target, size));

// This code is contributed by ipg2016107.
</script>

Output: 
 

5

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