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

• Last Updated : 17 Jun, 2021

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:
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:
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++

 // C++ program for the above approach#include using namespace std; // Function to count minimum stepsint 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 Codeint 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);     // Function call    cout << countMinSteps(arr, target, size);     return 0;}

Java

 // Java program implementation// of the above approachimport 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 codepublic 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

 # Python3 program for the above approach # Function to count minimum stepsdef 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 valuetarget = 11 size = len(arr)     # Function callprint(countMinSteps(arr, target, size)) # This code is contributed by Stuti Pathak

C#

 // C# program implementation// of the above approachusing 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 codepublic 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

Javascript



Output:

5

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

My Personal Notes arrow_drop_up