Minimum number of array indices required to be selected to make the sum of a set greater than another

Given two arrays arr[] and brr[] of size N and an integer K. Consider two sets A, contains K initially, and B, initially empty. In each operation, an index is required to be selected. For every selected index, say i, arr[i] and brr[i] is added to B. For every unselected index, arr[i] is added to A. The task is to find the minimum number of indices required to be selected to make sum of B greater than sum of A. If it is not possible to do so, then print -1.

Examples:

Input: arr[] = {3, 2, 5, 6}, brr[] = {4, 4, 2, 3}, K = 12   
Output: 3
4 3 1
Explanation: Indices 2, 3 and 0 are selected. Sum of B = arr[0] + brr[0] + arr[2] + brr[2] + arr[3] + brr[3] = 3 + 4 + 5 + 2 + 6 + 3 = 23. Sum of A = K + arr[1] = 12 + 2 = 14.

Input: arr[] = {3, 2, 5, 6}, brr[] = {4, 4, 2, 3}, K = 34
Output: -1

Approach: The idea is to use a Greedy Approach. Follow the steps below to solve the problem:

• Initialize a vector of pair, B[] to keep track of indexes.
• Initialize a variable, A with K to store the value of set A.
• Traverse the array arr[] using the variable i,
  • Add the value arr[i] to A, if the index was not chosen.
  • Insert {brr[i] + 2 * arr[i], i} as a pair in vector B.
• Sort vector B in decreasing order.
• Initialize a vector, ans to store the indexes that are chosen.
• Run a while loop and keep on choosing the indices until A’s value is bigger than B.
• If all indexes are chosen but B’s value is still less than A, print “-1”.
• Otherwise, print the size of the vector ans as the minimum number of moves.
• Traverse the vector, ans, and print the chosen indices.

Below is the implementation of the above approach:

C++

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

// Function to print the minimum number
// of indices required to be selected
void numberOfIndexes(int arr[], int brr[],
                     int N, int K)
{
    // Declare vector to keep track of indexes
    vector<pair<int, int> > B;

    // Set A contains K
    int A = K;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // Adding value that set A can
        // get if no index was chosen
        A += arr[i];

        // Insert as B's value
        B.push_back({ brr[i] + 2 * arr[i], i });
    }

    // Sort the vector
    sort(B.begin(), B.end());

    // Reverse the vector
    reverse(B.begin(), B.end());

    int tot = 0, idx = 0;

    // Stores chosen indexes
    vector<int> ans;

    // Keep on choosing more indices until
    // B's value is bigger than A or stop
    // incase all the indexes is chosen
    while (A >= tot && idx < B.size()) {

        // Update tot
        tot += B[idx].first;

        // Update ans
        ans.push_back(B[idx].second);

        // Increment idx
        idx++;
    }

    // If all indices are selected and
    // sum of B is less than sum of A
    if (tot <= A) {
        cout << -1 << endl;
        return;
    }

    // Print the minimum number of indices
    cout << ans.size() << endl;

    // Print chosen indices
    for (auto c : ans)
        cout << c + 1 << " ";
}

// Driver Code
int main()
{
    // Given arrays
    int arr[] = { 3, 2, 5, 6 };
    int brr[] = { 4, 4, 2, 3 };

    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);

    // Given value of K
    int K = 12;

    // Function Call
    numberOfIndexes(arr, brr, N, K);

    return 0;
}

Java

// Java program for the above approach

import java.util.*;

public class GFG {
    // Function to print the minimum number
    // of indices required to be selected
    public static void numberOfIndexes(int[] arr, int[] brr, int N, int K) {
        // Declare list to keep track of indexes
        List<Pair<Integer, Integer>> B = new ArrayList<>();

        // Set A contains K
        int A = K;

        // Traverse the array
        for (int i = 0; i < N; i++) {
            // Adding value that set A can
            // get if no index was chosen
            A += arr[i];

            // Insert as B's value
            B.add(new Pair<>(brr[i] + 2 * arr[i], i));
        }

        // Sort the list
        B.sort((pair1, pair2) -> pair2.getKey() - pair1.getKey());

        int tot = 0, idx = 0;

        // Stores chosen indexes
        List<Integer> ans = new ArrayList<>();

        // Keep on choosing more indices until
        // B's value is bigger than A or stop
        // incase all the indexes is chosen
        while (A >= tot && idx < B.size()) {
            // Update tot
            tot += B.get(idx).getKey();

            // Update ans
            ans.add(B.get(idx).getValue());

            // Increment idx
            idx++;
        }

        // If all indices are selected and
        // sum of B is less than sum of A
        if (tot <= A) {
            System.out.println(-1);
            return;
        }

        // Print the minimum number of indices
        System.out.println(ans.size());

        // Print chosen indices
        for (int c : ans) {
            System.out.print((c + 1) + " ");
        }
        System.out.println();
    }

    public static void main(String[] args) {
        // Given arrays
        int[] arr = {3, 2, 5, 6};
        int[] brr = {4, 4, 2, 3};

        // Size of the array
        int N = arr.length;

        // // Given value of K
        int K = 12;


        // Function Call
        numberOfIndexes(arr, brr, N, K);
    }
}

class Pair<K, V> {
    private K key;
    private V value;

    public Pair(K key, V value) {
        this.key = key;
        this.value = value;
    }

    public K getKey() {
        return key;
    }

    public V getValue() {
        return value;
    }
}

Python3

# Python 3 program for the above approach

# Function to print the minimum number
# of indices required to be selected
def numberOfIndexes(arr, brr, N, K):

    # Declare vector to keep track of indexes
    B = []

    # Set A contains K
    A = K

    # Traverse the array
    for i in range(N):

        # Adding value that set A can
        # get if no index was chosen
        A += arr[i]

        # Insert as B's value
        B.append([brr[i] + 2 * arr[i], i])

    # Sort the vector
    B.sort()

    # Reverse the vector
    B = B[::-1]
    tot = 0
    idx = 0

    # Stores chosen indexes
    ans = []

    # Keep on choosing more indices until
    # B's value is bigger than A or stop
    # incase all the indexes is chosen
    while (A >= tot and idx < len(B)):

        # Update tot
        tot += B[idx][0]

        # Update ans
        ans.append(B[idx][1])

        # Increment idx
        idx += 1

    # If all indices are selected and
    # sum of B is less than sum of A
    if (tot <= A):
        print(-1)
        return

    # Print the minimum number of indices
    print(len(ans))

    # Print chosen indices
    for c in ans:
        print(c + 1,end = " ")

# Driver Code
if __name__ == '__main__':

    # Given arrays
    arr = [3, 2, 5, 6]
    brr = [4, 4, 2, 3]

    # Size of the array
    N = len(arr)

    # Given value of K
    K = 12

    # Function Call
    numberOfIndexes(arr, brr, N, K)

    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# code for the above approach
using System;
using System.Collections.Generic;

class GFG
{
  // Function to print the minimum number
  // of indices required to be selected
  static void numberOfIndexes(int[] arr, int[] brr, int N, int K)
  {
    // Declare list to keep track of indexes
    List<Tuple<int, int>> B = new List<Tuple<int, int>>();

    // Set A contains K
    int A = K;

    // Traverse the array
    for (int i = 0; i < N; i++)
    {
      // Adding value that set A can
      // get if no index was chosen
      A += arr[i];

      // Insert as B's value
      B.Add(Tuple.Create(brr[i] + 2 * arr[i], i));
    }

    // Sort the list
    B.Sort((pair1, pair2) => pair2.Item1 - pair1.Item1);

    int tot = 0, idx = 0;

    // Stores chosen indexes
    List<int> ans = new List<int>();

    // Keep on choosing more indices until
    // B's value is bigger than A or stop
    // incase all the indexes is chosen
    while (A >= tot && idx < B.Count)
    {
      // Update tot
      tot += B[idx].Item1;

      // Update ans
      ans.Add(B[idx].Item2);

      // Increment idx
      idx++;
    }

    // If all indices are selected and
    // sum of B is less than sum of A
    if (tot <= A)
    {
      Console.WriteLine(-1);
      return;
    }

    // Print the minimum number of indices
    Console.WriteLine(ans.Count);

    // Print chosen indices
    foreach (int c in ans)
    {
      Console.Write((c + 1) + " ");
    }
    Console.WriteLine();
  }

  static void Main(string[] args)
  {
    // Given arrays
    int[] arr = { 3, 2, 5, 6 };
    int[] brr = { 4, 4, 2, 3 };

    // Size of the array
    int N = arr.Length;

    // Given value of K
    int K = 12;

    // Function Call
    numberOfIndexes(arr, brr, N, K);
  }
}

// This code is contributed by phasing17.

Javascript

<script>

// JavaScript program for the above approach

// Function to print the minimum number
// of indices required to be selected
function numberOfIndexes(arr, brr, N, K) {
  // Declare vector to keep track of indexes
  let B = [];

  // Set A contains K
  let A = K;

  // Traverse the array
  for (let i = 0; i < N; i++) {
    // Adding value that set A can
    // get if no index was chosen
    A += arr[i];

    // Insert as B's value
    B.push([brr[i] + 2 * arr[i], i]);
  }

  // Sort the vector

  // Reverse the vector
  B.sort((a, b) => a[0] - b[0]).reverse();


  let tot = 0,
    idx = 0;

  // Stores chosen indexes
  let ans = [];

  // Keep on choosing more indices until
  // B's value is bigger than A or stop
  // incase all the indexes is chosen
  while (A >= tot && idx < B.length) {
    // Update tot
    tot += B[idx][0];

    // Update ans
    ans.push(B[idx][1]);

    // Increment idx
    idx++;
  }

  // If all indices are selected and
  // sum of B is less than sum of A
  if (tot <= A) {
    document.write(-1 + "<br>");
    return;
  }

  // Print the minimum number of indices
  document.write(ans.length + "<br>");
  // Print chosen indices
  for (let c of ans) document.write(Number(c) + 1 + " ");
}

// Driver Code

// Given arrays
let arr = [3, 2, 5, 6];
let brr = [4, 4, 2, 3];

// Size of the array
let N = arr.length;

// Given value of K
let K = 12;

// Function Call
numberOfIndexes(arr, brr, N, K);

</script>

Output:

3
4 3 1

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