Rearrange the given array to minimize the indices with prefix sum at least arr[i]

Given an array arr[] consisting of N positive integers, rearrange the array to minimize the number of indices i such that the prefix sum from index 1 to index i-1 is greater than or equal to the current element arr[i] i.e. arr[1]+arr[2]+...+arr[i-1] >= arr[i].

Examples:

Input: arr[] = [4, 2, 1]
Output:
0
1 2 4
Explanation: If we arrange array as [1, 2, 4], there will be no indices following the criteria.
At index 1, there is no prefix sum.
At index 2, prefix sum = 1 < 2 hence not included in answer.
At index 3, prefix sum= 3 < 4 hence not included in answer.

Input: arr[] = [4, 7, 5, 10]
Output:
1
4 5 10 7
Explanation: If we arrange array as [4 5 10 7],
At index 1, there is no prefix sum.
At index 2, prefix sum=4 < 5 hence not included in answer.
At index 3, prefix sum= 9<10 hence not included in answer.
At index 4, prefix sum= 19>=7 hence included in answer.
There can be no other combination which can minimize answer.

Approach: To solve the problem, follow the below idea,

While the initial intuition might suggest sorting the array, it is not an optimal strategy. For example, consider the array [4, 5, 7, 10]. A direct sort could result in two indices following the condition. However, by rearranging the array as [4, 5, 10, 7], the number of indices is minimized to one.

A better approach involves systematically checking each index element that it is possible to add the current element to ans[] vector such that it exceeds its prefix sum. Additionally, any skipped elements between successful additions are included in the final result.

Step-by-step algorithm:

• Sort the array and initialize count = 0.
• Add first element to ans vector.
• Now you just have to iterate over the sorted array.
• If the current element is greater than prefix_sum then you have to just add that element into the answer array otherwise you have to skip that element.
• The ans[] vector contains all the elements not satisfying the condition. Hence the remaining elements are included in count i.e. n-ans.size().
• Push back all the skipped elements (use boolean array for tracking skipped elements).
• Finally print the count and ans array.

Below is the implementation of the above approach:

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

// Function to find number of
// required indices and print the rearranged array
int Rearrange_array(int arr[], int N)
{

    // initialising prefix sum
    long long int prefix = 0;

    vector<int> ans;
    vector<bool> checked(N, true);

    // sort the array
    sort(arr, arr + N);
    ans.push_back(arr[0]);
    prefix += arr[0];
    for (int i = 1; i < N; i++) {
        if (arr[i] > prefix) {
            prefix += arr[i];
            ans.push_back(arr[i]);
            checked[i] = false;
        }
    }

    // the count of required numbers
    int count = N - ans.size();
    for (int i = 1; i < N; i++) {
        if (checked[i])
            ans.push_back(arr[i]);
    }
    cout << count << endl;
    for (auto x : ans) {
        cout << x << " ";
    }
    cout << endl;
}

// Driver Code
int main()
{
    int arr[] = { 4, 7, 5, 10 };
    int N = sizeof(arr) / sizeof(arr[0]);
    Rearrange_array(arr, N);

    return 0;
}

// java program for the above approach

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Solution {

    // Function to find the number of required indices
    // and print the rearranged array
    static void rearrangeArray(int[] arr, int N) {
        // Initializing prefix sum
        long prefix = 0;

        List<Integer> ans = new ArrayList<>();
        boolean[] checked = new boolean[N];
        Arrays.fill(checked, true);

        // Sort the array
        Arrays.sort(arr);
        ans.add(arr[0]);
        prefix += arr[0];

        for (int i = 1; i < N; i++) {
            if (arr[i] > prefix) {
                prefix += arr[i];
                ans.add(arr[i]);
                checked[i] = false;
            }
        }

        // The count of required numbers
        int count = N - ans.size();

        for (int i = 1; i < N; i++) {
            if (checked[i]) {
                ans.add(arr[i]);
            }
        }

        System.out.println(count);

        for (int x : ans) {
            System.out.print(x + " ");
        }
        System.out.println();
    }

    // Main function
    public static void main(String[] args) {
        int[] arr = {4, 7, 5, 10};
        int N = arr.length;
        rearrangeArray(arr, N);
    }
}

# Python program for the above approach
def rearrange_array(arr, N):
    # Initializing prefix sum
    prefix = 0

    ans = []
    checked = [True] * N

    # Sort the array
    arr.sort()
    ans.append(arr[0])
    prefix += arr[0]

    for i in range(1, N):
        if arr[i] > prefix:
            prefix += arr[i]
            ans.append(arr[i])
            checked[i] = False

    # The count of required numbers
    count = N - len(ans)

    for i in range(1, N):
        if checked[i]:
            ans.append(arr[i])

    print(count)

    for x in ans:
        print(x, end=" ")

    print()

# Main function
if __name__ == "__main__":
    arr = [4, 7, 5, 10]
    N = len(arr)
    rearrange_array(arr, N)

//// c# program for the above approach
using System;
using System.Collections.Generic;

public class Solution
{
    // Function to find the number of required indices
    // and print the rearranged array
    static void RearrangeArray(int[] arr, int N)
    {
        // Initializing prefix sum
        long prefix = 0;

        List<int> ans = new List<int>();
        bool[] checkedArray = new bool[N];
        Array.Fill(checkedArray, true);

        // Sort the array
        Array.Sort(arr);
        ans.Add(arr[0]);
        prefix += arr[0];

        for (int i = 1; i < N; i++)
        {
            if (arr[i] > prefix)
            {
                prefix += arr[i];
                ans.Add(arr[i]);
                checkedArray[i] = false;
            }
        }

        // The count of required numbers
        int count = N - ans.Count;

        for (int i = 1; i < N; i++)
        {
            if (checkedArray[i])
            {
                ans.Add(arr[i]);
            }
        }

        Console.WriteLine(count);

        foreach (int x in ans)
        {
            Console.Write(x + " ");
        }
        Console.WriteLine();
    }

    // Main function
    public static void Main(string[] args)
    {
        int[] arr = { 4, 7, 5, 10 };
        int N = arr.Length;
        RearrangeArray(arr, N);
    }
}

// Javascript program for the above approach
// Function to find the number of required indices
// and print the rearranged array
function rearrangeArray(arr, N) {
    // Initializing prefix sum
    let prefix = 0;

    let ans = [];
    let checkedArray = new Array(N).fill(true);

    // Sort the array
    arr.sort((a, b) => a - b);
    ans.push(arr[0]);
    prefix += arr[0];

    for (let i = 1; i < N; i++) {
        if (arr[i] > prefix) {
            prefix += arr[i];
            ans.push(arr[i]);
            checkedArray[i] = false;
        }
    }

    // The count of required numbers
    let count = N - ans.length;

    for (let i = 1; i < N; i++) {
        if (checkedArray[i]) {
            ans.push(arr[i]);
        }
    }

    console.log(count);
    console.log(ans.join(" "));
}

// Main function
let arr = [4, 7, 5, 10];
let N = arr.length;
rearrangeArray(arr, N);

1
4 5 10 7 

Time Complexity: O(N*logN), where N is the size of input array arr[].
Auxiliary Space: O(N)