Check if all people can vote on two machines

There are n people and two identical voting machines. We are also given an array a[] of size n such that a[i] stores time required by i-th person to go to any machine, mark his vote and come back. At one time instant, only one person can be there on each of the machines. Given a value x, defining the maximum allowable time for which machines are operational, check whether all persons can cast their vote or not.

Examples:

Input  : n = 3, x = 4
         a[] = {2, 4, 2}
Output: YES
There are  n = 3 persons say and maximum
allowed time is x = 4 units. Let the persons
be P0, P1, and P2 and two machines be M0 and M1.
At t0: P0 goes to M0
At t0: P1 goes to M1
At t2: M0 is free, P2 goes to M0
At t4: both M0 and M1 are free and all 3 have
        given their vote.

Method 1
Let sum be the total time taken by all n people. If sum <=x, then answer will obviously be YES. Otherwise, we need to check whether the given array can be split into two parts such that the sum of the first part and sum of the second part are both less than or equal to x. The problem is similar to the knapsack problem. Imagine two knapsacks each with capacity x. Now find, maximum people who can vote on any one machine i.e. find maximum subset sum for a knapsack of capacity x. Let this sum be s1. Now if (sum-s1) <= x, then answer is YES else answer is NO.

C++

// C++ program to check if all people can
// vote using two machines within limited
// time
#include<bits/stdc++.h>
using namespace std;

// Returns true if n people can vote using
// two machines in x time.
bool canVote(int a[], int n, int x)
{
    // dp[i][j] stores maximum possible number
    // of people among arr[0..i-1] can vote
    // in j time.
    int dp[n+1][x+1];
    memset(dp, 0, sizeof(dp));

    // Find sum of all times
    int sum = 0;
    for (int i=0; i<=n; i++ )
        sum += a[i];

    // Fill dp[][] in bottom up manner (Similar
    // to knapsack).
    for (int i=1; i<=n; i++)
        for (int j=1; j<=x; j++)
            if (a[i] <= j)
                dp[i][j] = max(dp[i-1][j],
                        a[i] + dp[i-1][j-a[i]]);
            else
                dp[i][j] = dp[i-1][j];

    // If remaining people can go to other machine.
    return (sum - dp[n][x] <= x);
}

// Driver code
int main()
{
    int n = 3, x = 4;
    int a[] = {2, 4, 2};
    canVote(a, n, x)? cout << "YES\n" :
                      cout << "NO\n";
    return 0;
}

Java

// Java program to check if all people can
// vote using two machines within limited
// time
public class Main
{
    // Returns true if n people can vote using
    // two machines in x time.
    static boolean canVote(int[] a, int n, int x)
    {
        // dp[i][j] stores maximum possible number
        // of people among arr[0..i-1] can vote
        // in j time.
        int[][] dp = new int[n + 1][x + 1];
        for(int i = 0; i < n + 1; i++)
        {
            for(int j = 0; j < x + 1; j++)
            {
                dp[i][j] = 0;
            }
        }

        // Find sum of all times
        int sum = 0;
        for (int i = 0; i < n; i++ )
            sum += a[i];

        // Fill dp[][] in bottom up manner (Similar
        // to knapsack).
        for (int i = 1; i < n; i++)
            for (int j = 1; j <= x; j++)
                if (a[i] <= j)
                    dp[i][j] = Math.max(dp[i - 1][j], a[i] + dp[i - 1][j - a[i]]);
                else
                    dp[i][j] = dp[i - 1][j];

        // If remaining people can go to other machine.
        return (sum - dp[n][x] >= x);
    }

    public static void main(String[] args) {
        int n = 3, x = 4;
        int[] a = {2, 4, 2};
        if(canVote(a, n, x))
        {
            System.out.println("YES");
        }
        else{
            System.out.println("NO");
        }
    }
}

// This code is contributed by mukesh07.

Python3

# Python3 program to check if all people can
# vote using two machines within limited
# time

# Function returns true if n people can vote
# using two machines in x time.
def canVote(a, n, x):

    # dp[i][j] stores maximum possible number
    # of people among arr[0..i-1] can vote
    # in j time.
    dp = [[0] * (x + 1) for _ in range(n + 1)]
    a = a[:]
    a.append(0)

    # Find sum of all times
    sm = 0
    for i in range(n + 1):
        sm += a[i]

    # Fill dp[][] in bottom up manner
    # (Similar to knapsack).
    for i in range(1, n + 1):
        for j in range(1, x + 1):
            if a[i] <= j:
                dp[i][j] = max(dp[i - 1][j], a[i] +
                               dp[i - 1][j - a[i]])
            else:
                dp[i][j] = dp[i - 1][j]

    # If remaining people can go to other machine.
    return (sm - dp[n][x]) <= x

# Driver Code
if __name__ == "__main__":
    n = 3
    x = 4
    a = [2, 4, 2]
    print("YES" if canVote(a, n, x) else "NO")

# This code is contributed
# by vibhu4agarwal

C#

// C# program to check if all people can
// vote using two machines within limited
// time
using System;
class GFG {

    // Returns true if n people can vote using
    // two machines in x time.
    static bool canVote(int[] a, int n, int x)
    {
        // dp[i][j] stores maximum possible number
        // of people among arr[0..i-1] can vote
        // in j time.
        int[,] dp = new int[n + 1, x + 1];
        for(int i = 0; i < n + 1; i++)
        {
            for(int j = 0; j < x + 1; j++)
            {
                dp[i, j] = 0;
            }
        }

        // Find sum of all times
        int sum = 0;
        for (int i = 0; i < n; i++ )
            sum += a[i];

        // Fill dp[][] in bottom up manner (Similar
        // to knapsack).
        for (int i = 1; i < n; i++)
            for (int j = 1; j <= x; j++)
                if (a[i] <= j)
                    dp[i, j] = Math.Max(dp[i - 1, j],
                            a[i] + dp[i - 1, j - a[i]]);
                else
                    dp[i, j] = dp[i - 1, j];

        // If remaining people can go to other machine.
        return (sum - dp[n, x] >= x);
    }

  // Driver code
  static void Main()
  {
    int n = 3, x = 4;
    int[] a = {2, 4, 2};
    if(canVote(a, n, x))
    {
        Console.Write("YES");
    }
    else{
        Console.Write("NO");
    }
  }
}

// This code is contributed by rameshtravel07.

Javascript

<script>
    // Javascript program to check if all people can
    // vote using two machines within limited
    // time

    // Returns true if n people can vote using
    // two machines in x time.
    function canVote(a, n, x)
    {
        // dp[i][j] stores maximum possible number
        // of people among arr[0..i-1] can vote
        // in j time.
        let dp = new Array(n+1);
        for(let i = 0; i < n + 1; i++)
        {
            dp[i] = new Array(x + 1);
            for(let j = 0; j < x+1; j++)
            {
                dp[i][j] = 0;
            }
        }

        // Find sum of all times
        let sum = 0;
        for (let i=0; i<n; i++ )
            sum += a[i];

        // Fill dp[][] in bottom up manner (Similar
        // to knapsack).
        for (let i = 1; i <= n; i++)
            for (let j = 1; j <= x; j++)
                if (a[i] <= j)
                    dp[i][j] = Math.max(dp[i-1][j],
                            a[i] + dp[i-1][j-a[i]]);
                else
                    dp[i][j] = dp[i-1][j];

        // If remaining people can go to other machine.
        return (sum - dp[n][x] <= x);
    }

    let n = 3, x = 4;
    let a = [2, 4, 2];
    if(canVote(a, n, x))
    {
        document.write("YES");
    }
    else{
        document.write("NO");
    }

// This code is contributed by decode2207.
</script>

Output

YES

Time Complexity: O(x * n) 
Auxiliary Space: O(x * n)

Method 2
The above method uses O(x * n) extra space, here we give a method that uses O(n) extra space. 
Idea is to sort the array and then check if we can get any two array that can be a subarray which can be between the start and end of array such that its sum less than or equal to x and the sum of remaining elements is also less than x. 
We use prefix sum for this purpose. We set i and j and check if sorted array between i to j – 1 gives sum <= x and the remaining elements also sum to sum <= x.

C++

// C++ program to check if all people can
// vote using two machines within limited
// time
#include<bits/stdc++.h>
using namespace std;

// Returns true if n people can vote using
// two machines in x time.
bool canVote(vector<int> a, int n, int x)
{
    // calculate total sum i.e total
    // time taken by all people
    int total_sum = 0;
    for(auto x : a){
        total_sum += x;
    }

    // if total time is less than x then
    // all people can definitely vote
    // hence return true
    if(total_sum <= x)
        return true;

    // sort the vector
    sort(a.begin(), a.end());

    // declare a vector presum of same size
    // as that of a and initialize it with 0
    vector<int> presum(a.size(), 0);

    // prefixsum for first element
    // will be element itself
    presum[0] = a[0];
    // fill the array
    for(int i = 1; i < presum.size(); i++){
        presum[i] = presum[i - 1] + a[i];
    }

    // Set i and j and check if array
    // from i to j - 1 gives sum <= x
    for(int i = 0; i < presum.size(); i++){
        for(int j = i + 1; j < presum.size(); j++){
            int arr1_sum = (presum[i] +
                           (total_sum - presum[j]));
            if((arr1_sum <= x) &&
                        (total_sum - arr1_sum) <= x)
                return true;
        }
    }

    return false;
}

// Driver code
int main()
{
    int n = 3, x = 4;
    vector<int>a = {2, 4, 2};
    canVote(a, n, x)? cout << "YES\n" :
                      cout << "NO\n";
    return 0;
}

Java

// Java program to check if all people can
// vote using two machines within limited
// time
import java.io.*;
import java.util.*;

class GFG {
    // Returns true if n people can vote using
    // two machines in x time.
    public static boolean canVote(int a[], int n, int x)
    {
        // calculate total sum i.e total
        // time taken by all people
        int total_sum = 0;
        for (int z : a) {
            total_sum += z;
        }

        // if total time is less than x then
        // all people can definitely vote
        // hence return true
        if (total_sum <= x)
            return true;

        // sort the array
        Arrays.sort(a);

        // declare a array presum of same size
        // as that of a and initialize it with 0
        int presum[] = new int[a.length];

        // prefixsum for first element
        // will be element itself
        presum[0] = a[0];
        // fill the array
        for (int i = 1; i < presum.length; i++) {
            presum[i] = presum[i - 1] + a[i];
        }

        // Set i and j and check if array
        // from i to j - 1 gives sum <= x
        for (int i = 0; i < presum.length; i++) {
            for (int j = i + 1; j < presum.length; j++) {
                int arr1_sum
                    = (presum[i] + (total_sum - presum[j]));
                if ((arr1_sum <= x)
                    && (total_sum - arr1_sum) <= x)
                    return true;
            }
        }

        return false;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int n = 3, x = 4;
        int a[] = { 2, 4, 2 };
        if (canVote(a, n, x) == true)
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

// This code is contributed by Rohit Pradhan

Python3

# Python3 program to check if all people can
# vote using two machines within limited
# time

# Returns true if n people can vote using
# two machines in x time.
def canVote(a, n, x):

    # calculate total sum i.e total
    # time taken by all people
    total_sum = 0
    for i in range(len(a)):
        total_sum += a[i]

    # if total time is less than x then
    # all people can definitely vote
    # hence return true
    if(total_sum <= x):
        return True

    # sort the list
    a.sort()

    # declare a list presum of same size
    # as that of a and initialize it with 0
    presum = [0 for i in range(len(a))]

    # prefixsum for first element
    # will be element itself
    presum[0] = a[0]

    # fill the array
    for i in range(1, len(presum)):
        presum[i] = presum[i - 1] + a[i]

    # Set i and j and check if array
    # from i to j - 1 gives sum <= x
    for i in range(0, len(presum)):
        for j in range(i + 1, len(presum)):
            arr1_sum = (presum[i] + (total_sum - presum[j]))
            if((arr1_sum <= x) and
               (total_sum - arr1_sum) <= x):
                return True

    return False

# Driver code
n = 3
x = 4
a = [2, 4, 2]
if(canVote(a, n, x)):
    print("YES")
else:
    print("NO")

# This code is contributed by ashutosh450

C#

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

class GFG
{
    // Returns true if n people can vote using
    // two machines in x time.
    public static bool CanVote(int[] a, int n, int x)
    {
        // calculate total sum i.e total
        // time taken by all people
        int total_sum = 0;
        foreach (int z in a) {
            total_sum += z;
        }

        // if total time is less than x then
        // all people can definitely vote
        // hence return true
        if (total_sum <= x)
            return true;

        // sort the array
        Array.Sort(a);

        // declare a array presum of same size
        // as that of a and initialize it with 0
        int[] presum = new int[a.Length];

        // prefixsum for first element
        // will be element itself
        presum[0] = a[0];
        // fill the array
        for (int i = 1; i < presum.Length; i++) {
            presum[i] = presum[i - 1] + a[i];
        }

        // Set i and j and check if array
        // from i to j - 1 gives sum <= x
        for (int i = 0; i < presum.Length; i++) {
            for (int j = i + 1; j < presum.Length; j++) {
                int arr1_sum
                    = (presum[i] + (total_sum - presum[j]));
                if ((arr1_sum <= x)
                    && (total_sum - arr1_sum) <= x)
                    return true;
            }
        }

        return false;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int n = 3;
        int x = 4;
        int[] a = { 2, 4, 2 };
        if (CanVote(a, n, x) == true)
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}

//This code is contributed by ik_9

Javascript

// Returns true if n people can vote using
// two machines in x time.
function canVote(a, n, x)
{

    // calculate total sum i.e total
    // time taken by all people
    let total_sum = 0;
    for(let x of a){
        total_sum += x;
    }

    // if total time is less than x then
    // all people can definitely vote
    // hence return true
    if(total_sum <= x)
        return true;

    // sort the array
    a.sort((a, b) => a - b);

    // declare a array presum of same size
    // as that of a and initialize it with 0
    let presum = new Array(a.length).fill(0);

    // prefixsum for first element
    // will be element itself
    presum[0] = a[0];

    // fill the array
    for(let i = 1; i < presum.length; i++){
        presum[i] = presum[i - 1] + a[i];
    }

    // Set i and j and check if array
    // from i to j - 1 gives sum <= x
    for(let i = 0; i < presum.length; i++){
        for(let j = i + 1; j < presum.length; j++){
            let arr1_sum = (presum[i] +
                           (total_sum - presum[j]));
            if((arr1_sum <= x) &&
                        (total_sum - arr1_sum) <= x)
                return true;
        }
    }

    return false;
}

// Driver code
let n = 3, x = 4;
let a = [2, 4, 2];
console.log(canVote(a, n, x) ? "YES" : "NO");

// This code is contributed by unstoppablepandu.

Output

YES

Time Complexity: O(n * n) // since two loops are nested inside each other the have a time complexity of n*n
Auxiliary Space: O(n) // since an extra vector of size equal to the length of the array