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: P2 goes to M1 At t2: M0 is free, p3 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;` `}` |

## 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` |

**Output:**

YES

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

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**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 reamining 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; ` `} ` |

## 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` |

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

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