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C++ code to find card spread way to make all sum equal for each player
Suppose we have an array A with n elements. Here n is even. A[i] is a number written on ith card. There are n/2 people who want to play a game. At the beginning, each player will take two cards. We have to find the way to distribute cards in such a way that sum of values written on the cards will be same for each player.
So, if the input is like A = [1, 5, 7, 4, 4, 3], then the output will be [(0, 2), (5, 1), (3, 4)], because A[0] + A[2] = 8, A[5] + A[1] = 8 and A[3] + A[4] = 8.
Steps
To solve this, we will follow these steps −
n := size of A Define one array of pairs p of size n for initialize i := 0, when i < n, update (increase i by 1), do: first element of p[i] := A[i] second element of p[i]:= i sort the array p, p + n for initialize i := 0, when i < n / 2, update (increase i by 1), do: print second element of p[i] and second element of p[n - i - 1]
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
void solve(vector<int> A){
int n = A.size();
pair<int, int> p[n];
for (int i = 0; i < n; i++){
p[i].first = A[i];
p[i].second = i;
}
sort(p, p + n);
for (int i = 0; i < n / 2; i++)
cout << "(" << p[i].second << ", " << p[n - i - 1].second <<"), ";
}
int main(){
vector<int> A = { 1, 5, 7, 4, 4, 3 };
solve(A);
}Input
{ 1, 5, 7, 4, 4, 3 }Output
(0, 2), (5, 1), (3, 4),
Updated on: 15-Mar-2022
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