Find Winner of Unique Bidding Game in C++

Suppose we have an array A with n elements. There are n participants, the i-th participant chose the number A[i]. The winner of the bidding game is such a participant that the number he chose is unique and is minimal. We have to find the index of the participant who won the game. If not possible, return -1.

Problem Category

Various problems in programming can be solved through different techniques. To solve a problem, we have to devise an algorithm first and to do that we have to study the particular problem in detail. A recursive approach can be used if there is a recurring appearance of the same problem over and over again; alternatively, we can use iterative structures also. Control statements such as if-else and switch cases can be used to control the flow of logic in the program. Efficient usage of variables and data structures provides an easier solution and a lightweight, low-memory-requiring program. We have to look at the existing programming techniques, such as Divide-and-conquer, Greedy Programming, Dynamic Programming, and find out if they can. This problem we can solve by some basic logics or brute-force approach. Follow the following contents to understand the approach better.

So, if the input of our problem is like A = [2, 3, 2, 4, 2], then the output will be 1, because the participant who has chosen 3 can win the game.

Steps

To solve this, we will follow these steps −

Define one map cnt, and another map pos, both for integer type key and integer type value
n := size of A
for initialize i := 0, when i < n, update (increase i by 1), do:
   x := A[i]
   increase cnt[x + 1] by 1
   pos[x + 1] = i
ans := -1
for each key-value pair it in cnt, do
   if value of it is same as 1, then:
      ans := pos[key of it]
      Come out from the loop
return ans

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
int solve(vector<int> A){
   map<int, int> cnt, pos;
   int n = A.size();
   for (int i = 0; i < n; i++){
      int x = A[i];
      cnt[x + 1]++;
      pos[x + 1] = i;
   }
   int ans = -1;
   for (auto it : cnt){
      if (it.second == 1){
         ans = pos[it.first];
         break;
      }
   }
   return ans;
}
int main(){
   vector<int> A = { 2, 3, 2, 4, 2 };
   cout << solve(A) << endl;
}

Input

{ 2, 3, 2, 4, 2 }

Output

1